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TN State Board 11th Maths Model Question Paper 2 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 90

PART – I

Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
The range of the function \(\frac{1}{1-2sin x}\) is………..
(a) (-∞, -1) ∪(\(\frac{1}{3}\), ∞)
(b) (-1, -1)
(c) (-1, \(\frac{1}{3}\))
(d) (-∞, -1] ∪[\(\frac{1}{3}\), ∞)
Solution:
(d) (-∞, -1] ∪[\(\frac{1}{3}\), ∞)

Question 2.
The value of log_√2 512 is……….
(a) 16
(b) 18
(c) 9
(d) 12
Solution:
(b) 18

Question 3.
If a and b are the roots of the equation x² – kx + 16 = 0 and satisfy a² + b² = 32 then the value of k is………..
(a) 10
(b) -8
(c) -8, 8
(d) 6
Solution:
(c) -8, 8

Question 4.
The value of log₉ 27 is …………
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{4}{3}\)
Solution:
(b) \(\frac{3}{2}\)

Question 5.
The value of \(\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}\) = ………….
(a) tan 3θ
(b) tan 6θ
(c) cot 3θ
(d) cot 6θ
Solution:
(b) tan 6θ

Question 6.
In 3 fingers the number of ways 4 rings can be worn in ways.
(a) 4³ – 1
(b) 3⁴
(c) 68
(d) 64
Solution:
(d) 64

Question 7.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is……….
(a) 11
(b) 12
(c) 10
(d) 6
Solution:
(b) 12

Question 8.
The H.M of two positive number whose AM and G.M. are 16, 8 respectively is
(a) 10
(b) 6
(c) 5
(d) 4
Solution:
(d) 4

Question 9.
The co-efficient of the term independent of x in the expansion of (2x + \(\frac{1}{3x}\))⁶ is……….
(a) \(\frac{160}{27}\)
(b) \(\frac{160}{37}\)
(c) \(\frac{80}{3}\)
(d) \(\frac{80}{9}\)
Solution:
(a) \(\frac{160}{27}\)

Question 10.
The value of \(\left|\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right|^{2}\) is……….
(a) abc
(b) -abc
(c) 0
(d) a²b²c²
Solution:
(d) a²b²c²

Question 11.

(a) Δ
(b) kΔ
(c) 3kΔ
(d) k³Δ
Solution:
(d) k³Δ

Question 12.
A vector makes equal angle with the positive direction of the co-ordinate axes then each angle is equal to………..
(a) cos^-1 (\(\frac{1}{3}\))
(b) cos^-1 (\(\frac{2}{3}\))
(c) cos^-1 (\(\frac{1}{√3}\))
(d) cos^-1 (\(\frac{2}{√3}\))
Solution:
(c) cos^-1 (\(\frac{1}{√3}\))

Question 13.
If the centroids of AABC and A’B’C’ are respectively G and G’ then \(\bar{AA’}\) + \(\bar{BB’}\) + \(\bar{CC’}\)…………
(a) \(\bar{GG’}\)
(b) 3\(\bar{GG’}\)
(c) 2\(\bar{GG’}\)
(d) 0
Solution:
(b) 3\(\bar{GG’}\)

Question 14.
If f(x) = \(\left\{\begin{aligned} k x^{2} & \text { for } \quad x \leq 2 \\ 3 & \text { for } \quad x>2 \end{aligned}\right.\) is continuous at x = 2 then the value of k is ………..
(a) \(\frac{3}{4}\)
(b) 0
(c) 1
(d) \(\frac{4}{3}\)
Solution:
(d) \(\frac{4}{3}\)

Question 15.
If x = \(\frac{1-t²}{1+t²}\) and y = \(\frac{2t}{1+t²}\) then \(\frac{dx}{dy}\) = ……….
(a) \(\frac{y}{x}\)
(b) \(\frac{-y}{x}\)
(c) –\(\frac{x}{y}\)
(d) \(\frac{x}{y}\)
Solution:
(c) –\(\frac{x}{y}\)

Question 16.
\(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+5 x+3}{x^{2}+x+3}\right)^{x}\) is ……….
(a) e⁴
(b) e²
(c) e³
(d) 1
Solution:
(a) e⁴

Question 17.
\(\int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1}\) dx is ……….
(a) e^xtan^-1(x + 1) + c
(b) tan^-1(e^x) + c
(c) e^x \(\frac{tan^{-1}x}{2}\) + c
(d) e^xtan^-1x + c
Solution:
(d) e^xtan^-1x + c

Question 18.
\(\int \frac{\sec x}{\sqrt{\cos 2 x}}\) dx = ………
(a) tan^-1(sin x) + c
(b) 2 sin^-1(tan x) + c
(c) tan^-1(cos x) + c
(d) sin^-1(tan x) + c
Solution:
(d) sin^-1(tan x) + c

Question 19.
\(\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\)
(a) x + c
(b) \(\frac{x³}{3}\) + c
(c) \(\frac{3}{x³}\) + c
(d) \(\frac{1}{x²}\) + c
Solution:
(b) \(\frac{x³}{3}\) + c

Question 20.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\) and P(B/A) = \(\frac{2}{3}\) then P(B) = …………
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{2}\)
Solution:
(b) \(\frac{1}{3}\)

PART- II

II. Answer any seven questions. Question No. 30 is compulsory.

Question 21.
Find x such that -π ≤ x ≤ π and cos 2x = sin x
Solution:
We have cos 2x = sin x which gives
2sin² x + sin x – 1 = 0
The roots of the equation are sin x = \(\frac{-1±3}{4}\) = -1 (or) \(\frac{1}{2}\)
Now, sin x = \(\frac{1}{2}\) ⇒ x = \(\frac{π}{6}\), \(\frac{5π}{6}\)
Also sin x = -1 ⇒ x = \(\frac{π}{2}\)
Thus x = \(\frac{π}{2}\), \(\frac{π}{6}\), \(\frac{5π}{6}\)

Question 22.
If ^(n-1)P₃ : ⁿP₄ = 1 : 10 find n.
Solution:

Question 23.
Find the 18^th and 25^th terms of the sequence defined by

Solution:
When n = 18 (even)
a_n = n(n + 2) = 18(18 + 2) = 18 (20) = 360
When n = 25 (odd)

Question 24.
Show that the lines are 3x + 2y + 9 = 0 and 12x + 8y – 15 = 0 are parallel lines.
Solution:

Here m₁ = m₂ ⇒ the two lines are parallel.

Question 25.
Prove that

Solution:

= 0 + 2(0) (∴ R₁ = R₃)
= 0 = RHS

Question 26.
Find the value of λ for which the vectors \(\vec{a}\) = 2\(\vec{i}\) + λ\(\vec{j}\) + \(\vec{k}\) and \(\vec{b}\) = \(\vec{i}\) – 2\(\vec{j}\) + 3\(\vec{k}\) are perpendicular.
Solution:
When \(\vec{a}\) and \(\vec{b}\) are ⊥^r If then \(\vec{a}\).\(\vec{b}\) = 0
\(\vec{a}\) ⊥^r \(\vec{b}\) ⇒ \(\vec{a}\).\(\vec{b}\) = 0
(2) (1) + (λ) (-2) + (1) (3) = 0 ⇒ λ = 5/2

Question 27.
If y = \(\frac{tan x}{x}\) and \(\frac{dx}{dy}\)
Solution:
now y = \(\frac{u}{v}\) ⇒ y’ =\(\frac{vu’-uv’}{v²}\)
u = tan x ⇒ u’ = sec²x
v = x ⇒ v’ = 1

Question 28.
Evaluate \(\int \sqrt{25 x^{2}-9}\) dx
Solution:

Question 29.
If A and B are two independent events such that
P(A) = 0.4 and P(A ∪ B) = 0.9. Find P(B).
Solution:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∪ B) = P(A) + P(B) – P(A).P(B)(since A and B are independent)
That is, 0.9 = 0.4 + P(B) – (0.4) P(B)
0.9 – 0.4 = (1 – 0.4) P(B)
Therefore, P(B) = \(\frac{5}{6}\)

Question 30.
A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed.
Solution:
The largest triangle will be an equilateral triangle
∴ side of the triangle = \(\frac{12}{3}\) = 4 m = a
Area of the triangle = \(\frac{a²√3}{4}\) = \(\frac{4²√3}{4}\) = 4√3 sq.m

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it

1. reflexive
2. symmetric
3. transitive
4. equivalence

Solution:

1. (c, c)
2. (c, a)
3. nothing
4. (c, c) and (c, a)

Question 32.
Solve 2|x + 1| – 6 ≤ 7 and graph the solution set in a number line.
Solution:
2|x + 1| – 6 ≤ 7
⇒ 2|x + 1| ≤ 7 + 6 (= 13)
⇒ |x + 1| ≤ \(\frac{13}{2}\)
⇒ x + 1 > \(\frac{-13}{2}\)
x + 1 > \(\frac{-13}{2}\)
⇒ x > \(\frac{-13}{2}\) -1 (= \(\frac{-15}{2}\)) ……….(1)
(or) x + 1 > \(\frac{13}{2}\)
x + 1 < \(\frac{-13}{2}\)
⇒ x < \(\frac{13}{2}\) -1 (= \(\frac{11}{2}\)) ……….(2)
From (1) and (2) \(\frac{-15}{2}\) ≤ x ≤ \(\frac{11}{2}\)

Question 33.
If the different permutations of all letters of the word BHASKARA are listed as in a dictionary, how many strings are there in this list before the first word starting with B?
Solution:
The required number of strings is the total number of strings starting with A and using the letters A, A, B, H, K, R, S = \(\frac{7!}{2}\) = 2520

Question 34.
Find the sum up to n terms of the series: 1 + \(\frac{6}{7}\) + \(\frac{11}{49}\) + \(\frac{16}{343}\) + ….
Solution:
Here a = 1, d = 5 and r = \(\frac{1}{7}\)

Question 35.
Area of the triangle formed by a line with the coordinate axes, is 36 square units. Find the equation of the line if the perpendicular drawn from the origin to the line makes an angle of 45° with positive the x-axis.
Solution:
Let p be the length of the perpendicular drawn from the origin to the required line.
The perpendicular makes 45° with the x-axis.
The equation of the required line is of the form,
x cos α + y sin α = p
⇒ x cos 45° + y sin 45° = p
⇒ x + y= √2 P
This equation cuts the coordinate axes at A(√2p, 0) and B (o, √2p).
Area of the ΔOAB is \(\frac{1}{2}\) × √2p × √2p = 36 ⇒ p = 6 (∵ p is positive)
Therefore the equation of the required line is x + y = 6 √2

Question 36.
If A^T = \(\left(\begin{array}{cc} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{array}\right)\) and B = \(\left(\begin{array}{ccc} 2 & -1 & 1 \\ 7 & 5 & -2 \end{array}\right)\) verify that (A – B)^T = A^T – B^T
Solution:
(A – B)^T = A^T – B^T

Here (1) and (2) ⇒ (A – B)^T = A^T – B^T

Question 37.
For any vector \(\vec{a}\) prove that \(|\vec{a} \times \hat{i}|^{2}+|\vec{a} \times \hat{j}|^{2}+|\vec{a} \times \hat{k}|^{2}=2|\vec{a}|^{2}\)
Solution:
Let = \(\vec{a}\) = a₁\(\hat{i}\) + a₂\(\hat{j}\) + a₃\(\hat{k}\)

Question 38.
Given y = cos^-1 \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\) find \(\frac{dx}{dy}\)
Solution:
put x = tan θ
so \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\) = \(\left(\frac{1-tan^{2}θ}{1+tan{2}θ}\right)\) = 2 cos θ
y = cos^-1 (cos 2θ) = 2θ
⇒ \(\frac{dy}{dθ}\)
Now x = tan θ
⇒ \(\frac{dx}{dθ}\) = sec²θ
= 1 + tan² θ
= 1 + x²
so \(\frac{dx}{dy}\) = \(\frac{dy}{dθ}\)/\(\frac{dx}{dθ}\) = \(\frac{2}{1+x²}\)

Question 39.
A wound is healing in such a way that t days since Sunday the area of the wound has been decreasing at a rate of –\(\frac{3}{(t+2)²}\) cm² per day. If on Monday the area of the wound was 2 cm²
(i) What was the area of the wound on Sunday?
(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the same rate ?
Solution:
Let A be the area of wound at time ‘ t ‘

By the given, condition area of the wound on monday is 2 cm²
⇒ A = 2, t = 1
⇒ 2 = \(\frac{3}{1+2}\) + c
c = 1
∴ Area of wound at any day.
⇒ 1 ⇒ A = \(\frac{3}{1+2}\) + 1
(i) The area of the wound on Sunday
t = 0 ⇒ A = \(\frac{3}{2}\) + 1 = \(\frac{5}{2}\) = 2.5 cm²
(ii) The area of the wound on Thursday
t = 4 ⇒ A = \(\frac{3}{6}\) + 1 = \(\frac{1}{2}\) + 1 = 1.5 cm²

Question 40.
An integer is chosen at random from the first fifty positive integers. What is probability that the integer chosen is a prime or multiple of 4?
Solution:
S = {1, 2, 3, ……. ,50} ∴ n(S) = 50
Let A be the event of getting prime number.
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
n(A) = 15, so P(A) = 15/50
Let B be the event of getting number multiple of 4
∴B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}
n(B) = 12, so P(B) = 12/50
Here A and B are mutually exclusive. (i.e.,) A ∩ B = Φ
∴ P(A ∪ B) = P(A) + P(B) = 15/50+ 12/50 = 27/50

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
The formula for converting from Fahrenheit to Celsius temperatures is y = \(\frac{5x}{9}\) – \(\frac{160}{9}\) Find the inverse of this function and determine whether the inverse is also a function.
Solution:

[OR]

(b) If f : R → R is defined by f(x) = 2x – 3 prove that f is a injection and find its inverse.
Solution:
Method 1:
One-to-one: Let f(x) = f(y). Then 2x – 3 = 2y – 3; this implies that x = y. That is, f(x) = f(y) implies that x = y. Thus f is one-to-one.

Onto: Let y ∈ R. Let x =\(\frac{y+3}{2}\) Then f(x) = 2(\(\frac{y+3}{2}\)) -3 = y. Thus f is onto. This also can be proved by saying the following statement. The range of f is R (how?) which is equal to the co-domain and hence f is onto.

Inverse: Inverse: Let y = 2x – 3. Then y + 3 = 2x and hence x = \(\frac{y+3}{2}\) Thus f^-1 (y) = \(\frac{y+3}{2}\). By replacing y as x, we get f^-1 (x) = \(\frac{y+3}{2}\)

Method 2:
Let y = 2x – 3 then x = \(\frac{y+3}{2}\). Let, g(y) = \(\frac{y+3}{2}\)
Now (g o f) (x) = g(f(x)) = g(2x – 3) = \(\frac{(2x-3)+3}{2}\) = x
(f o g)(y) = f(g(y)) = f(\(\frac{y+3}{2}\)) = 2(\(\frac{y+3}{2}\)) -3 = y
This implies that f and g are bijections and inverses to each other. Hence f is a bijection and f^-1(y) = \(\frac{y+3}{2}\). Replacing y by x we get, f^-1(x) = \(\frac{x+3}{2}\)

Question 42 (a).
If the equations x² – ax + b = 0 and x² – ex + f = 0 have one root in common and if the second equation has equal roots then prove that ae = 2(b + f).
Solution:
Let a be the common root
then a² – aα + b = 0 . . . .(1)
we are given that
x² – ex + f = 0 has equal roots.
So the roots will be α, β
Now sum of roots = 2α
= -(-e) ⇒ a = e/2
product of the roots
α × α = α²f
substituting a and α², values in (1) we get
f – a (\(\frac{e}{2}\)) + b = 0
f – \(\frac{ae}{2}\) + b = 0
\(\frac{ae}{2}\) = b + f ⇒ ae = 2(b + f)

[OR]

(b) Prove that
cos θ + cos(\(\frac{2π}{2}\)– θ) + (cos\(\frac{2π}{3}\) + θ) = 0
Solution:
we have

cos θ – cos θ = 0 = RHS

Question 43 (a).
Find the number of strings that can be made using all letters of the word THING. If these words are written as in a dictionary, what will be the 85th string?
Solution:
(i) Number of words formed = 5! = 120
(ii) The given word is THING
Taking the letters in alphabetical order G H I N T
To find the 85th word
The No. of words starting with G = 4! = 24
The No. of words starting with H = 4! = 24
The No. of words starting with 1 = 4! = 24
The No. of words starting with NG = 3! = 6
The No. of words starting with NH = 3! = 6
The No. of words starting with NIGH = 1! = 1
Total = 85
So the 85th word is NIGHT

[OR]

(b) A straight line passes through a fixed point (6, 8). Find the locus of the foot of the perpendicular drawn to it from the origin O.
Solution:
Let the point (x₁, y₁) be (6, 8). and P (h, k) be a point on thr lequired locus.
Family of equations of the straight lines passing through the fixed point (x₁, y₁) is y – y₁ = m (x – x₁) ⇒ y – 8 = m(x – 6)
Since OP is perpendicular to the line (6.25)
m × (\(\frac{k-0}{h-0}\)) = -1 ⇒ m = –\(\frac{h}{k}\)
Also P(h, k) lies on (6.25)
⇒ k – 8 = – \(\frac{h}{k}\)(h – 6) ⇒ k (k – 8) = -h(h – 6) ⇒ h² + k² – 6h – 8k = 0
Locus of P (h, k) is x² + y² – 6x – 8y = 0

Question 44 (a).
If p is a real number and if the middle term in the expansion of (\(\frac{π}{2}\) + 2)⁸ is 1120, find p.
Solution:
In the equation of (\(\frac{π}{2}\) + 2)⁸, number of terms = 8 + 1 = 9 (odd)
∴ There is only one middle term i.e. (\(\frac{9+1}{2}\))^th or 5^th term

[OR]

(b) Express the matrix A = \(\left[\begin{array}{rrr} 1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5 \end{array}\right]\) the sum Of skew-symmetric matrices.
Solution:

Thus A is expressed as the sum of symmetric and skew-symmetric matrices.

Question 45 (a).
For any vector a prove that \(|\vec{a} \times \hat{i}|^{2}+|\vec{a} \times \hat{j}|^{2}+|\vec{a} \times \hat{k}|^{2}=2|\vec{a}|^{2}\)
Solution:

[OR]

(b) If y = e^tan-1x show that (1 + x²) y” + (2x – 1) y’ = 0
Solution:
y = e^tan-1x
y’ = e^tan-1x (\(\frac{1}{1+x²}\))
y’ = \(\frac{y}{1+x²}\) ⇒ y’ (1 + x²) = y
differentiating w.r.to x
y’ (2x) + (1 + x²)(y”) = y’
(i.e.) (1 + x²) y” + y’(2x) – y’ = 0
(i.e.) (1 + x²)y” + (2x – 1) y’ = 0

Question 46 (a).
Evaluate \(\lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\ldots+(x+100)^{10}}{x^{10}+10^{10}}\)
Solution:

[OR]

(b) Evaluate

Solution:

Question 47 (a).
Evaluate \(\int \sin ^{-1}\)x dx
Solution:
Let 1 = sin^-1x dx
u = sin^-1x dv = dx

[OR]

(b) A factory has two machines I and II. Machine I produces 40% of items of the output and Machine II produces 60% of the items. Further 4% of items produced by Machine I are defective and 5% produced by Machine II are defective. An item is drawn at random. If the drawn item is defective, find the probability that it was produced by Machine II.
Solution:
Let A₁ be the event that the items are produced by Machine-I, A₂ be the event that items are produced by Machine-II, Let B be the event of drawing a defective item. Now we are asked to find the conditional probability P(A₂/B). Since A₁, A₂ are mutually exclusive and exhaustive events, by Bayes theorem,

Students can Download Tamil Nadu 11th Maths Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Maths Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
Let A and B be subsets of the universal set N, the set of natural numbers. Then A’∪[(A ∩ B) ∪B’] is
(a) A
(b) A’
(c) B
(d) N
Solution:
(d) N

Question 2.
For any two sets A and B if (A – B) ∪ (B – A) = ………..
(a) (A – B) ∪ A
(b) (B – A) ∪ B
(c) (A ∪ B) – (A ∩ B)
(d) (A ∪ B) ∩ (A ∩ B)
Solution:
(c) (A ∪ B) – (A ∩ B)

Question 3.
The equations whose roots are numerically equal but opposite in sign to the roots of 3x² – 5x – 7 = 0 is ……….
(a) 3x² – 5x – 7 = 0
(b) 3x² + 5x – 7 = 0
(c) 3x² – 5x + 7 = 0
(d) 3x² + x – 7 = 0
Solution:
(b) 3x² + 5x – 7 = 0

Question 4.
The value of sin(45° + θ) – cos (45° – θ) is
(a) 2 cos θ
(b) 1
(c) 0
(d) 2 sin θ
Solution:
(c) 0

Question 5.
If tan α sin β = 840, are the roots of x² + ax + b = 0 then \(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}\) is equal to ……..
(a) \(\frac{b}{a}\)
(b) \(\frac{a}{b}\)
(c) –\(\frac{a}{b}\)
(d) –\(\frac{b}{a}\)
Solution:
(c) –\(\frac{a}{b}\)

Question 6.
If a² – aC₂ = a² – aC₄ then the a value of a is ………
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(b) 3

Question 7.
If ⁿP_r = 840, ⁿC_r = 35 then n = …………
(a) 7
(b) 6
(c) 5
(d) 4
Solution:
(a) 7

Question 8.
If 2x² + 3xy – cy² = 0 represents a pair of perpendicular lines then c = ……….
(a) -2
(b) \(\frac{1}{2}\)
(c) –\(\frac{1}{2}\)
(d) 2
Solution:
(d) 2

Question 9.
If the nth term of an A.P is 2n – 1 then sum to n terms of that A.P. is……….
(a) n²
(b) n² + 1
(c) 2n – 1
(d) n² – 1
Solution:
(a) n²

Question 10.
If A = \(\left(\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right)\), B = \(\left(\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right)\)
(a) a = 4, b = 1
(b) a = 1, b = 4
(c) a = 0, 6 = 4
(d) a = 2, 6 = 4
Solution:
(b) a = 1, b = 4

Question 11.
If the points (x – 2), (5, 2), (8, 8) are collinear then x is equal to………..
(a) -3
(b) \(\frac{1}{3}\)
(c) 1
(d) 3
Solution:
(d) 3

Question 12.
In a regular hexagon ABCDEF if \(\vec { AB }\) and \(\vec { BC }\) are represented by \(\vec { a}\) and \(\vec { b }\) respectively then \(\vec { EF }\) =
(a) \(\vec { a }\) – \(\vec { b }\)
(b) \(\vec { a}\)
(c) –\(\vec { b }\)
(d) \(\vec { a }\) + \(\vec { b }\)
Solution:
(c) –\(\vec { b }\)

Question 13.
If |\(\vec { a }\) + \(\vec { b }\)| = 60, |\(\vec { a }\) – \(\vec { b }\)| = 40 and |\(\vec { b }\)| = 46, then |\(\vec { a }\)| is………….
(a) 42
(b) 12
(c) 22
(d) 32
Solution:
(c) 22

Question 14.
For \(\vec { a }\) = \(\vec { i }\) + \(\vec { j }\) – 2\(\vec { k }\), \(\vec { b }\) = –\(\vec { i }\) + 2\(\vec { j }\) + \(\vec { k }\) and \(\vec { c }\) = \(\vec { i }\) – 2\(\vec { j }\) + 2\(\vec { k }\), the unit vector parallal to \(\vec { a }\) + \(\vec { b }\) + \(\vec { c }\) is ………….
(a) \(\frac{\vec{i}+\vec{j}-\vec{k}}{\sqrt{3}}\)
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)
(c) \(\frac{\vec{i}+\vec{j}+\vec{k}}{3}\)
(d) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{6}}\)
Solution:
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)

Question 15.
The differential co-efficient of log₁₀x with respect to log₁₀ x is……….
(a) 1
(b) -(log₁₀x)²
(c) (log_x10)²
(d) \(\frac{x²}{100}\)
Solution:
(b) -(log₁₀x)²

Question 16.
\(\frac{d}{dx}\)(e_x+5logx) is………
(a) e₁₀x₁₀(x + 5)
(b) e_xx(x + 5)
(c) e_x + \(\frac{5}{x}\)
(d) e_x – \(\frac{5}{x}\)
Solution:
(a) e₁₀x₁₀(x + 5)

Question 17.
If f(x) = x tan_-1x then f'(1) = ……………
(a) 1 + \(\frac{π}{4}\)
(b) \(\frac{1}{2}\) + \(\frac{π}{4}\)
(c) \(\frac{1}{2}\) – \(\frac{π}{4}\)
(d) 2
Solution:
(b) \(\frac{1}{2}\) + \(\frac{π}{4}\)

Question 18.
∫ cosec x dx = ………..
(a) log tan \(\frac{π}{2}\) + c
(b) -log (cosec x + cot x) + c
(c) -log (cosec x + cot x) + c
(d) All of them
Solution:
(d) All of them

Question 19.
If A and B are two events such that A⊂B and P(B) ≠ 0, then which of the following is correct?
(a) P(A / B) = \(\frac{p(A)}{p(B)}\)
(b) P(A/B) < P(A)
(c) P(A/B ≥ P(A))
(d) P(A/B) > P(B)
Solution:
(c) P(A/B ≥ P(A))

Question 20.
A number x is chosen at random from the first 100 natural numbers. Let A be the event of numbers which satisfies \(\frac{(x-10)(x-50)}{x-30}\) ≥ 0, then P(A) is ………..
(a) 0.20
(b) 0.51
(c) 0.71
(d) 0.70
Solution:
(c) 0.71

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Write the values of f at -4, 1, -2, 7, 0 if

Solution:
f (- 4) = 4 + 4 = 8
f(1) = 1 – 1² = 0
f(-2) = 4 + 2 = 6
f(7) = 0
f(0) = 0

Question 22.
Solve 23x < 100 when
(i) x is a natural number
(ii) x is an integer
Solution:
23x < 100
⇒ \(\frac{23x}{23}\) < \(\frac{100}{23}\) (i.e.,) x > 23
(i) x = 1, 2, 3, 4 (x ∈ N)
(ii) x = -3, -2, -1, 0, 1, 2, 3, 4 (x ∈ Z)

Question 23.
Expand \(\frac{1}{5+x}\) in ascending powers of x.
Solution:

Question 24.
Find the nearest point on the line 2x +.y = 5 from the origin.
Solution:
The required point is the foot of the perpendicular from the origin on the line 2x + y = 5.
The line perpendicular to the given line, through the origin is x – 2y = 0.
Solving the equations 2x + y = 5 and x – 2y = 0, we get x = 2, y = 1.
Hence the nearest point on the line from the origin is (2, 1).
Alternate method: Using the formula

Question 25.
Determine 3B + 4C – D if B, C and D are given by

Solution:

Question 26.
Find the constant b that makes g continuous on (-∞, ∞) g(x) = \(\left\{\begin{array}{l} x^{2}-b^{2}, \text { if } x<4 \\ b x+20, \text { if } x \geq 4 \end{array}\right.\)
Solution: Since g(x) is continuous,
\(lim _{x \rightarrow 4^{-}}\) g(x) =\(lim _{x \rightarrow 4^{+}}\) g(x)
\(lim _{x \rightarrow 4^{-}}\)(x² – b²) = \(lim _{x \rightarrow 4^{+}}\) bx + 20
16 – b² = 4b + 20
b² + 4b + 4 = 0
(b + 2)² = 0
b = -2

Question 27.
Find \(\frac{dx}{dy}\) if x² + y² = 1
Solution:
We differentiate both sides of the equation.

Solving for the derivative yields
\(\frac{dx}{dy}\) = –\(\frac{x}{y}\)

Question 28.
Evaluate ∫ \(\frac{1}{sin²x cos²x}\) dx
Solution:

Question 29.
If P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8 find P(A/B) and P(A ∪ B)
Solution:
Given P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8
p(B/A) = \(\frac{p(A ∩ B)}{p(A)}\) = 0.8 (given)
⇒ \(\frac{p(A ∩ B)}{0.5}\) = 0.8
⇒ p(A ∩ B) = 0.8 × 0.5 = 0.4
(i.e.,) p(A ∩ B) = 0.4
(i) P(A/B) = \(\frac{p(A ∩ B)}{p(B)}\) = \(\frac{0.4}{0.8}\) = \(\frac{4}{8}\) = 0.5
(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.5 + 0.8 – 0.4 = 0.9
So, P(A/B) = 0.5 and P(A ∩ B) = 0.9.

Question 30.
Find the angle between the vectors 2\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\) and \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{i}\) using vector product.
Solution:
The angle between \(\vec{b}\) and \(\vec{b}\) using vector product is given by

PART – III

Answer any seven questions. Question No. 40 is compulsory.

Question 31.
If (x^1/2 + x^-1/2)² = \(\frac{9}{2}\) find the value of (x^1/2 – x ^-1/2) for x > 1
Solution:

Question 32.
If \(\frac{n!}{3!(n-4)!}\) and \(\frac{n!}{5!(n-5)!}\) are in the ratio 5 : 3 find the value of n.
Solution:

Question 33.
Expand (1 + x)^{\(\frac{2}{3}\)} up to four terms for |x| < 1.
Solution:

Question 34.
Find the equation of the line if the perpendicular drawn from the origin makes an angle 30° with x axis and its length is 12.
Solution:
The equation of the line is x cos a + y sin a = p
here a = 30°, cos a = cos 30° = \(\frac{√3}{2}\) ; sin a = sin 30° = 1/2; p = 12.
So equation of the line is x\(\frac{√3}{2}\)+ v\(\frac{1}{2}\) = 12
(i.e) √3x + y = 12 × 2 = 24 ⇒ √3x + y – 24 = 0

Question 35.
Prove that

Solution:

Question 36.
Find \(\lim _{t \rightarrow 0} \frac{\sqrt{t^{2}+9}-3}{t^{2}}\)
Solution:
We can’t apply the quotient theorem immediately. Use the algebra technique of rationalising the numerator.

Question 37.
Find \(\frac{dy}{dx}\) where x = \(\frac{1-t²}{1+t²}\), y = \(\frac{2t}{1+t²}\)
Solution:

Question 38.
Evaluate ∫(5x² – 4 + \(\frac{7}{x}\) + \(\frac{2}{√x}\))dx
Solution:

Question 39.
What is the chance that leap year should have fifty three Sundays?
Solution:
Leap Year: In 52 weeks we have 52 Sundays. We have to find the probability of getting one Sunday form the remaining 2 days the remaining 2 days can be a combination of the following S = {Saturday and Sunday, Sunday and Monday, Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday}.
(i.e) n(s) = 7
In this n(A) = {Saturday and Sunday, Sunday and Monday}
(i.e) n(A) = 2
So, P(A) = \(\frac{2}{7}\)

Question 40.
Find x from the equation cosec (90° + A) + x cos A cot (90° + A) = sin (90° + A).
Solution:
cosec (90° + A) = sec A, cot (90° + A) = – tan A
LHS = sec A + x cos A (-tan A)

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
From the curve y = x, draw
(i) y = -x
(ii) y = 2x
(iii) y = x + 1
(iv) y= \(\frac{1}{2}\)x + 1
(v) 2x + y + 3 = 0
Solution:

[OR]

(b) Solve √3 sin θ – cos θ = √2
Solution:
√3 sin θ – cos θ = √2
Here a = -1; b = √3 ; c = √2 ; r = \(\sqrt{a²+b²}\) = 2
Thus, the given equation can be rewritten as

Question 42 (a).
Solve \(\frac{x^{2}-4}{x^{2}-2x-15}\) ≤ 0
Solution:
\(\frac{x^{2}-4}{x^{2}-2x-15}\) ≤ 0 ⇒ \(\frac{(x-2)(x+2)}{(x+3)(x-5)}\) ≤ 0
x – 2 ⇒ x = 2; x + 2 = 0 ⇒ x = -2
x + 3 = 0 ⇒ x = -3; x – 5 = 0 ⇒ x = 5
plotting the points -3, -2, 2, 5 in the number line and taking the intervals

So the solution for the inequality \(\frac{x^{2}-4}{x^{2}-2x-15}\) ≤ 0 are (-3, -2) ∪ (2, 5)

[OR]

(b) Solve \(\frac{x+1}{x-1}\) > 0
Solution:
\(\frac{x+1}{x-1}\) > 0 ⇒ \(\frac{(x+1)(x-1)}{(x-1)²}\) > 0
(x + 1)(x – 1) > 0 (∵(x – 1)² >0 for all x ≠ l)

(x + 1) (x – 1) > 0
⇒ x ∈ (-∞, -1) ∪ (1, ∞)

Question 43 (a).
Use the principle of mathematical induction to prove that for every natural number n.

Solution:
Let P(n) be the given statement

For n = 1, LHS = 1 + \(\frac{3}{1}\) = 4
RHS = (1 + 1)² = 2² = 4
LHS = RHS
∴ ⇒ P(1) is true.
We note than P(n) is true for n = 1.
Assume that P(k) is true.

= k² +2k+ 1 + 2k + 3 = k² + 4k + 4 = (k + 2)²
= (k + 1 + 1)²
∴ p(k + 1)is also true whenever P(k) is true Hence, by the principle of mathematical induction, P(n) is also true for all n ∈ N.

[OR]

(b) If cos 2θ = 0 determine
Solution:
Given cos 2θ = 0
⇒ 2θ = π/2 ⇒ θ = π/4
∴ cos θ = cos π/4 = 1/√2
and
sin θ = sin π/4 = 1/√2

Question 44 (a).
Find the distance of the line 4x – y = 0 from the point P(4,1) measured along the line making an angle 135° with the positive x axis.
Solution:
The equation in distance form of the line passing through P (4, 1) and making an angle of 135° with the positive x – axis is
\(\frac{x-4}{cos135°}\) = \(\frac{y-1}{sin135°}\)
Suppose it cuts 4x – y – 0 at Q such that PQ = r then the coordinates of Q are given by

Hence, required distance is 3√2 units.

[OR]

(b) Evaluate \(\lim _{x \rightarrow \infty} x\left[3^{\frac{1}{x}}+1-\cos \left(\frac{1}{x}\right)-e^{\frac{1}{x}}\right]\)
Solution:

Question 45 (a).
Prove that \(\sqrt[3]{x^{3}+7}-\sqrt[3]{x^{3}+4}\) is approximately equal to \(\frac{1}{x²}\) when x is large.
Solution:

Since x is large, \(\frac{1}{x}\) is very small and hence higher powers of \(\frac{1}{x}\) are negligible. Thus

[OR]

(b) Evaluate sec³ 2x
I = ∫sec³ 2x dx = ∫sec 2x sec² 2x dx
Let u = sec 2x; du = 2 sec 2x tan 2x dx
sec²2x dx = dv

Question 46 (a).
Evaluate y = sin(tan(\(\sqrt{sin x}\)))
Solution:

[OR]

(b) Evaluate y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
Solution:

Question 47 (a).
Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.
Solution:
In ΔABC,

[OR]

(b) Given P(A) = 0.4 and P(A ∪ B) = 0.7 Find P(B) if
(i) A and B are mutually exclusive
(ii) A and B are independent events
(iii) P(A / B) = 0.4
(iv) P(B / A) = 0.5
Solution:
P(A) = 0.4, P(A ∪ B) = 0.7
(i) When A and B are mutually exclusive
P(A ∪ B) = P(A) + P(B)
(i.e.,) 0.7 = 0.4 + P(B)
0.7 – 0.4 = P(B)
(i.e.,) P(B) = 0.3

(ii) Given A and B are independent
⇒ P(A ∩ B) = P(A). P(B)
Now, P(A ∪ B) = P(A) + P(B) – P (A ∩ B)
(i.e.,) 0.7 = 0.4 + P(B) – (0.4) (P(B)
(i.e.,) 0.7 – 0.4 = P(B)(1 – 0.4)
0.3 = P (B) 0.6
⇒ P(B) = \(\frac{0.3}{0.6}\) = \(\frac{3}{6}\) = 0.5

(iii) P(A/B) = 0.4
(i.e.,) \(\frac{P(A ∩ B)}{P(B)}\) = 0.4
⇒ P(A ∩ B) = 0.4 P(B) …..(i)
But We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
⇒ P(A ∩ B) = 0.4 + P(B) – 0.7
= P(B) – 0.3
from (i) and (it) (equating RHS) we get
0-4 [P(B)] = P(B) – 0.3
0.3 = P(B)(1 – 0.4)
0.6 (P(B)) = 0.3 ⇒ P(B) = \(\frac{0.3}{0.6}\) = \(\frac{3}{6}\) = 0.5

(iv) P(B/A) = 0.5
(i.e.,) \(\frac{P(A ∩ B)}{P(A)}\) = 0.5
(i.e.,) P(A ∩ B) = 0.5 × P(A)
= 0.5 × 0.4 = 0.2
Now P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.7 = 0.4 + P(B) – 0.2
⇒ 0.7 = P(B) + 0.2
⇒ P(B) =0.7 – 0.2 = 0.5

Students can Download Tamil Nadu 11th Maths Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Maths Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 90

Part – I

Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
The number of students who take both the subjects Mathematics and Chemistry is 70. This represents 10% of the enrollment in Mathematics and 14% of the enrollment in Chemistry. The number of students take at least one of these two subjects, is
(a) 1120
(b)1130
(c) 1100
(d) insufficient data
Answer:
(b)1130

Question 2.
If 8 and 2 are the roots of x² + ax + c = 0 and 3, 3 are the roots of x² + dx + b = 0, then the roots of the equation x² + ax + b = 0 are
(a) 1, 2
(b) -1, 1
(c) 9, 1
(d) -1, 2
Answer:
(c) 9, 1

Question 3.
If tan 40° = λ then \(\frac{tan 140° – tan 130°}{1+tan 140° tan 130°}\) = ………….
(a) \(\frac{1-λ²}{λ}\)
(b) \(\frac{1+λ²}{λ}\)
(c) \(\frac{1+λ²}{2λ}\)
(d) \(\frac{1-λ²}{2λ}\)
Answer:
(d) \(\frac{1-λ²}{2λ}\)

Question 4.
The value of 2 sin A cos³ A – 2 cos A sin³ A is………..
(a) sin 4A
(b) cos 4A
(c) \(\frac{1}{2}\) sin 4A
(d) \(\frac{1}{2}\) cos 4A
Answer:
(c) \(\frac{1}{2}\) sin 4A

Question 5.
In a triangle ABC, sin²A + sin²B + sin²C = 2 then the triangle is ………. triangle.
(a) equilateral
(b) isosceles
(c) right
(d) scalene
Answer:
(c) right

Question 6.
The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is ………….
(a) 2 × 11C₇ + 10C₈
(b) 11C₇+ 10C₈
(c) 12C₈ – 10C₆
(d) 10C₆ + 2!
Answer:
(c) 12C₈ – 10C₆

Question 7.
If a is the arithmetic mean and g is the geometric mean of two numbers then…………
(a) a ≤ g
(b) a ≥ g
(c) a = g
(d) a > g
Answer:
(b) a ≥ g

Question 8.
The number of rectangles that a chessboard has…………
(a) 81
(b) 9⁹
(c) 1296
(d) 6561
Answer:
(c) 1296

Question 9.
The intercepts of the perpendicular bisector of the line segment joining (1,2) and (3,4) with coordinate axes are………….
(a) 5, -5
(b) 5, 5
(c) 5, 3
(d) 5, -4
Answer:
(b) 5, 5

Question 10.
The coordinates of the four vertices of a quadrilateral are (-2, 4), (-1, 2), (1, 2) and (2, 4) taken in order.The equation of the line passing through the vertex (-1, 2) and dividing the quadrilateral in the equal areas is…………
(a) x + 1 = 0
(b) x + y= 1
(c) x + y + 3 = 0
(d) x – y + 3 = 0
Answer:
(c) x + y + 3 = 0

Question 11.
The vectors \(\vec{a}\) – \(\vec{b}\), \(\vec{b}\) – \(\vec{c}\), \(\vec{c}\) – \(\vec{a}\) are …………. vectors.
(a) parallel
(b) unit
(c) mutually perpendicular
(d) coplanar
Answer:
(d) coplanar

Question 12.
If |\(\vec{a}\) + \(\vec{b}\)| = 60, |\(\vec{a}\) – \(\vec{b}\)| = 40, |\(\vec{b}\)| = 46 then |\(\vec{a}\)| is ……………
(a) 42
(b) 12
(c) 22
(d) 32
Answer:
(c) 22

Question 13.
Given \(\vec{a}\) = 2\(\vec{i}\) + \(\vec{j}\) – 8\(\vec{k}\) and \(\vec{b}\) = \(\vec{i}\) + 3\(\vec{j}\) – 4\(\vec{k}\) then |\(\vec{a}\) + \(\vec{b}\)|= …………..
(a) 13
(b) \(\frac{13}{3}\)
(c) \(\frac{4}{13}\)
(d) \(\frac{3}{13}\)
Answer:
(a) 13

Question 14.
If \(\vec{r}\) = \(\frac{9\vec{a}+7\vec{b}}{16}\) then the point P whose position vector \(\vec{r}\) divides the line joining the points with position vectors \(\vec{a}\) and \(\vec{b}\) in the ratio……….

(a) 7 : 9 internally
(b) 9 : 7 internally
(c) 9 : 7 externally
(d) 7 : 9 externally
Answer:
(a) 7 : 9 internally

Question 15.
If f(x) = x + 2 then f’ (f(x)) at x = 4 is…………
(a) 8
(b) 1
(c) 4
(d) 5
Answer:
(b) 1

Question 16.
The derivative of (x + \(\frac{1}{x}\))² w.r.to. x is ……….
(a) 2x – \(\frac{2}{x³}\)
(b) 2x + \(\frac{2}{x³}\)
(c) 2(x + \(\frac{1}{x}\) )
(d) 0
Answer:
(a) 2x – \(\frac{2}{x³}\)

Question 17.
If y = \(\frac{1}{a-z}\) then \(\frac{dz}{dy}\) is………..
(a) (a – z)²
(b) – (z – a)²
(c) (z + a)²
(d) -(z + a)²
Answer:
(a) (a – z)²

Question 18.
∫sin7x cos5x dx = ……….

Answer:
(b) –\(\frac{1}{2}\) [\(\frac{cos 12x}{2}\) + \(\frac{cos 2x}{2}\)] + c

Question 19.
∫\(\frac{1}{e^x}\) dx = …………..
(a) log e^x + c
(b) x + c
(c) \(\frac{1}{e^x}\) + c
(d) \(\frac{-1}{e^x}\) + c
Answer:
(d) \(\frac{-1}{e^x}\) + c

Question 20.
A letter is taken at random from the letters of the word ‘ASSISTANT’ and another letter is taken at random from the letters of the word ‘ STATISTICS ’. The probability that the selected letters are the same………..
(a) \(\frac{7}{45}\)
(b) \(\frac{17}{90}\)
(c) \(\frac{29}{90}\)
(d) \(\frac{19}{90}\)
Answer:
(d) \(\frac{19}{90}\)

PART- II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
For a set A, A × A contains 16 elements and two of its elements are (1,3) and (0,2). Find the elements of A.
Answer:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

Question 22.
Find the area of the triangle whose sides are 13 cm, 14 cm and 15 cm.
Answer:

Question 23.
If \(\frac{1}{7!}\) + \(\frac{1}{9!}\) = \(\frac{x}{10!}\) find x.
Answer:
Here \(\frac{1}{7!}\) + \(\frac{1}{9!}\) = \(\frac{x}{10!}\)

Question 24.
Find \(\sqrt[3]{1001}\) approximately (two decimal places)
Answer:

Question 25.
The slope of one of the straight lines ax² + 2hxy + by² = 0 is three times the other, show that 3h² = 4 ab.
Answer:
Let the slopes be m and 3m.
Now m + 3m = 4m = –\(\frac{2h}{b}\)
⇒ m = –\(\frac{2h}{4b}\) = –\(\frac{h}{2b}\) …….. (1)
m × 3m = \(\frac{a}{b}\) ⇒ 3m² = \(\frac{a}{b}\) ⇒ m² = \(\frac{a}{3b}\) ………. (2)
Eliminating m from (1) and (2)
we get

Question 26.
Simplify

Answer:
If we denote the given expression by A, then using the scalar multiplication rule, we get

Question 27.
Find a point whose position vector has magnitude 5 and parallel to the vector 4\(\hat{i}\) – 3\(\hat{j}\) + 10\(\hat{k}\).
Answer:
Let \(\vec{a}\) be the vector 4\(\hat{i}\) – 3\(\hat{j}\) + 10\(\hat{k}\)
The unit vector \(\hat{a}\) along the direction of \(\hat{a}\) is \(\frac{\vec{a}}{|\vec{a}|}\) which is equal to \(\frac{4 \hat{i}-3 \hat{j}+10 \hat{k}}{5 \sqrt{5}}\). the vector whose magnitude is 5 and parallel to

Question 28.
Evaluate \(\lim _{x \rightarrow-1}\left(x^{2}-3\right)^{10}\)
Answer:

Question 29.
Evaluate \(\int \frac{e^{2 x}+e^{-2 x}+2}{e^{x}}\)
Answer:

Question 30.
If A and B are mutually exclusive events P(A) = 3/8 and P(B) = 1/8 then find
(i) P(\(\bar{A}\) ∩ B)
(ii) p(\(\bar{A}\) ∪ \(\bar{B}\))
Answer:
(i) P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= 3/8 + 1/8 – 1/2
P(A ∩ B) = \(\frac{4-4}{8}\) = 0
P(\(\bar{A}\) ∩ B) = P(B) – P (A ∩ B) = 1/8 – 0 = 1/8

(ii) p(\(\bar{A}\) ∪ \(\bar{B}\)) = P[(A ∩B)’] = 1 – P(A ∩ B)
= 1 – 0 = 1

PART- III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
Find the range of the function f(x) = \(\frac{1}{1-3cos x}\)
Answer:
Clearly, -1 ≤ cos x ≤ 1
⇒ 3 ≥ -3 cos x ≥ -3
⇒ -3 ≤ -3 cos x ≤ 3
⇒ 1 – 3 ≤ 1 – 3 cos x ≤ 1 + 3
Thus we get -2 ≤ 1 – 3 cos x and 1 – 3 cos x ≤ 4.
By taking reciprocals, we get \(\frac{1}{1-3 cos x}\) ≤ –\(\frac{1}{2}\) and \(\frac{1}{1-3 cos x}\) ≥ \(\frac{1}{4}\).
Hence the range of f is (-∞, –\(\frac{1}{2}\) ] ∪ [\(\frac{1}{4}\), ∞)

Question 32.
Solve for x, -x² + 3x – 2 ≥ 0
Answer:
-x² + 3x – 2 ≥ 0 ⇒ x² – 3x + 2 ≤ 0
(x – 1)(x – 2) ≤ 0
[(x – 1) (x – 2) = 0 ⇒ x = 1 or 2. Here α = 1 and β = 2. Note that α < β]
So for the inequality (x – 1) (x – 2) ≤ 2
x lies between 1 and 2
(i.e.) x ≥ 1 and x ≤ 2 or x ∈ [1, 2] or 1 ≤ x ≤ 2

Question 33.
Solve the following equation for which solutions lies in the interval 0° ≤ θ ≤ 360°. sin⁴x = sin²x
Answer:
sin²x – sin⁴x = 0
sin² x (1 – sin² x) = 0
sin²x (cos² x) = 0
[\(\frac{1}{2}\) (2 sinx cos x)]² = 0
⇒ (sin2 x)² = 0
⇒ sin 2x = 0 = 0, π, 2π, 3π, 4π
x = 0, \(\frac{π}{2}\), π, \(\frac{π}{2}\), 2π

Question 34.
Find n if n – 1 P₃: nP₄ = 1 : 9
Answer:
Here ^{n – 1}P₃ : ⁿP₄ = 1 : 9

Question 35.
A family is using Liquefied petroleum gas (LPG) of weight 14.2 kg for consumption. (Full weight 29.5 kg includes the empty cylinders tare weight of 15.3 kg.). If it is used with constant rate then it lasts for 24 days. Then the new cylinder is replaced (i) Find the equation relating the quantity of gas in the cylinder to the days. (ii) Draw the graph for first 96 days.
Answer:
Since the usage is in constant rate and it is the slope m = \(\frac{14.2}{24}\)

which is the equation relating the quantity.
y – f(x) is a periodic function with period 24. (i.e.) f(x) = f(x + 24)

Question 36.
If cos 2θ = 0, determine
Answer:
Given cos 2θ = 0
⇒ 2θ = π/2 ⇒ θ = π/4

Question 37.
Show that the points (4, -3, 1), (2, -4, 5) and (1, -1, 0) form a right angled triangle.
Answer:
Trivially they form a triangle. It is enough to prove one angle is \(\frac{π}{2}\) . So find the sides of the triangle.
Let O be the point of reference and A, B, C be (4, -3, 1), (2, -4, 5) and (1, -1, 0) respectively.
\(\vec{OA}\) = 4\(\hat{i}\) -3\(\hat{j}\) + \(\hat{k}\), \(\vec{OB}\) = 2\(\hat{i}\) – 4\(\hat{j}\) + 5\(\hat{k}\), \(\vec{OC}\) = \(\hat{i}\) – \(\hat{j}\)
Now, \(\vec{AB}\) = \(\vec{OB}\) – \(\vec{OA}\) = -2\(\hat{i}\) – \(\hat{j}\) + 4\(\hat{k}\)
Similarly, \(\vec{BC}\) = –\(\hat{i}\) + 3\(\hat{j}\) – 5\(\hat{k}\); \(\vec{CA}\) = 3\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\)
Clearly \(\vec{AB}\) . \(\vec{CA}\) = 0
Thus one angle is \(\frac{π}{2}\). Hence they form a right angled triangle.

Question 38.
Compute

Answer:

= (0 + 1) + (0 + 3)
= 4

Question 39.
If for two events A and B, P(A) = \(\frac{3}{4}\), P(B) = \(\frac{2}{5}\) and A∪B conditional probability P(A/B).
Answer:
Given P(A) = \(\frac{3}{4}\), P(B) = \(\frac{2}{5}\) and P(A ∪ B) = l
Now P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
1 = \(\frac{3}{4}\) + \(\frac{2}{5}\) – P(A ∩ B)
P(A ∩ B) = \(\frac{3}{4}\) + \(\frac{2}{5}\) – 1 = \(\frac{15+8-20}{20}\)
P(A ∩ B) = 3/20
so P(A/B) = \(\frac{P(A ∪ B)}{P(B)}\) = \(\frac{3/20}{2/5}\) = \(\frac{3}{20}\) × \(\frac{5}{2}\) = \(\frac{3}{8}\)

Question 40.
Evaluate \(\int \frac{(x-1)^{2}}{x^{3}+x}\) dx
Answer:

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
Find the range of the function
Answer:
The range of cos x is – 1 to 1
– 1 < cos x < 1
(x by 2) – 2 < 2 cos x < 2
adding -1 throughout
-2 – 1 < 2 cos x – 1 < 2 – 1
(i.e.,) -3 < 2 cos x -1 < 1
so 1 < \(\frac{1}{2cosx – 1}\) < \(\frac{-1}{3}\)
The range is outside \(\frac{-1}{3}\) and 1
i.e., range is (-∞, \(\frac{-1}{3}\)]∪[1, ∞)

[OR]

(b) In any triangle ABC prove that a² = (b + c)² sin² \(\frac{A}{2}\) + (b – c)² cos² \(\frac{A}{2}\) Answer:
RHS = (b + c)² sin² \(\frac{A}{2}\) + (b – c)² cos² \(\frac{A}{2}\)
= (b² + c² + 2bc) sin² \(\frac{A}{2}\) + (b² + c² – 2bc) cos² \(\frac{A}{2}\)
= (b² + c²) [sin² \(\frac{A}{2}\) + cos² \(\frac{A}{2}\)] + 2bc [sin² \(\frac{A}{2}\) – cos² \(\frac{A}{2}\)]
= b² + c² – 2bc cos A = a² = LHS

Question 42 (a).
Find all values of x for which \(\frac{x³(x-1)}{x-2}\) > 0
Answer:
\(\frac{x³(x-1)}{x-2}\) > 0
Now we have to find the signs of
x³, x – 1 and x – 2 as follows
x³ = 0 ⇒ x = 0; x – 1 = 0 ⇒ x = 1; x – 2 = 0 ⇒ x = 2.
Plotting the points in a number line and finding intervals

So the solution set = (0, 1) ∪ (2, ∞)

[OR]

(b) resolve \(\frac{x}{(x²+1)(x-1)(x+2)}\) Into partial fractions.
Answer:

Equating numerator on both sides
x = A (x + 2) (x² + 1) + B (x – 1) (x² + 1) + (Cx + D) (x – 1) (x + 2)
This equations is true for any value of x to find A, B, C and D.
put x = 1
1 = A (3) (2) + B (0) + (0)
6A = 1 ⇒ A= 1/6
put x = -2
-2 = + 0 + B (-3) (5) + 0
⇒ -15B = -2 ⇒ B = 2/15
put x = 0
⇒ 2A – B – 2D = 0
(i.e.,) \(\frac{2}{6}\) – \(\frac{2}{15}\) – 2D = 0

Question 43 (a).
7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife has also 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of mens relative and 3 of the wifes relatives?
Answer:

We need 3 ladies and 3 gentlemen for the party which consist of 3 Husbands relative and 3 wifes relative.
This can be done as follows

⁴C₀ = ⁴C₄ = 1; ³C₀ = ³C₃ = 1
⁴C₁ = ⁴C₃ = 4; ³C₁ = ³C₂ = 3
⁴C₂ = \(\frac{4×3}{2×1}\) = 6
(4) (1) (1) (4) + (6) (3) (3) (6) + (4) (3) (3) (4) + (1) (1) (1) (1)
= 16 + 324 + 144 + 1 = 485 ways.

[OR]

(b) Show that the points (1, 3), (2, 1) and (\(\frac{1}{2}\), 4) are collinear, by using
(i) Concept of slope
(ii) Using a straight line and
(iii) Any other method.
Answer:
Let the given points be A (1, 3), B (2, 1), and C(\(\frac{1}{2}\), 4)

Slope of AB = Slope of BC ⇒ AB parallel to BC but B is a common point.
⇒ The points A, B, C are collinear.

(ii) Equation of the line passing through A and B is \(\frac{y-1}{3-1}\) = \(\frac{x-2}{1-2}\) ⇒ \(\frac{y-1}{2}\) = \(\frac{x-2}{-1}\)
1 – y = 2x – 4
2x +y = 5 ……….(1)
Substituting C\(\frac{1}{2}\), 4 in (1),
We get LHS = 2(\(\frac{1}{2}\)) + 4 = 1 + 4 = 5 = RHS
C is a point on AB
⇒ The points A, B, C lie on a line.
⇒ The points A, B, C are collinear.

(iii) Area of ΔABC = \(\frac{1}{2}\)(x₁) (y₂ y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)
= \(\frac{1}{2}\) {1(1, -4) + 2(4 – 3) +\(\frac{1}{2}\)(3 – 1)} = \(\frac{1}{2}\)(-3 + 2 + 1) = 0
⇒ The points A, B, C are collinear.

Question 44 (a).
Prove by vector Method’s that the Medians of a triangle are concurrent.
Answer:
Theorem: The medians of a triangle are concurrent.
Proof: Let ABC be a triangle and let D, E, F be the mid points of its sides BC, CA and AB respectively. We have to prove that the medians AD, BE, CF are concurrent.
Let O be the origin and \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be the position vectors of A, B, and C respectively.
The position vectors of D, E and F are respectively.
\(\frac{\vec{b}}{\vec{c}}\), \(\frac{\vec{c}}{\vec{a}}\), \(\frac{\vec{a}}{\vec{b}}\)
Let G₁, be the point on AD dividing it internally in the ratio 2 : 1.

From (1), (2) and (3) we find that the position vectors of the three points G₁, G₂, G₃ are one and the same. Hence they are not different points. Let the common point be G.
Therefore the three medians are concurrent and the point of concurrence is G.

[OR]

(b) If y = Ae^6x + Be^-x prove that \(\frac{d²y}{dx²}\) – 5\(\frac{dy}{dx}\) – 6y = 0
Answer:

Question 45 (a).
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer. [Hint: write aⁿ = (a-b + b)ⁿ and expand]
Answer:
a = a – b + b
So, aⁿ = [a – b + b]ⁿ =[(a – b) + b]ⁿ
= ⁿC₀ (a – b)ⁿ + ⁿC₁ (a -b)^n-1b¹ + ⁿC₂ (a – b)^n-2b² + ……… + ⁿC_n-1(a – b) b^n-1
+ ⁿC_n(bⁿ)
⇒ aⁿ – bⁿ = (a – b)ⁿ + ⁿC₁ (a – b)^n-1b + _nC₂ (a – b)^n-2b² + ……. + ⁿC_n-1 (a – b) b^n-1
= (a – b) [(a – b)^n-1 + ⁿC₁(a – b)^n-2b + ⁿC₂ (a – b)^n-3b² + ……… + ⁿC_n-1, b^n-1]
= (a – b) [an integer]
⇒ aⁿ – 6ⁿ is divisible by (a – b)

[OR]

(b) Verify the property A (B + C) = AB + AC when the matrices A, B and C are given by

Answer:

Question 46 (a).
Evaluate \(\lim _{x \rightarrow \infty} \sqrt{x^{2}+x+1}-\sqrt{x^{2}+1}\)
Answer:
Here the expression assumes the form ∞ to – ∞ as x → ∞. SO, we first reduce it to the rational form \(\frac{f(x)}{g(x)}\)

[OR]

(b) Evaluate \(\lim _{x \rightarrow \infty} x\left[3^{\frac{1}{x}}+1-\cos \left(\frac{1}{x}\right)-e^{1 / x}\right]\)
Answer:
Let y = \(\frac{1}{x}\) as x → ∞, y → 0

Question 47 (a).
Evaluate \(\int \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\) dx.
Answer:

[OR]

(b) Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?
Answer:
Given P(X) = 3/4, P(X’) = 1 – 3/4 = 1/4
P(Y) = 4/5, P(Y’) = 1 – 4/5 = 1/5
P(Z) = \(\frac{2}{3}\) P(Z’) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Question 1.
Fill in the blanks.
(i) Ratio of ₹ 3 to ₹ 5 = ____
(ii) Ratio of 3 m to 200 cm = ______
(iii) Ratio of 5 km 400 m to 6 km = ____
(iv) Ratio of 75 paise to ₹ 2 = ____
Solution:
(i) 3 : 5
(ii) 3 : 2
Hint: 3m = 300 cm
(iii) 9 : 10
Hint: 5km 400 m = 5400m and 6 km = 6000 m
(iv) 3 : 8
Hint: ₹ 2 = 200 paise

Question 2.
Say whether the following statements are True or False.
(i) The ratio of 130 cm to 1 m is 13 : 10.
(ii) One of the terms in a ratio cannot be 1.
Solution:
(i) True
(ii) False

Question 3.
Find the simplified form of the following ratios.
(i) 15 : 20
(ii) 32 : 24
(iii) 7 : 15
(iv) 12 : 27
(v) 75 : 100
Solution:

Question 4.
Akilan walks 10 km in an hour while Selvi walks 6 km in an hour. Find the simplest ratio of the distance covered by Akilan to that of Selvi.
Solution:
Ratio of the distance covered by Akilan to that of Selvi = 10 km : 6 km
= \(\frac{10}{6}\)
= \(\frac{5}{3}\)
= 5 : 3

Question 5.
The cost of parking a bicycle is ₹ 5 and the cost of parking a scooter is ₹ 15. Find the simplest ratio of the parking cost of a bicycle to that of a scooter.
Solution:
Parking cost of bicycle = ₹ 5
Parking cost of Scooter = ₹ 15
\(\frac{\text { Parking cost of bicyle }}{\text { Parking cost of scooter }}=\frac{5}{15}=\frac{1}{3}\)
Parking cost of bicycle : scooter = 1 : 3

Question 6.
Out of 50 students in class 30 are boys. Find the ratio of
(i) number of boys to the number of girls
(ii) number of girls to the total number of students
(iii) number of boys to the total number of students
Solution:
The total number of students = 50.
The number of boys = 30.
Then number of girls = 50 – 30 = 20.

Objective Type Questions

Question 7.
The ratio of Rs 1 to 20 paise is
(a) 1 : 5
(b) 1 : 2
(c) 2 : 1
(d) 5 : 1
Solution:
(d) 5 : 1

Question 8.
The ratio of 1 m to 50 cm is _____
(a) 1 : 50
(b) 50 : 1
(c) 2 : 1
(d) 1 : 2
Solution:
(c) 2 : 1
Hint: 1 m = 100 cm

Question 9.
The length and breadth of a window are in 1 m and 70 cm respectively. The ratio of the length to the breadth is
a) 1 : 7
(b) 7 : 1
(c) 7 : 10
(d) 10 : 7
Solution:
(d) 10 : 7

Question 10.
The ratio of the number of sides of a triangle to the number of sides of a rectangle is ____
(a) 4 : 3
(b) 3 : 4
(c) 3 : 5
(d) 3 : 2
Solution:
(b) 3 : 4
Triangle has three sides and a rectangle has four sides.

Question 11.
If Azhagan is 50 years old and his son is 10 years old then the simplest ratio between the age of Azhagan to his son is
(a) 10 : 50
(b) 50 : 10
(c) 5 : 1
(d) 1 : 5
Solution:
(c) 5 : 1

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 5 Information Processing Ex 5.2

Miscellaneous Practice problems

Question 1.
Find HCF of 188 and 230 by Euclid’s game.
Solution:
By Euclid’s game HCF (a, b) = HCF (a, a – b) if a > b.
Here HCF (188, 230) = HCF (230, – 188) because 230 > 188
= HCF (188, 42) = HCF (146, 42)
= HCF (104, 42) = HCF (62, 42)
= HCF (42, 20) = HCF (22, 20)
= HCF (20,2) = HCF (18, 2) = 2
∴ HCF (230, 188) = 2

Question 2.
Write the numbers from 1 to 50. From that find the following.
(i) The numbers which are neither divisible by 2 nor 7.
(ii) The prime numbers between 25 and 40.
(iii) All square number upto 50.
Solution :
(i) 9, 11, 13, 15, 17, 19, 23, 25, 27, 29, 31, 33, 37, 39, 41, 43, 45, 47
(ii) 29, 31, 37
(iii) 1, 4, 9, 16, 25, 36, 49

Question 3.
Complete the following pattern




Solution:
(i)

(ii)

Question 4.
Complete the table by using the following instructions.
A : It is the 6th term in the Fibonacci sequence.
B : The predecessor of 2.
C : LCM of 2 and 3.
D : HCF of 6 and 20.
E : The reciprocal of 1/5.
F : The opposite number of -7.
G : The first composite number.
H : Area of a square of side 3 cm.
I : The number of lines of symmetry of an equilateral triangle. After completing the table, what do you observe? Discuss.

Solution:
A : 6th term in Fibonacci sequence is 8.
B : Predecessor of 2 is 1.
C : LCM of 2 and 3 is 6.
D : HCF of 6 and 20 is 2.
E : Reciprocal of \(\frac{1}{5}\) is 5.
F : Opposite number of – 7 is 7.
G : The first composite number is 4.
H : Area of square of side 3 cm is 3 × 3 = 9 cm².
I : The number of lines of symmetry of an equilateral triangle is 3.
∴ The table becomes

From the table we observe that the numbers are from 1 to 9

Question 5.

Solution:

Question 6.
Replace the letter by symbols as + for A, – for B, × for C and ÷ for D. Find the answer for the pattern 4B3C5A30D2 by doing the given operations.
Solution:
4

Question 7.
Observe the pattern and find the word by hiding the numbers
1H2O3W 4A5R6E 7Y809U?
Solution:
When hiding the numbers we get

HOW ARE YOU?

Question 8.
Arranging from eldest to the youngest. What do you get ?

Solution:
Arranging from eldest to the youngest we get
F – refers to grandparents
A – refers to parents
M – refers to uncle
I – refers to elder sister
L – refers to me
Y – refers to younger brother
So we get FAMILY

Challenge Problems

Question 9.
Prepare a daily time schedule for evening study at home.
Solution:
5.30 pm – arrival
5.30 pm – 6.30 pm – Tea, Tv programme
6.30 pm – 7.30 pm – Maths
7.30 pm – 8.30 pm – Supper, Tv news
8.30 pm – 9.00 pm – English
9.00 pm – 9.30 pm – Science
9.30 pm – 10.00 pm – Social science
10.00 pm – Going to bed.

Question 10.
Observe the geometrical pattern and answer the following questions

i) Write down the number of sticks used in each of the iterative pattern.
ii) Draw the next figure in the pattern also find the total number sticks used in it.
Solution:
Number of sticks used in first pattern = 3
Number of sticks in second pattern = 9
Number of sticks in third pattern =18
ii) Next pattern

Question 11.
Find HCF of 28y, 35, 42 by Euclid’s game.
Solution:

Question 12.
Follow the given instructions to fill your name in the OMR sheet.
* The name should be written in capital letters from left to right.
* One alphabet is to be entered in each box.
* If any empty boxes are there at the end they should be left blank.
* Ball point pen is to be used for shading the bubbles for the corresponding alphabets.

Solution:

Question 13.
Consider the Postal Index Number (PIN) written on the letters as follows: 604506; 604516; 604560; 604506; 604516; 604516; 604560; 604516; 604505; 604470; 604515; 604520; 604303; 604509; 604470. How the letters can be sorted as per Postal Index Numbers?
Solution:
604 is common for all postal index numbers. Compare the remaining 3 digits, 303, 470, 505, 506 (two) 509, 510. 515, 516 (Four), 520, 560 (two).

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.1

Question 1.
Fill in the blanks.
(i) A line through two endpoints ‘A’ and ‘B’ is denoted by ______
(ii) Aline segment from point ‘B’ to point ‘A’ is denoted by ______
(iii) A ray has ______ endpoint(s).
Solution:
(i) \(\overleftrightarrow { AB }\)
(ii) \(\bar { BA } \)
(iii) one

Question 2.
How many line segments are there in the given line? Name them.

Solution:
(i) There are 10 line segments.
(ii) They are \(\overline{\mathrm{PA}}, \overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CQ}}, \overline{\mathrm{PB}}, \overline{\mathrm{AC}}, \overline{\mathrm{BQ}}, \overline{\mathrm{PC}}, \overline{\mathrm{AQ}} \text { and } \overline{\mathrm{PQ}}\)

Question 3.
Measure the following line segments.

Solution:
\(\overline{\mathrm{XY}}=2.4 \mathrm{cm} ; \overline{\mathrm{AB}}=3.4 \mathrm{cm} ; \overline{\mathrm{EF}}=4 \mathrm{cm} ; \overline{\mathrm{PQ}}=3 \mathrm{cm}\)

Question 4.
Construct a line segment using ruler and compass.
(i) \(\overline{\mathrm{AB}}=7.5 \mathrm{cm}\)
(ii) \(\overline{\mathrm{CD}}=3.6 \mathrm{cm}\)
(iii) \(\overline{\mathrm{QR}}=10 \mathrm{cm}\)
Solution:
(i) \(\overline{\mathrm{AB}}=7.5 \mathrm{cm}\)

Construction:
Step 1: Drawn a line l and marked point A.
Step 2: Measured 7.5 cm using compass placing the pointer at ‘0’ and pencil pointer at 7.5 cm.
Step 3: Placing a compass pointer at A, drawn an arc on l with the pencil pointer. It cut l at B.
Step 4: AB is the required segment of length 7.5 cm.

(ii) \(\overline{\mathrm{CD}}=3.6 \mathrm{cm}\)

Construction:
Step 1: Drawn a line l and marked the point ‘C’ on it.
Step 2: Measured 3.6 cm using a compass by placing the pointer at ‘O’ and pencil at 3.6 cm
Step 3: Placing pointer at C drawn the arc on ‘l’ with pencil pointer
Step 4: CD is the required line segment of length 3.6 cm

(iii) \(\overline{\mathrm{QR}}=10 \mathrm{cm}\)

Construction:
Step 1: Drawn a line ‘l’ and marked the point Q on it
Step 2: Measured 10 cm using a compass by placing the pointer at ‘0’ and pencil pointer at 10 cm.
Step 3: Placing pointer at Q drawn the arc on ‘l’ with pencil pointer and named the point R
Step 4: QR is the required line segments of 10 cm.

Question 5.
From the given figure
(i) identify the parallel lines
(ii) identify the intersecting lines
(iii) name the points of intersection

Solution:
(i) Parallel lines:
(a) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{AB}}\)
(b) \(\overrightarrow{\mathrm{EF}} \text { and } \overrightarrow{\mathrm{GH}}\)
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{EF}}\)
(b) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{GH}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(iii) Point of Intersection:
P, Q, R and S are the points of Intersection.

Question 6.
From the given figure, name the
(i) parallel lines
(ii) intersecting lines
(iii) points of Intersection.

Solution:
(i) Parallel lines:
\(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{EF}}\); \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{IJ}}\); \(\overrightarrow{\mathrm{EF}} \text { and } \overrightarrow{\mathrm{IJ}}\) are parallel lines.
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CD}}\)
(b) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{IJ}}\)
(e) \(\overrightarrow{\mathrm{GH}} \text { and } \overrightarrow{\mathrm{IJ}}\)
Points of Intersection:
P, Q and R are the points of intersection.

Question 7.
From the given figure, name
(i) all pairs of parallel lines
(ii) all pairs of intersecting lines
(iii) pair of lines whose point of intersection is ‘ V’
(iv) point of intersection of the lines ‘l₂‘ and ‘l₃‘
(v) point of intersection of the lines ‘l₁‘ and ‘l₅‘

Solution:
(i) Pairs of parallel lines:

• l₃ and l₄
• l₄ and l₅
• l₃ and l₅

(ii) Pairs of intersecting lines:

• l₁ and l₂
• l₁ and l₃
• l₁ and l₄
• l₁ and l₅
• l₂ and l₃
• l₂ and l₄
• l₂ and l₅

(iii) l₁ and l₂ intersect at ‘V’
(iv) point of intersection of the lines ‘l₂‘ and; l₅‘ is ‘Q’
(v) point of intersection of the lines ‘l₁‘ and ‘l₅‘ is ‘U’

Objective Type Questions

Question 8.
The number of line segments in  is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3
Hint: \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{BC}}\)

Question 9.
A line is denoted as
(a) AB
(b) \(\overrightarrow{AB}\)
(c) \(\overleftrightarrow {AB} \)
(d) \(\overline { AB }\)
Solution:
(c) \(\overleftrightarrow {AB} \)

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.1

Question 1.
Suppose, you have two shorts, one is black and the other one is blue; three shirts which are in white, blue and red. You again wish to make different combinations, but you always want to make sure that the shorts and shirt that you wear are of different colours. List and check how many combinations are possible now.
Solution:
We have given two shorts which are black and blue in colour. Take it as T black and T blue.
Also, we have 3 shirts, coloured white, blue and red denoted by S white, S blue and S red.
Now fix T black and then T blue the different combinations are

Thus we get a total of 6 combinations as
Black short and White shirt
Black short and Blue shirt
Black short and Redshirt
Blue short and White shirt
Blue short and Blue shirt
Blue short and Redshirt.
But it is given short and the shirt is of different colours.
We give up Blue short and Blue shirt combination. So we have 5 different combinations.

Question 2.
You have two red and two blue blocks. How many different towers can you build that are four blocks high using these blocks? List all the possibilities.
Solution:
6 Possibilities,
R-B-R-B
R-R-B-B
B-R-R-B
B-R-B-R
B-B-R-R
R-B-B-R

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Ex 3.2

Miscellaneous Practice Problems

Question 1.
A shopkeeper buys three articles for ₹ 325, ₹ 450 and ₹ 510. He is able to sell them for ₹ 350, ₹ 425 and ₹ 525 respectively. Find the gain or loss to the shopkeeper on the whole.
Solution:
C.P of three articles = 325 + 450 + 510 = ₹ 1285
S.P of three articles = 350 + 425 + 525 = ₹ 1,300
Here S.P > C.P
Profit = S.P – C.P = 1,300 – 1285 = ₹ 15
The shopkeeper gained = ₹ 15

Question 2.
A stationery shop owner bought a scientific calculator for Rs 750. He had put a battery worth Rs 100 in it. He had spent Rs 50 for its outer pouch. He was able to sell it for Rs 850. Find his profit or loss.
Solution:
CP = Rs 750 + Rs 100 + Rs 50
= Rs 900
SP = Rs 850
SP < CP
Loss = CP – SP
= Rs 900 – Rs 850
= Rs 50

Question 3.
Nathan paid ₹ 800 and bought 10 bottles of honey from a village vendor. He sold them in a city for ₹ 100 per bottle. Find his profit or loss.
Solution:
C.P of 10 bottles of honey = ₹ 800
C.P of 1 bottle honey = 800/10 = ₹ 80
S.P of a bottle honey = ₹ 100
Here S.P > C.P
Profit per bottle = ₹ 100 – ₹ 80 = ₹ 20
Profit for 10 bottles = 20 × 10 = ₹ 200
Profit = ₹ 200

Question 4.
A man bought 400 metre of cloth for Rs 60,000 and sold it at the rate of Rs 400 per metre. Find his profit or loss.
Solution:
Cost of 400 m of cloth = Rs 60,000
CP = Rs 60,000
Selling price of 400 m of cloth = 400 × Rs 400
SP = Rs 1,60,000
SP > CP
Profit = SP – CP
= Rs 1,60,000 – Rs 60,000
= Rs. 1,00,000

Challenge Problems

Question 5.
A fruit seller bought 2 dozen bananas at ₹ 20 a dozen and sold them at ₹ 3 per banana. Find his gain or loss.
Solution:
Cost of one dozen banana = ₹ 20
Cost of 2 dozen bananas = ₹ 20 × 2 = ₹ 40
C.P = ₹ 40
S.P per banana = ₹ 3
S.P for 2 dozen banana = ₹ 3 × 24 = ₹ 72
Here S.P > C.P
Profit = S.P – C.P = 72 – 40 = 32
Profit = ₹ 32

Question 6.
A store purchased pens at Rs 216 per dozen. He paid Rs 58 for conveyance and sold the pens at the discount of Rs 2 per pen and made an overall profit of Rs 50. Find the M.P. of each pen.
Solution:
Cost of 1 dozen pens = Rs 216 + Rs 58
CP = Rs 274
Discount for each pen = Rs 2
Overall profit = Rs 50
Total discount for 12 pens = Rs 2 × 12
= Rs 24
Selling price of 12 pens = Mp – Discount
SP = Mp – Rs 24
Profit = SP – CP
Rs 50 = Mp – Rs 24 – Rs 274
MP = Rs 50 + Rs 24 + Rs 274
= Rs 348 (1 dozen)
MP of each pen = Rs 348 / 12
= Rs 174/6
= Rs 29

Question 7.
A vegetable vendor buys 10 kg of tomatoes per day at ₹ 10 per kg, for the first three days of a weak. 1 kg of tomatoes got smashed on every day for those 3 days. For the remaining 4 days of the week, he buys 15 kg of tomatoes daily at ₹ 8 per kg. If for the entire week he sells tomatoes at ₹ 20 per kg, then find his profit or loss for the week.
Solution:
Total tomatoes bought for 3 days = 3 × 10 = 30 kg
Cost of 1 kg = ₹ 10
Cost of 30kg tomatoes = 30 × 10 = ₹ 300
Total tomatoes bought for other 4 days = 4 × 15 = 60 kg
Cost of 1 kg = ₹ 8
Cost of 60 kg tomatoes = 60 × 8 = ₹ 480
Total cost of 90 kg tomatoes = 300 + 480 = ₹ 780
C.P = ₹ 780
Tomatoes smashed = 3 kg
Total kg of Tomatoes for sale = 90 – 3 = 87 kg
S.P of 1 kg tomatoes = ₹ 20
S.P of 87 kg tomatoes = 87 × 20 = ₹ 1740
Here S.P > C.P
Profit = S.P – C.P = 1740 – 780 = ₹ 960
Profit = ₹ 960

Question 8.
An electrician buys a used T.V. for Rs 12,000 and a used Fridge for Rs 11,000. After spending Rs 1,000 on repairing the T.V. and Rs 1500 on painting the Fridge, he fixes up the M.P. of T.V. as Rs 15,000 and that of Fridge as Rs 15,500. If he gives Rs 1000 discount on each find his profit or loss.
Solution:
Total cost price of TV and Fridge = Rs 12,000 + Rs 11,000 + Rs 1,000 + Rs 1,500 = Rs 25,500
Selling price of TV = MP – Discount = Rs 15,000 – Rs 1,000 = Rs 14,000
Selling price of Fridge = MP – Discount = Rs 15,500 – Rs 1,000 = Rs 14,500
Total Selling price of TV and Fridge = Rs 14,000 + Rs 14,500 = Rs 28,500
SP > CP
Profit = SP – CP = Rs 28,500 – Rs 25,500 = Rs 3,000

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2

Miscellaneous Practice Problems

Question 1.
A piece of wire is 36 cm long. What will be the length of each side if we form
i) a square
ii) an equilateral triangle.
Solution:
Given the length of the wire = 36 cm
i) When a square is formed out of it
The perimeter of the square = 36 cm
4 × side = 36
side = \(\frac{36}{4}\) = 9 cm
Side of the square

ii) When an equilateral triangle is formed out of it, its perimeter = 36 cm
i.e., side + side + side = 36 cm .
3 × side = 36 cm
side = \(\frac{36}{3}\) = 12 cm
One side of an equilateral triangle = 12 cm

Question 2.
From one vertex of an equilateral triangle with side 40 cm, an equilateral triangle with 6 cm side is removed. What is the perimeter of the remaining portion?
Solution:

If an equilateral triangle of side 6 cm is removed the perimeter = (40 + 34 + 6 +34) cm = 114 cm
Perimeter of the remaining portion = 114 cm

Question 3.
Rahim and peter go for a morning walk, Rahim walks around a square path of side 50 m and Peter walks around a rectangular path with length 40 m and breadth 30 m. If both of them walk 2 rounds each, who covers more distance and by how much?
Solution:
Distance covered by Rahim
= 50 × 4 m
= 200 m
If he walks 2 rounds, distance covered = 2 × 200 m
= 400 m
Distance covered by peter
= 2 (40 + 30) m
= 2(70)m
= 140 m
If he walks 2 rounds, distance covered = 2 × 140 m
= 280 m
∴ Rahim covers more distance by (400 – 280) = 120 m

Question 4.
The length of a rectangular park is 14 m more than its breadth. If the perimeter of the park is 200 m, what is its length? Find the area of the park?
Solution:
Given length of rectangular park is 14m more than its breadth.
Let the breadth be b m .
∴ Length of the park will be l = b + 14 m
Given perimeter = 200 m
2 × (l + b) = 200 m
2 × (b + 14 + b) = 200 m [∵ l = b + 14]
2 × (2b + 14) = 200 m
2b + 14 = \(\frac{200}{2}\) m
2b + 14 = 100 m
2b = 100 – 14 m
2b = 86 m
b = \(\frac{86}{2}\) m
b 43 m
Length Length of the park = 57 m
Area of a rectangle = (length × breadth) unit²
= (57 × 43) m² = 2,451 m²
Area of the park = 2,451 m²

Question 5.
Your garden is in the shape of a square of side 5 m. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at Rs 10 per meter.
Solution:
a = 5 m
Perimeter of the garden
= 4 a units
= 4 × 5 m
= 20 m
For 1 row
Amount needed to fence l m= Rs 10
Amount needed to fence 20 m
= Rs 10 × 20
= Rs 200
For 2 rows
Total amount needed = 2 × Rs 200
= Rs 400

Challenge Problems

Question 6.
A closed shape has 20 equal sides and one of its sides is 3 cm. Find its perimeter.
Solution:
Number of equal sides in the shape = 20
One of its side = 3 cm
Perimeter = length of one side × Number of equal sides
∴ Perimeter = (3 × 20) cm = 60 cm
∴ Perimeter = 60 cm

Question 7.
A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it.
Solution:
l = 40 cm, b = 20 cm
Area of the rectangle = l × b sq units
= 40 × 20 cm²
= 800 cm²
a = 10 cm
Area of the square = a × a sq. units
= 10 × 10 cm²
= 100 m²
No of squares formed = \(\frac{800}{100}\) cm²
= 8

Question 8.
The length of a rectangle is three times its breadth. If its perimeter is 64 cm, find the sides of the rectangle.
Solution:
Given perimeter of a rectangle = 64 cm
Also given length is three times its breadth.
Let the breadth of the rectangle = b cm
∴ Length = 3 × b cm
Perimeter = 64 m
i.e., 2 × (l + b) = 64 m
2 × (3b + b) = 64 m
2 × 4b = 64m
4b = \(\frac{64}{2}\) = 32 m
b = \(\frac{32}{4}\) = 8 m
l = 3 × b = 3 × 8 = 24 m
∴ Breadth of the rectangle = 8 m
Length of the rectangle = 24 m

Question 9.
How many different rectangles can be made with a 48 cm long string? Find the possible pairs of length and breadth of the rectangles.
Solution:
12 rectangles
(1, 23), (2, 22), (3, 21), (4, 20), (5, 19), (6, 18), (7, 17), (8, 16), (9, 15), (10, 14), (11, 13), (12, 12)

Question 10.
Draw a square B whose side is twice of the square A. Calculate the perimeters of the squares A and B.
Solution:

Perimeter of A = s + s + s + s units = 4 s units
Perimeter of B = (2s + 2s + 2s + 2s) units
= 8s units = 2 (4s) units.
∴ Perimeter of B is twice perimeter of A

Question 11.
What will be the area of a new square formed if the side of a square is made one – fourth?
Solution:
Area of the new square is reduced to \(\frac{1}{16}\) th times to that of the original area.

Question 12.
Two plots have the same perimeter. One is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?
Solution:

Given perimeter of square = perimeter of rectangle
4 × side = 2 (length + breadth)
(4 × 10) m = 2(l + 8)m
\(\frac{4 \times 10}{2}\) = l + 8
20 = l + 8
l = 20 – 8
l = 12 m
∴ length of the rectangle = 12 m
Area of the square plot – side × side = 10 × 10 m² = 100 m²
Area of the rectangular plot = length × breadth = (12 × 8) m² = 96 m²
100 m² > 96 m²
∴ Square plot has greater area by 4m²

Question 13.
Look at the picture of the house given and find the total area of the shaded portion.

Solution:
Total area of the shaded region = Area of a right triangle + Area of a rectangle
= (\(\frac{1}{2}\) × b × h) + (l × b) cm²
= [(\(\frac{1}{2}\) × 3 × 4) + (9 × 6)] cm²
= (6 + 54) cm² = 60 cm²

Question 14.
Find the approximate area of the flower in the given square grid.

Solution:
Approximate area = Number of full squares + Number of more than half squares + \(\frac{1}{2}\) × Number of half squares.
= 10 + 5 + (\(\frac{1}{2}\) × 1) Sq units. = 10 + 5 + \(\frac{1}{2}\) sq. units
= 15 \(\frac{1}{2}\) sq. units = 15.5 sq. units.

Students can Download Tamil Nadu 11th Maths Previous Year Question Paper June 2019 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Maths Previous Year Question Paper June 2019 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
The range of the function \(\frac{1}{1-2sin x}\) is………….
(a) (-∞, -1) ∪(\(\frac{1}{3}\), ∞)
(b) (-1, \(\frac{1}{3}\))
(c) [-1, \(\frac{1}{3}\)]
(d) (-∞, -1] ∪(\(\frac{1}{3}\), ∞)
Answer:
(d) (-∞, -1] ∪(\(\frac{1}{3}\), ∞)

Question 2.
If the function f : [-3, 3] → S defined by f(x) = x² is onto, then S is…………
(a) [-9, 9]
(b) R
(c) [-3, 3]
(d) [0, 9]
Answer:
(d) [0, 9]

Question 3.
The number of solutions of x² + |x – 1| = 1 is ………..
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

Question 4.
cos 1° + cos 2° + cos 3° + …. + cos 179° =……………..
(a) 0
(b) 1
(c) -1
(d) 89
Answer:
(a) 0

Question 5.
If tan α and tan β are the roots of x² + ax + b = 0, then \(\frac{sin(α+β)}{sin α sin β}\) is equal to…………
(a) \(\frac{b}{a}\)
(b) \(\frac{a}{b}\)
(c) –\(\frac{a}{b}\)
(d) –\(\frac{b}{a}\)
Answer:
(b) \(\frac{a}{b}\)

Question 6.
The number of sides of a polygon having 44 diagonals is…………
(a) 4
(b) 4
(c) 11
(d) 22
Answer:
(d) 22

Question 7.
The H.M. of two positive numbers whose A.M. and G.M. are 16, 8 respectively is………….
(a) 10
(b) 6
(c) 5
(d) 4
Answer:
(d) 4

Question 8.
The nth term of the sequence \(\frac{1}{2}\), \(\frac{3}{4}\), \(\frac{7}{8}\), \(\frac{15}{16}\) is……………
(a) 2ⁿ – n – 1
(b) 1 – 2^-n
(c) 2^-n+ n – 1
(d) 2^n-1
Answer:
(b) 1 – 2^-n

Question 9.
The intercepts of the perpendicular bisector of the line segment joining (1,2) and (3,4) with coordinate axes are
(a) 5, -5
(b) 5, 5
(c) 5, 3
(d) 5, -4
Answer:
(b) 5, 5

Question 10.
The image of the point (2, 3) in the line y = -x is
(a) (-3, -2)
(b) (-3, 2)
(c) (-2, -3)
(d) (3, 2)
Answer:
(a) (-3, -2)

Question 11.
If A = \(\left[\begin{array}{cc} \lambda & 1 \\ -1 & -\lambda \end{array}\right]\), then for what value of λ, A² = 0?………
(a) 0
(b) ± 1
(c) -1
(d) 1
Answer:
(b) ± 1

Question 12.
If \(\vec{a}\) and \(\vec{b}\) are having same magnitute and angle between them is 60° and their scalar product is \(\frac{1}{2}\), then |\(\vec {a}\)| is
(a) 2
(b) 3
(c) 7
(d) 1
Answer:
(d) 1

Question 13.
\(\lim _{x \rightarrow \infty} \frac{a^{x}-b^{x}}{x}\) = …………..
(a) log ab
(b) log (\(\frac{a}{b}\))
(c) log (\(\frac{b}{a}\))
(d) \(\frac{a}{b}\)
Answer:
(b) log (\(\frac{a}{b}\))

Question 14.
If f(x) = \(\left\{\begin{array}{ccc}x, & x \text { is irrational } \\ 1-x, & x \text { is rational }\end{array}\right.\) then f is………..
(a) discontinuous at x = \(\frac{1}{2}\)
(b) continuous at x = \(\frac{1}{2}\)
(c) continuous everywhere
(d) discontinuous everywhere
Answer:
(b) continuous at x = \(\frac{1}{2}\)

Question 15.
The derivative of f(x) = x |x| at x = -3 is…………..
(a) 6
(b) -6
(c) does not exist
(d) 0
Answer:
(a) 6

Question 16.
If f(x) = x² – 3x, then the points at which f(x) = f'(x) are…………..
(a) both positive integers
(b) both negative integers
(c) both irrational
(d) one rational and another irrational
Answer:
(c) both irrational

Question 17.
\(\int \tan ^{-1}(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}})\) dx ………….
(a) x² + c
(b) 2x² + c
(c) \(\frac{x²}{2}\) + c
(d) –\(\frac{x²}{2}\) + c
Answer:
(c) \(\frac{x^2}{2}\) + c

Question 18.
e^-7x sin 5x dx is…………
(a) \(\frac{e^{-7x}}{74}\) [-7 sin 5x – 5 cos 5x] + c
(b) \(\frac{e^{-7x}}{74}\) [7 sin 5x + 5 cos 5x] + c
(c) \(\frac{e^{-7x}}{74}\) [7 sin 5x – 5 cos 5x] + c
(d) \(\frac{e^{-7x}}{74}\) [-7 sin 5x + 5 cos 5x] + c
Answer:
(a) \(\frac{e^{-7x}}{74}\) [-7 sin 5x – 5 cos 5x] + c

Question 19.
If A and B are any two events then the probability that exactly one of them occur is
(a) P(A ∪\(\bar { B }\)) + P(\(\bar { A }\) ∪B)
(b) P(A ∩\(\bar { B }\)) + P(\(\bar { A }\) ∩B)
(c) P(A) + P(B) – P(A ∩ B)
(d) P(A) + P(B) + 2P(A ∩ B)
Answer:
(a) P(A ∪\(\bar { B }\)) + P(\(\bar { A }\) ∪B)

Question 20.
In a certain college 4% of the boys and 1 % of the girls are taller than 1.8 meter. Further 60% of the students are girls. If a student is selected at random and is taller than 1.8 meters, then the probability that the student is a girl is ………….
(a) \(\frac{2}{11}\)
(b) \(\frac{3}{11}\)
(c) \(\frac{5}{11}\)
(d) \(\frac{7}{11}\)
Answer:
(b) \(\frac{3}{11}\)

PART- II

II. Answer any seven questions. Question No: 30 is compulsory. [7 × 2 = 14]

Question 21.
Resolve \(\frac{3x+1}{(x-2)(x+1)}\) into partial fractions.
Answer:

Equating numerator parts
3x + 1 = A (x + 1) + B (x – 2)
This equation is true for any value of x.
To find A and B
Put x = -1
-3 + 1 = A (0) + B (-1 -2)
-3 B = -2 ⇒ B = 2/3
Put x = 2
3(2) + 1 = A(2 + 1) + B (0)
3A = 7 ⇒ A = 7/3
Hence \(\frac{3x+1}{(x-2)(x+1)}\) = \(\frac{7}{3(x-2)}\) + \(\frac{2}{3(x+1)}\)

Question 22.
Prove that

Answer:
cot (180° + θ) = cot θ
sin (90° – θ) = cos θ
cos (- θ) = cos θ
sin (270 + θ) = – cos θ
tan (-θ) = – tan θ
cosec (360° + θ) = cosec θ

Question 23.
If cos θ = \(\frac{1}{2}\) (a + \(\frac{1}{a}\)), show that cos 3θ = \(\frac{1}{2}\) (a³ + \(\frac{1}{a³}\))
Answer:

Question 24.
If the letters of the word IITJEE are permuted in all possible ways and the strings thus formed are arranged in the lexicographic order, find the rank of the word IITJEE.
Answer:
The lexicographic order of the letters of given word is E, E, I, I, J, T. In the lexicographic order, the strings which begin with E come first. If we fill the first place with E, remaining 5 letters (E, I, I, J, T) can be arranged in \(\frac{5!}{2!}\) ways. On proceeding like this we get,
E – – – – = \(\frac{5!}{2!}\) = 60 ways
IIE – – – = 3! = 6 ways
IIJ – – – = \(\frac{3!}{1!}\) = 3 ways
IITE – – =2! = 2 ways
IITJEE = 1 way
The rank of the word IITJEE = 60 + 6 + 3 + 2 +1 = 72.

Question 25.
Prove that \(\frac{(2n)!}{n!}\) = 2ⁿ (1, 3, 5 ……..(2n – 1)).
Answer:

Question 26.
Write the equation of the line passing through the point (1, -1) and parallel to the line x + 3y – 4 = 0
Answer:
Equation of a line parallel to x + 3y – 4 = 0 will be of the form x + 3y + k = 0
It passes through (1,-1)
⇒ (1) + 3(-1) + k = 0
-2 + k = 0 ⇒ k = 2
So the required equation is x + 3y + 2 = 0

Question 27.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
Answer:
Area of Δ with vertices (k, 2) (2, 4) and (3, 2) = \(\frac{1}{2}\) \(\left|\begin{array}{lll} k & 2 & 1 \\ 2 & 4 & 1 \\ 3 & 2 & 1 \end{array}\right|\) = 4 (given)
⇒ \(\left|\begin{array}{lll} k & 2 & 1 \\ 2 & 4 & 1 \\ 3 & 2 & 1 \end{array}\right|\) = 2(4) = 8
(i.e.,) k (4 – 2) -2(2 – 3) + 1 (4 – 12) = ±8
(i.e.,) 2k – 2(-1) + 1(-8) = ± 8
(i.e.,) 2k + 2 – 8 = 8
(i.e.,) 2k = 8 + 8 – 2 = 14
k = 14/2 = 7
∴ k = 7
or
2k + 2 – 8 = -8
⇒ 2k = -8 + 8 – 2
2k = -2
k = -1
So k = 7 (or) k = -1

Question 28.
Find λ, when the projection of \(\vec {a}\) = λ\(\hat{j}\) + \(\hat{j}\) + 4\(\hat{k}\) on \(\vec {b}\) = 2\(\hat{i}\) + 6\(\hat{j}\) + 2\(\hat{k}\) is 4 units.
Answer:
\(\vec {a}\) = λ\(\hat{j}\) + \(\hat{j}\) + 4\(\hat{k}\) and \(\vec {b}\) = 2\(\hat{i}\) + 6\(\hat{j}\) + 2\(\hat{k}\)
Now \(\vec {a}\) – \(\vec {b}\) = (λ) (2) + (1) (6) + (4) (3)
= 2λ + 6 + 12 = 2λ +18
|\(\vec {a}\)| = \(\sqrt{4+36+9}\) = \(\sqrt{49}\) = 7
Here \(\frac{2λ+18}{7}\) = 4
⇒ 2λ+ 18 = 4 × 7 = 28
2λ = 28 – 18 = 10
λ = 10/2 = 5

Question 29.
Find \(\frac{dy}{dx}\) if y = e^x sin x
Answer:
y = e^x sin x
⇒ y’ = uv’ + vu’
Now u = e^x ⇒ u’ = \(\frac{du}{dx}\) = e^x
v = sin x ⇒ v’ = \(\frac{dv}{dx}\) = cos x
y’ = e^x (cos x) + sin x (e^x)
= e^x [sin x + cos x]

Question 30.
Find \(\frac{dy}{dx}\) if x² + y² = 1
Answer:
We differentiate both sides of the equation.
\(\frac{d}{dx}\) (x)² + \(\frac{d}{dx}\) (y)² = \(\frac{d}{dx}\) (1)
2x + 2y \(\frac{dy}{dx}\) = 0
Solving for the derivative yields
\(\frac{dy}{dx}\) = –\(\frac{x}{y}\)

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
From the curve y = sin x, draw y = sin |x|
Answer:

Question 32.
If one root of k (x – 1)² = 5x – 7 is double the other root, show that k = 2 or -25
Answer:
k(x – 1)² = 5x – 7
(i.e.,) k (x² – 2x + 1) – 5x + 7 = 0
x² (k) + x(-2k – 5) + k + 7 = 0
kx² – x(2k + 5) + (k + 7) = 0
Here it is given that one root is double the other.
So let the roots to α and 2α
Sum of the roots = α + 2α = 3α = \(\frac{2k+5}{k}\) α \(\frac{2k+5}{3k}\) ……..(1)
Product of the roots = α(2α) = 2α² = \(\frac{k+7}{k}\)
⇒ α² = \(\frac{k+7}{2k}\) = ……..(2)
Substituting a value from (1) in (2)

2(4k² + 25 + 20k) = 9k (k + 7)
2(4k² + 25 + 20k) = 9k² + 63k
8k² + 50 + 40k – 9k² – 65k = 0
-k² – 25k + 50 = 0
k² + 23k – 50 = 0
(k + 25) (k – 2) = 0
k = -25 or 2

Question 33.
If θ + ∅ = α and tan θ = k tan ∅, then prove that sin(θ – ∅) =\(\frac{k-1}{k+1}\)
Answer:

Question 34.
If a, b, c are in geometric progression, and if a^{\(\frac{1}{x}\)} = b^{\(\frac{1}{y}\)} = c^{\(\frac{1}{z}\)} then prove that x, y, z are in arithmetic progression.
Answer:
Given a, b, c are in G.P.
⇒ b² = ac
⇒ log b² = log ac
(i.e.) 2 log b = log a + log c …(1)
We are given a^{\(\frac{1}{x}\)} = b^{\(\frac{1}{y}\)} = c^{\(\frac{1}{z}\)} = k (say)
⇒ log a^k = \(\frac{1}{x}\) = b^k \(\frac{1}{y}\) = c^k \(\frac{1}{z}\)
⇒ a^k = \(\frac{1}{x}\) ⇒ x = log k^a
Similarly y = log k^b
z = log k^c
Substituting these values in equation (1) we get 2y = x + z ⇒ x, y, z are in A.P.

Question 35.
Show that the points (1,3), (2,1) and (\(\frac{1}{2}\), 4) are collinear.
Answer:
Let the given points be A (1, 3), B (2, 1), and C (\(\frac{1}{2}\), 4)
Slope of AB = \(\frac{1-3}{2-1}\) = \(\frac{-2}{1}\) = -2 = m₁
Slope of BC = \(\frac{4-1}{1/2-1}\) = \(\frac{3}{-3/2}\) = -2 = m₂
Slope of AB = Slope of BC ⇒ AB parallel to BC but B is a common point.
⇒ The points A, B, C are collinear.

Question 36.
If A and B are symmetric matrices of same order, prove that AB + BA is a symmetric matrix.
Answer:
Given A and B are symmetric matrices
⇒ A^T = A and B^T = B
To prove AB + BA is a symmetric matrix.
Proof: Now(AB + BA)^T = (AB)^T + (BA)^T = B^TA^T + A^TB^T
= BA + AB = AB + BA
i.e. (AB + BA)^T = AB + BA
⇒ (AB + BA) is a symmetric matrix.

Question 37.
Let \(\vec {a}\), \(\vec {b}\), \(\vec {c}\) be three vectors such that |\(\vec {a}\)| = 3, |\(\vec {b}\)| = 4, |\(\vec {c}\)| = 5 and each one of them being perpendicular to the sum of the other two, find |\(\vec {a}\) + \(\vec {b}\) + \(\vec {c}\)|.
Answer:
Given |\(\vec {a}\)| = 3; |\(\vec {b}\)| = 4; |\(\vec {c}\)| = 5
Now, (\(\vec {a}\) + \(\vec {b}\) + \(\vec {c}\))² = a^-2 +b ^-2+c^-2 + 2(\(\vec {a}\).\(\vec {b}\) + \(\vec {b}\).\(\vec {c}\) + \(\vec {a}\).\(\vec {c}\))
= 3² + 4² + 5² + 2 \(\vec {a}\).\(\vec {b}\) + 2\(\vec {b}\).\(\vec {c}\) + 2\(\vec {a}\).\(\vec {c}\)
= 9 + 16 + 25 + \(\vec {a}\).\(\vec {b}\) + \(\vec {a}\).\(\vec {b}\) + \(\vec {b}\).\(\vec {c}\) + \(\vec {b}\).\(\vec {c}\) + \(\vec {a}\).\(\vec {c}\) + \(\vec {a}\).\(\vec {c}\)
= 50 + \(\vec {a}\).(\(\vec {b}\) + \(\vec {c}\)) + \(\vec {b}\).(\(\vec {c}\) + \(\vec {a}\)) + \(\vec {c}\).(\(\vec {a}\) + \(\vec {b}\))
= 50 + 0 + 0 + 0 = 50
(one vector is ⊥^r to the sum of other two vectors)
|\(\vec {a}\) + \(\vec {b}\) + \(\vec {c}\)| = \(\sqrt {50}\) = \(\sqrt {25×2}\) = 5√2

Question 38.
Evaluate \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-1}{x}\)
Answer:

Question 39.
Find the derivatives of the following functions y = x^{cos x}
Answer:
y = x^{cos x}
Taking log on both sides, we get
log y = log x^{cos x} = cos x log x
Differentiating w.r.to x we get

Question 40.
Evaluate

Answer:

PART – IV

IV. Answer all the questions. [7 x 5 = 35]

Question 41 (a).
Let f, g: R → R be defined as f(x) = 2x – |x| and g(x) = 2x + |x|. Find fog.
Answer:

fog (x) = f(g(x))
= f(x) = 3x
For x > 0
fog (x) = f(g(x))
= f(3x) = 3x

[OR]

(b) Prove that \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}\) = 1
Answer:
Let θ → 0, thought positive values
∴ Let 0 < θ < \(\frac{π}{2}\)
Draw a circule center O and radius unity and let
∠AOB = θ radiAnswer:
Let the tagent at A meet OB product in the piont P.
Jion AB. From the figure it is clear that
Area of ΔAOB < Area of sector AOB < Area of ΔOAP

i.e., \(\frac{1}{2}\) OA. OB sin θ < 1 (radius)² θ < \(\frac{1}{2}\) OA. AP
or sin θ < θ < tan θ
[ ∵ \(\frac{AP}{OA}\) = tan θ or AP = tan θ as OA = 1
Dividing by sin θ, which is positive

Question 42 (a).
(i) If A × A has 16 elements, S = {(a, b) ∈ A × A: a < b} ; (-1, 2) and (0,1) are two elements of S, then find the remaining elements of S.
(ii) Find the range of the function \(\frac{1}{2 cos x – 1}\)
Answer:
(i) n(A × A) = 16
⇒ n(A) = 4
S = {(-1, 0), (-1, 1), (0, 2), (1, 2)}

(ii) The range of cos x is – 1 to 1
-1 ≤ cos x ≤ 1
(× by 2) -2 ≤ 2 cos x ≤ 2
adding -1 throughout
-2 -1 ≤ 2 cos x – 1 ≤ 2 – 1
(i.e.,) -3 ≤ 2 cos x – 1 ≤ 1
so 1 ≤ \(\frac{1}{2 cos x – 1}\) ≤ \(\frac{-1}{3}\)
The range is outside \(\frac{-1}{3}\) and 1
i.e., range is (-∞,\(\frac{-1}{3}\)] ∪ [1, ∞)

[OR]

(b) Evaluate: \(\int \frac{3 x+5}{x^{2}+4 x+7}\) dx
Answer:
Let I = \(\int \frac{3 x+5}{x^{2}+4 x+7}\) dx
3x + 5 = A \(\frac{d}{dx}\) (x² + 4x + 7) + B
3x + 5 = A(2x + 4) + B
Comparing the coefficients of like terms we got
2A = 3 ⇒ A = \(\frac{3}{2}\); 4A + B = 5 ⇒ B = -1

Question 43 (a).
Derive cosine formula using the law of sines in a ΔABC.
Answer:

[OR]

(b) Find \(\sqrt[3]{65}\) using binomial expansion upto two decimal places.
Answer:
We know that for |x| < 1,

≈ 4 + 0.02 ( Since \(\frac{1}{36864}\) + …….. is very small)
\(\sqrt[3]{65}\) = 4.02 (approximately)

Question 44 (a).
(i) Do the limit of the function \(\frac{sin |x|}{x}\) Li exist as x → 0? State reason for the answer.
Answer:

Hence the limit does ndt exist. Since that f(0^–) ≠ f(0⁺)

[OR]

(b)
Answer:

Question 45.
(i) Evaluate \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\) dx
Answer:
Let sin^-1 x = t
⇒ \(\frac{1}{\sqrt{1-x^{2}}}\) dx = dt
Now sin^-1 x = t ⇒ x = sin t
So I = ∫(sin t) (t) dt = ∫ t sin t dt
Now ∫t sin t dt = ∫ t d (-cos t)
= t(- cos t) – ∫(-cos t) dt
= -t cos t + ∫cos t dt
= – t cos t + sin t

[OR]

(b) Show that the points whose position vectors 4\(\hat{i}\) + 5\(\hat{j}\) + \(\hat{k}\),- \(\hat{j}\) – \(\hat{k}\), 3\(\hat{i}\) +9\(\hat{j}\) + 4\(\hat{k}\) and -4\(\hat{i}\) + 4\(\hat{j}\) + 4\(\hat{k}\) are coplanar.
Answer:
Let the given points be A, B, C and D. To prove that the point A, B, C, D are coplanar, we have to prove that the vectors \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AD}}\) are coplanar.

Equating \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) components we get
-4 = -m – 8n
⇒ m + 8n = 4 ….(i)
-6 = 4m – n
⇒ 4m – n = -6 ……. (ii)
-2 = 3m + 3n
⇒ 3m + 3n = -2 …….. (iii)
Solving (i) and (ii)

Substituting m = –\(\frac{4}{3}\) in (i) we get,
–\(\frac{4}{3}\) + 8n = 4

we are able to write one vector as a linear combination of the other two vectors ⇒ the given vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.
(i.e.,) The given points A, B, C, D are coplanar.

Question 46 (a).
If Q is a point on the locus of x² + y² + 4x – 3x + 7 = 0, then find the equation of locus of P which divides segment OQ externally in the ratio 3 : 4, where O is the origin.
Answer:
Let (h, k) be the moving-point O = (0, 0), Let PQ = (a, b) on x² + y² + 4x – 3y + 7 = 0
P divides OQ externally in the ratio 3 : 4

h² + k² – 12h + 9k + 63 = 0, Locus of (h, k) is x² + y² – 12x + 9y + 63 = 0

[OR]

(b) A consulting firm rents car from three agencies such that 50% from agency L, 30% from agency M and 20% from agency N. If 90% of the cars from L, 70% of cars from M and 60% of cars from N are in good conditions (i) what is the probability that the firm will get a car in good condition? (ii) if a car is in good condition, what is probability that it has come from agency N?
Answer:
Let A₁, A₂ and A₃ be the events that the cars are rented from the agencies X, Y and Z respectively.
Let G be the event of getting a car in good condition.
We have to find

(i) The total probability of event G that is, P(G).
(ii) Find the conditional probability A3, given G that is, P(A₃/G) We have
P(A₁) = 0.50, P(G/A₁) = 0.90
P(A₂) = 0.30, P(G/A₂) = 0.70
P(A₃) = 0.20, P(G/A₃) = 0.60.

(i) Since A₁, A₂ and A₃ are mutually exclusive and exhaustive events and G is an event in S, then the total probability of event G is P(G).
P(G) = P(A₁) P(G/A₁) + P(A₂) P(G/A₂) + P(A₃) P(G/A₃)
P(G) = (0.50) (0.90) + (0.30) (0.70) + (0.20) (0.60)
P(G) = 0.78.

(ii) The conditional probability A₃ given G is P(A₃/G)
By Bayes’ theorem,

Question 47 (a).
Using the Mathematical induction, show that for any natural number n,

Answer:


∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true.
Hence by principle of mathematical induction,
P(n) is true for all n ∈ Z.

(b) If 7 = (cos^-1 x)², prove that (1 – x²) \(\frac{d²y}{dx²}\) -x\(\frac{dx}{dy}\) -2 = 0. Hence find y₂, when x = 0.
Answer:

Squaring on both sides (1 – x²) (\(y_{1}^{2}\)) = 4(cos^-1 x)² = 4y
⇒ (1 – x²)(\(y_{1}^{2}\)) = 4y
Differentiating again w.r.to x we get
(1 – x²) (2y₁, y₂) + (\(y_{1}^{2}\)) (-2x) = 4y₁
⇒ (1 – x²) (2y₁, y₂) = 4y₁ + 2xy\(y_{1}^{2}\)
(i.e.,) (1 – x²) (2y₁, y₂) = 2y₁ (2 +xy₁)
(÷ by 2y₁)(1 – x²)y² = 2 + xy₁
So (1 – x²)y₂ – xy₁ – 2 = 0
When x = 0
(1 – 0) y₂ – 0y₁ – 2 = 0
y₂ – 2 = 0
y₂ = 2