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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)

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

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 nPr = 840, nCr = 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

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

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) (logx10)²
(d) \(\frac{x²}{100}\)
Solution:
(b) -(log10x)²

Question 16.
\(\frac{d}{dx}\)(ex+5logx) is………
(a) e10x10(x + 5)
(b) exx(x + 5)
(c) ex + \(\frac{5}{x}\)
(d) ex – \(\frac{5}{x}\)
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))

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

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
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 1
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:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 2

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
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 3

Question 25.
Determine 3B + 4C – D if B, C and D are given by
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 4
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 5

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

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

Question 27.
Find \(\frac{dx}{dy}\) if x² + y² = 1
Solution:
We differentiate both sides of the equation.
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 6
Solving for the derivative yields
\(\frac{dx}{dy}\) = –\(\frac{x}{y}\)

Question 28.
Evaluate ∫ \(\frac{1}{sin²x cos²x}\) dx
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 7

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
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 8

PART – III

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

Question 31.
If (x1/2 + x-1/2)² = \(\frac{9}{2}\) find the value of (x1/2 – x -1/2) for x > 1
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 9

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:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 10

Question 33.
Expand (1 + x)\(\frac{2}{3}\) up to four terms for |x| < 1.
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 11

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

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

Question 35.
Prove that
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 12
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 13

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.
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 14

Question 37.
Find \(\frac{dy}{dx}\) where x = \(\frac{1-t²}{1+t²}\), y = \(\frac{2t}{1+t²}\)
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 15

Question 38.
Evaluate ∫(5x² – 4 + \(\frac{7}{x}\) + \(\frac{2}{√x}\))dx
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 16

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)
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 17

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:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 18

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

[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
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 19

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
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 20
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)
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 21
(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.
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 22
Solution:
Let P(n) be the given statement
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 23
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.
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 24
= 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 Tamil Nadu 11th Maths Model Question Paper 3 English Medium 25
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
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 26

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

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
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 27
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:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 28

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:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 29
Since x is large, \(\frac{1}{x}\) is very small and hence higher powers of \(\frac{1}{x}\) are negligible. Thus
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 30

[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
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 31

Question 46 (a).
Evaluate y = sin(tan(\(\sqrt{sin x}\)))
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 32

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

[OR]

(b) Evaluate y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
Solution:
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 33

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,
Tamil Nadu 11th Maths Model Question Paper 3 English Medium 34

[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

Tamil Nadu 11th Maths Model Question Paper 3 English Medium

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