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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2

Question 1.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Solution:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities

Question 2.
A six sided die is marked ‘1’ on one face, ‘3’ on two of its faces, and ‘5’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find
(i) the probability mass function
(ii) the cumulative distribution function
(iii) P(4 ≤ X < 10)
(iv) P( X ≥ 6)
Solution:
Given that die is marked ‘ 1 ’ on one face, ‘3’ on two of its faces and ‘5’ on remaining three faces. i.e., {1, 3, 3, 5, 5, 5} in a single die.
When it is thrown twice, the number of sample points is 36, in which the sum of faces numbers are 2, ,4, 6, 8 and 10 are the value of random variable ‘X’.

(i) Probability mass function:

(ii) The Cumulative distribution function:

(iii) (4 ≤ 10) = P(X = 4) + P(X = 6) + P(X = 8)

(iv) P(X ≥ 6) = P (X = 6) + P (X = 8) + P (X = 10)

Question 3.
Find the probability mass function and cumulative distribution function of number of girl child in families with 4 children, assuming equal probabilities for boys and girls.
Solution:
Let ‘X’ be the random variable which denotes the number of girl children in the family of 4 children and X takes the values of 0, 1, 2, 3, 4.
Probability of child being a boy = P (B) = \(\frac{1}{2}\)
Probability of child being a girl = P (G) = \(\frac{1}{2}\)

Question 4.

Find
(i) the value of k
(ii) cumulative distribution function
(iii) P(X ≥ 1).
Solution:

Question 5.
The cumulative distribution function of a discrete random variable is given by

Find the (i) the probability mass function
(ii) P(X < 1)
(iii) P(X ≥ 2)
Solution:

For x = -1, f(x) = 0.15 – 0 = 0.15
For x = 0, f(x) = 0.35 – 0.15 = 0.20
For x = 1, f(x) = 0.60 – 0.35 = 0.25
For x = 2, f(x) = 0.85 – 0.60 = 0.25
For x = 3, f(x) = 1 – 0.85 = 0.15
(i) Probability mass function table

(ii) P (X < 1) = P(X = -1) + P(X = 0) = 0.15 + 0.20 = 0.35
(iii) P (X ≥ 2) = P (X = 2) + P (X = 3) = 0.25 + 0.15 = 0.40

Question 6.
A random variable X has the following probability mass function.

Solution:
Given probability mass function

(i) We know that \(\Sigma P_{i}\) = 1
i.e., k² + 2k² + 3k² + 2k + 3k = 1
6k² + 5k = 1
6k² + 5k – 1 = 0
(k + 1) (6k – 1) = 0

(ii) P (2 ≤ X < 5)
= P (X = 2) + P (X = 3) + P (X = 4)
= 2k² + 3k² + 2k = 5k² + 2k

(iii) P (3 < X) = P (X > 3)
= P (X = 4) + P (X = 5) = 2k + 3k = 5k = 5/6

Question 7.
The cumulative distribution function of a discrete random variable is given by.

Find (i) the probability mass function
(ii) P(X < 3) and
(iii) P(X ≥ 2).
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2 Additional Problems

Question 1.
Find the probability mass function, and the cumulative distribution function for getting ‘3’s when two dice are thrown.
Solution:
Two dice are thrown. Let X be the random variable of getting number of ‘3’s.

Cumulative distribution function:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

(i) P(0.5 < X < 0.75)
(ii) P(X ≤ 0.5)
(iii) P(X > 0.75)
Solution:

Question 5.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.1

Question 1.
Suppose X is the number of tails occurred when three fair coins are tossed once simultaneously. Find the values of the random variable X and number of points in its inverse images.
Solution:
Let X is the random variable denotes the number of tails when three coins are tossed simultaneously.
Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ ‘X’ takes the values 0,1, 2, 3
i.e., X (HHH) = 0 ; X (HHT, HTH, THH) = 1 ; X (HTT, THT, TTH) = 2 ; X (TTT) = 3

Question 2.
In a pack of 52 playing cards, two cards are drawn at random simultaneously. If the number of black cards drawn is a random variable, find the values of the random variable and number of points in its inverse images.
Solution:
Total number of playing cards = 52
Number of Black cards = 26
Number of Non-black (or) Red cards = 26
Let ‘X’ be the random variable denotes the number of black cards. Since two black cards are drawn,’X’ takes the values 0, 1, 2
X (Non-black Cards) = X (26C₁ × 25C₁) = X (650) = 0
X (1 Black Card) = X (26C₁ × 26C₀) = X (26) = 1
X (2 Black Cards) = X (26C₁ × 25C₁) = X (650) = 2

Question 3.
An urn contains 5 mangoes and 4 apples. Three fruits are taken at random. If the number of apples taken is a random variable, then find the values of the random variable and number of points in its inverse images.
Solution:
Number of mangoes = 5
Number of Apples = 4
Total number of fruits = 9
Let ‘X’ be the random variable denotes the number of apples taken, then it takes the values 0, 1, 2, 3
X (MMM) = 0
X (AMM (or) MAM (or) MMA) = 1
X (AAM (or) AMA (or) MAA) = 2
X (AAA) = 3

Question 4.
Two balls are chosen randomly from an urn containing 6 red and 8 black balls. Suppose that we win ₹ 15 for each red ball selected and we lose ₹ 10 for each black ball selected. X denotes the winning amount, then find the values of X and number of points in its inverse images.
Solution:
Number of red balls = 6
Number of black balls = 8
‘X’ is the random variable denotes the winning amount.
∴ The values of ‘X’ are 0, 15, 30
i.e., X (BB) = 0
X (RB) = 15 + 0 = 15
X (RR) = 15 + 15 = 30

Question 5.
A six sided die is marked ‘2’ on one face, ‘3’ on two of its faces, and ‘4’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images.
Solution:
Six sided die marked ‘2’ on one face, ‘3’ on two faces and ‘4’ on three faces.
When it is thrown twice, we get 36 sample points.
‘X’ denotes sum of the face numbers and the possible values of ‘X’ are 4, 5, 6, 7 and 8
For X = 4, the sample point is (2, 2)
For X = 5, the sample points are (2, 3), (3, 2)
For X = 6, the sample points are (3, 3), (2, 4), (4, 2)
For X = 7, the sample points are (3, 4), (4, 3)
For X = 8, the sample point is (4, 4)

Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.1 Additional Problems

Question 1.
Four defective oranges are accidentally mixed with sixteen good ones. Two oranges are drawn at random from the mixed lot. If the random variable ‘X’ denotes the number of defective oranges, then find the values of ‘X’ and number of points in its inverse image.
Solution:
Number of good oranges = 16
Number of bad oranges = 4
Total = 20
Let ‘X’ be the random variable denotes the number of bad oranges and it can take the values 0, 1, 2
X (GG) = 0
X (GB (or) BG) = 1
X (BB) = 2

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.9

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
The order and degree of the differential equation are respectively ………
(a) 2, 3
(b) 3, 3
(c) 2, 6
(d) 2, 4
Solution:
(a) 2, 3
Hint:

Order = 2,
degree = 3

Question 2.
The differential equation representing the family of curves y = A cos (x + B), where A and B
are parameters, is …….

Solution:
(b) \(\frac{d^{2} y}{d x^{2}}+y=0\)
Hint:

Question 3.
The order and degree of the differential equation is …..
(a) 1, 2
(b) 2, 2
(c) 1, 1
(d) 2, 1
Solution:
(c) 1, 1
Hint:

Since, the first order derivatives are involved, Order = 1 and degree 1.

Question 4.
The order of the differential equation of all circles with centre at (h, k) and radius ‘a’ is …….
(a) 2
(b) 3
(c) 4
(d) 1
Solution:
(a) 2
Hint:
Equation of circle is (x – h)² + (y – k)² = a²
Equation is to be differentiated twice as two parameters are given.
∴ Order = 2

Question 5.
The differential equation of the family of curves y = Ae^x + Be^-x, where A and B are arbitrary constants is ……

Solution:
\(\frac{d^{2} y}{d x^{2}}-y=0\)
Hint:

Question 6.
The general solution of the differential equation is …..
(a) xy = k
(b) y = k log x
(c) y = kx
(d) log y = kx
Solution:
(c) y = kx

Question 7.
The solution of the differential equation represents ……..
(a) straight lines
(b) circles
(c) parabola
(d) ellipse
Solution:
(c) parabola
Hint:

Question 8.
The solution of is …….

Solution:
(b) \(y=c e^{-\int \mathbf{P} d x}\)
Hint:

Question 9.
The integrating factor of the differential equation is …….

Solution:
(b) \(\frac{e^{x}}{x}\)
Hint:

Question 10.
The integrating factor of the differential equation is x, then P(x) …………

Hint:

Question 11.
The degree of the dififerential equation is ……..
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(c) 1
Hint:
Degree = 1

Question 12.
If p and q are the order and degree of the differential equation when …..
(a) p < q
(b) p = q
(c) p > q
(d) p exists and q does not exist
Solution:
(c) p > q
Hint:

Question 13.
The solution of the differential equation is …….

Solution:
(a) \(y+\sin ^{-1} x=c\)
Hint:

Question 14.
The solution of the differential equation \(\frac{d y}{d x}=2 x y\) is ………

Solution:
(a) \(y=c e^{x^{2}}\)
Hint:

Question 15.
The general solution of the differential equation is ……

Solution:
(b) \(e^{x}+e^{-y}=c\)
Hint:

Question 16.

Solution:
(c) \(\frac{1}{2^{x}}-\frac{1}{2^{y}}=c\)
Hint:

Question 17.
The solution of the differential equation is ……..

Solution:
(b) \(\phi\left(\frac{y}{x}\right)=k x\)
Hint:

Question 18.
If sin x is the integrating factor of the linear differential equation , then P is ……
(a) log sin x
(b) cos x
(c) tan x
(d) cot x
Solution:
(d) cot x
Hint:

Question 19.
The number of arbitrary constants in the general solutions of order n and n + 1 are respectively ……….
(a) n – 1, n
(b) n, n + 1
(c) n + 1, n + 2
(d) n + 1, n
Solution:
(b) n, n + 1

Question 20.
The number of arbitrary constants in the particular solution of a differential equation of third order is ………….
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) 0

Question 21.
Integrating factor of the differential equation is ……..

Solution:
(a) \(\frac{1}{x+1}\)
Hint:

Question 22.
The population P in any year t is such that the rate of increase in the population is proportional to the population. Then ……

Solution:
(a) \(\mathbf{P}=c e^{k t}\)
Hint:

Question 23.
P is the amount of certain substance left in after time t. If the rate of evaporation of the substance is proportional to the amount remaining, then ……
(a) P = ce^kt
(b) P = ce^-kt
(c) P = ckt
(d) Pt = c
Solution:
(b) P = ce^-kt
Hint:

Question 24.

(a) 2
(b) -2
(c )1
(d) -1
Solution:
(b) -2
Hint:

Question 25.
The slope at any point of a curve y =f (x) is given by and it passes through (-1, 1). Then the equation of the curve is ……..
(a) y = x³ + 2
(b) y = 3x² + 4
(c) y = 3x³ + 4
(d) y = x³ + 5
Solution:
(a) y = x³ + 2
Hint:

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.9 Additional Problems

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
The integrating factor of is ………
(a) log x
(b) x²
(c) e^x
(d) x
Solution:
(b) x²
Hint:

Question 2.
If cos x is an integrating factor of the differential equation then P = ……
(a) – cot x
(b) cot x
(c) tan x
(d) – tan x
Solution:
(d) – tan x

Question 3.
The integrating factor of dx + x dy = e^-y sec²y dy is ……….
(a) e^x
(b) e^-x
(c) e^y
(d) e^-y
Solution:
(c) e^y
Hint:

Question 4.
Integrating factor of
is …..
(a) e^x
(b) log x
(c) \(\frac{1}{x}\)
(d) e^-x
Solution:
(b) log x
Hint:

Question 5.
Solution of where m < 0 is ……

Solution:
\(x=c e^{-m y}\)
Hint:

Question 6.
y = cx – c² is the general solution of the differential equation …….

Solution:
(a) \(\left(y^{\prime}\right)^{2}-x y^{\prime}+y=0\)
Hint:

Question 7.
The differential equation of all non-vertical lines in a plane is ……

Solution:
\(\frac{d^{2} y}{d x^{2}}=0\)
Hint:
The equation of the straight line is y = mx + c

Question 8.
The differential equation of all circles with centre at the origin is
(a) x dy + y dx = 0
(b) x dy – y dx = 0
(c) x dx + y dy = 0
(d) x dx – y dy = 0
Solution:
(c) x dx + y dy = 0
Hint:
The equation of family of circle with the centre at the origin is x² + y² = a²

Question 9.
The differential equation of the family of lines y = mx is ……..

Solution:
(d) 6
Hint:

Question 10.
The degree of the differential equation
(a) 1
(b) 2
(c) 3
(d) 6
Solution:
(d) 6
Hint:

Question 11.

(a) 1
(b) 3
(c) -2
(d) 2
Solution:
(b) 3
Hint:

Question 12.
The amount present in a radio active element disintegrates at a rate proportional to its amount. The differential equation corresponding to the above statement is (k is negative)

Solution:
(c) \(\frac{d p}{d t}=k p\)
Hint:
Let p be the amount present in a radio active element

Question 13.
On putting y = vx, the homogeneous differential equation x²dy + y (x + y)dx = 0 becomes …….
(a) xdv + (2v + v²) dx = 0
(b) vdx + (2x + x²)dv = 0
(c) v²dx – (x + x²) dv = 0
(d) vdv + (2x + x²) dx = 0
Solution:
(a) xdv + (2v + v²) dx = 0
Hint:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.5

Question 1.
Find the value, if it exists. If not, give the reason for non-existence.
(i) sin^-1(cos π)
(ii) \(\tan ^{-1}\left(\sin \left(-\frac{5 \pi}{2}\right)\right)\)
(iii) sin^-1[sin 5]
Solution:
(i) sin^-1(cos π) = sin^-1(-1) = \(-\frac{\pi}{2}\)
(ii) \(\tan ^{-1}\left(\sin \frac{5 \pi}{2}\right)=\tan ^{-1}\left(-\sin \frac{\pi}{2}\right)=\tan ^{-1}(-1)=-\frac{\pi}{4}\)
(iii) sin^-1(sin 5) = sin^-1[sin (5 – 2π)] = 5 – 2π

Question 2.
Find the value of the expression in terms of x, with the help of a reference triangle.
(i) sin(cos^-1(1 – x))
(ii) cos(tan^-1(3x – 1))
(iii) \(\tan \left(\sin ^{-1}\left(x+\frac{1}{2}\right)\right)\)
Solution:
(i) Let cos^-1(1 – x) = θ
1 – x = cos θ
We know sin² θ = 1 – cos² θ

Question 3.
Find the value of
(i) \(\sin ^{-1}\left(\cos \left(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)\right.\)
(ii) \(\cot \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}\right)\)
(iii) \(\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
Solution:
(i) \(\sin ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{3} \text { and } \cos \left(\frac{\pi}{3}\right)=\frac{1}{2} \text { and }\)

Question 4.
Prove that
(i) \(\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}\)
(ii) \(\sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{16}{65}\)
Solution:

Question 5.
Prove that tan^-1 x + tan^-1 y + tan^-1 z = tan^-1 \(\left[\frac{x+y+z-x y z}{1-x y-y z-z x}\right]\)
Solution:

Question 6.
If tan^-1 x + tan^-1 y + tan^-1 z = π, show that x + y + z = xyz.
Solution:
Given tan^-1 x + tan^-1 y + tan^-1 z = π

Question 7.
Prove that \(\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\tan ^{-1} \frac{3 x-x^{3}}{1-3 x^{2}},|x|<\frac{1}{\sqrt{3}}\)
Solution:

Question 8.
Simplify: \(\tan ^{-1} \frac{x}{y}-\tan ^{-1} \frac{x-y}{x+y}\)
Solution:

Question 9.
Find the value of
(i) \(\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}\)
(ii) \(2 \tan ^{-1} x=\cos ^{-1} \frac{1-a^{2}}{1+a^{2}}-\cos ^{-1} \frac{1-b^{2}}{1+b^{2}}\), a > 0, b > 0
(iii) 2 tan^-1(cos x) = tarn^-1 (2 cosec x)
(iv) cot^-1 x – cot^-1 (x + 2) = \(\frac{\pi}{12}\), x > 0
Solution:

Question 10.
Find the number of solution of the equation tan^-1(x – 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1(3x).
Solution:
tan^-1(x – 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1(3x)
tan^-1(x – 1) + tan^-1 (x + 1) = tan^-1 3x – tan^-1 x

LHS = RHS
⇒ \(\tan ^{-1} \frac{2 x}{2-x^{2}}=\tan ^{-1} \frac{2 x}{1+3 x^{2}}\)
⇒ \(\frac{2 x}{2-x^{2}}=\frac{2 x}{1+3 x^{2}}\)
⇒ 2 – x² = 1 + 3x²
⇒ 4x² = 1
⇒ x² = \(\frac{1}{4}\)
⇒ x = ±\(\frac{1}{2}\)
So, the equation has 2 solutions.

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.5 Additional Questions

Question 1.
Solve the following equation: sin^-1(1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:

Question 2.
Solve: tan^-1 2x + tan^-1 3x = \(\frac{\pi}{4}\)
Solution:

Question 3.

Solution:
Do it yourself

Question 4.

Solution:
Do it yourself

Question 5.

Solution:
Do it yourself

Question 6.

Solution:
Do it yourself

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2

Question 1.
Write the first 6 terms of the sequences whose n^th terms are given below and classify them as arithmetic progression, geometric progression, arithmetico-geometric progression, harmonic progression and none of them.

Solution:


It is not a G.P. or A.P. or H.P. or A.G.P.

It is not an A.P. or G.P. or H.P. or A.G.P

It is not an A.P. or G.P. or H.P. or A.G.P.


It is a A.G.P.

Question 2.
Write the first 6 terms of the sequences whose n^th term a_n is given below.

Solution:
a₁ = 1 + 1 = 2 ; a₂ = 2
a₃ = 3 + 1 = 4 ; a₄ = 4
a₅ = 5 + 1 = 6 ; a₆ = 6
So, the first 6 terms are 2, 2, 4, 4, 6, 6

Solution:
a₁ = 1 ; a₂ = 2, a₃ = 3
a₄ = a₃ + a₂ + a₁ = 3 + 2 + 1 = 6 ⇒ a₄ = 6
a₅ = a₄ + a₃ + a₂ = 6 + 3 + 2 = 11 ⇒ a₅ = 11
a₆ = a₅ + a₄ + a₃ = 11 + 6 + 3 = 20 ⇒ a₆ = 20
So the first 6 terms are 1, 2, 3, 5, 8, 13.

Solution:

Question 3.
Write the n_th term of the following sequences.
Solution:
(i) 2, 2, 4, 4, 6, 6……
Solution:

Solution:
Nr: 1, 2, 3, ……t_n = n
Dr: 2, 3, 4, …..t_n = n + 1


Solution:
Nr: 1, 3, 5, 7, . . .which is an A.P. a = 1, d = 3 – 1 = 2
t_n = a + (n – 1)d
t_n = 1 + (n – 1)2 = 1 + 2n – 2 = 2n – 1.
Dr : 2, 4, 6, 8, . . .
So the n^th term is 2 + (n – 1)2 = 2 + 2n – 2 = 2n.

(iv) 6, 10, 4, 12, 2, 14, 0, 16, -2,….
Solution:
t₁ = 6 ; t₂ = 10
t₃ = 4 ; t₄ = 12
t₅ = 2 ; t₆ = 14
t₇ = 0 ; t₈ = 16
When n is odd, the sequence is 6, 4, 2, 0,…
(i.e.) a = 6 and d = 4 – 6 = -2.
So, t_n = 6 + (n – 1)(-2) = 6 – 2n + 2 = 8 – 2n
When n is even, the sequence is 10, 12, 14, 16,…
Here a = 10 and d = 12 – 10 = 2
t_n = 10 + (n – 1)2 = 10 + 2n – 2 = 2n + 8 (i.e.) 8 + 2n

Question 4.
The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP.
Solution:
The 3 numbers in a G.P. is taken as \(\frac{a}{r}\), a, ar
Their product is 5832.

6r² + 6 = 13
6r² – 13r + 6 = 0
(3r – 2)(2r – 3) = 0
r = 2/3 or 3/2

Question 5.

Solution:

Question 6.
If t_k is the k^th term of a G.P., then show that t_{n – k}, t_n, t_{n + k} also form a GP for any positive integer k.
Solution:
Let a be the first term and r be the common ratio.
We are given t_k = ar^{k – 1}
We have to Prove : t_{n – k}, t_n, t_{n + k} form a G.P.

Question 7.
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.
Solution:
Given a, b, c are in G.P.
⇒ b² = ac
⇒ log b² = log ac
(i.e.) 2log b = log a + log c …(1)

Substituting these values in equation (1) we get 2y = x + z ⇒ x, y z are in A.P.

Question 8.
The AM of two numbers exceeds their GM by 10 and HM by 16. Find the numbers.
Solution:
Let the two numbers be a and b.

So, (a + b – 20)² = 4ab
(i.e.) (a + b)² + 400 – 40(a + b) = 4ab
(a + b)² – 4ab = 40(a + b) – 400
from(3) (a + b)² – 4ab = 32(a + b)
⇒ 32(a + b) = 40(a + b) – 400
(÷ by 8) 4(a + b) = 5(a + b) – 50
4a + 4b = 5a + 5b – 50
a + b = 50
a = 50 – b
Substituting a = 50 – b in (3) we get
(50 – b – b)² = 32(50)
(50 – 2b)² = 32 × 50

When b = 5, a = 50 – 5 = 45
When b = 45, a = 50 – 45 = 5
So the two numbers are 5 and 45.

Question 9.
If the roots of the equation (q – r)x² + (r – p)x + p – q = 0 are equal, then show that p, q and r are in AP.
Solution:
The roots are equal ⇒ ∆ = 0
(i.e.) b² – 4ac = 0
Hence, a = q – r ; b = r – p ; c = p – q
b² – 4ac = 0
⇒ (r – p)² – 4(q – r)(p – q) = 0
r² + p² – 2pr – 4[qr – q² – pr + pq] = 0
r² + p² – 2pr – 4qr + 4q² + 4pr – 4pq = 0
(i.e.) p² + 4q² + r² – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r)² = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.

Question 10.
If a, b, c are respectively the p^th, q^th and r^th terms of a G.P., show that (q – r) log a + (r – p) log b + (p – q) log c = 0.
Solution:
Let the G.P. be l, lk, lk²,…
We are given t_p = a, t_q = b, t_r = c

LHS = (q – r) log a + (r – p) log b + (p – q) log c

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2 Additional Questions Solved

Question 1.

Solution:
Here a₁ = 1
Substituting n = 2, we obtain a₂ = a₁ + 2 = 1 + 2 = 3
Substituting n = 3, 4 and 5, we obtain respectively
a₃ = a₂ + 2 = 3 + 2 = 5, a₄ = a₃ + 2 = 5 + 2 = 7
a₅ = a₄ + 2 = 7 + 2 = 9
Thus, the first five terms are 1, 3, 5, 7 and 9.

Question 2.
Find the 18^th and 25^th term 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 3.
Write the first six terms of the sequences given by
(i) a₁ = a₂= 1 = a_{n – 1} + a_{n – 2} (n ≥ 3)
(ii) a₁ = 4, a_{n + 1} = 2na_n
Solution:
(i) Here a₁ = a₂= 1 = a_{n – 1} + a_{n – 2} (n ≥ 3)
Putting n = 3, a₃= a₂ + a₁ = 1 + 1 = 2
Putting n = 4, a₄ = a₃ + a₂ = 2 + 1 = 3
Putting n = 5, a₅ = a₄ + a₂ = 3 + 2 = 5
Putting n = 6, a₆ = a₅ + a₄ = 5 + 3 = 8
∴ First six terms of the sequence are 1, 1, 2, 3, 5, 8

(ii) Here a₁ = 4 and a_{n + 1} = 2na_n
Putting n = 1, a₂ = 2 × 1 × a₁ = 2 × 1 × 4 = 8
Putting n = 2, a₃ = 2 × 2 × a₂ = 4 × 8 = 32
Putting n = 3, a₄ = 8 × 192 = 1536
Putting n = 4, a₅ = 2 × 4 × a₄ = 8 × 192 = 1536
Putting n = 5, a₆ = 2 × 5 × a₅ = 10 × 1536 = 15360

Question 4.
An A.P. consists of 21 terms. The sum of the three terms in the middle is 129 and of the last three is 237. Find the series.
Solution:
Let a₁ be the first term and d, be the common difference. Here n = 21.
∴ The three middle terms are a₁₀,a₁₁, a₁₂
Now, a₁₀ + a₁₁ + a₁₂ = 129 [Given]
∴ (a₁ + 9d) + (a₁ + 10d) + (a₁ + 11d) = 129
⇒ 3a₁ + 30d = 129 ⇒ a₁ + 10d = 43 ……(i)
The last three terms are a₁₉, a₂₀, a₂₁
a₁₉, a₂₀, a₂₁ = 237 [Given]
∴ (a₁ + 18d)+(a₁ + 19d) + (a₁ + 20d) = 237
(i.e.,) 3a₁ + 57d = 237 ⇒ a₁ + 19d = 79 … (2)
Subtracting (i) from (ii), we get 9d = 36, ⇒ d = 4
∴ From (i), a₁ + 40 = 43 ⇒ a₁ = 3
Hence, the series is 3, 7, 11, 15 …….

Question 5.
Prove that the product of the 2^nd and 3^rd terms of an arithmetic progression exceeds the product of the first and fourth by twice the square of the difference between the 1^st and 2^nd.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of A.P.
Then, a₁ = a, a₂ = a + (2 – 1)d = a + d
a₃ = a + (3 – 1)d = a + 2d, a₄ = a + (4 – 1)d = a + 3d
We have to show that a₂.a₃ – a₁.a₄ = 2(a₂ – a₁)²
LHS = a₂.a₃ – a₁.a₄ = (a + d)(a + 2d) – a(a + 3d)
= a₂ + 3ad + 2d₂ – a₂ – 3ad = 2d₂
RHS = 2(a₂ – a₁)² = 2d₂
Since LHS = RHS. Hence proved.

Question 6.
If the p^th, q^th and r^th terms of an A.P. are a, b, c respectively, prove that a(q – r) + b (r – p) + c(p – q) = 0.
Solution:
Let A be the first term and D be the common difference of A.P.
a_p = a, ∴ A + (p – 1)D = a ….. (1)
a_q = b, ∴ A + (q – 1)D = b ……. (2)
a_r = c, ∴ A + (r – 1)D = c …….. (3)
∴ a (q – r) + b (r – p) + c (p – q) = [A + (p – l) D] (q – r) + [A + (q – 1) D]
(r – p) + [A + (r – 1) D] (p – q) [Using (1), (2) and (3)]
= (q – r + r – p + p – q)A + [(p – l)(q – r) + (q – l)(r – p) + (r – l)(p – q)]D
= (0) A + (pq – pr – q + r + qr – pq – r + p + pr – p – qr + q)D
= (0)A + (0)D = 0.

Question 7.
If a, b, c are in A.P. and p is the A.M. between a and b and q is the A.M. between b and c, show that b is the A.M. between p and q.
Solution:
a, b, c are in A.P.
2b = a + c …… (1)
p is the A.M. between a and b

q is the A.M. between b and c


Hence, b is the A.M. between p and q

Question 8.
If x, y, z be respectively the p^th, q^th and r^th terms of a G.P. show that x^{q – r}, y^{r – p} ,z^{p – q} = 1
Solution:
Let A be the first term and R be the common ratio of G.P.
a_p = x ⇒ x = AR^{p – 1} ……… (1)
a_q = y ⇒ x = AR^{q – 1} ……… (2)
a_r = z ⇒ x = AR^{r – 1} ……… (3)
Raising (1), (2), (3) to the powers q – r, r – p, p – q respectively, we get

Multiplying (4), (5) and (6), we get

Hence, x^{q – r}, y^{r – p}, z^{p – q} = 1

You can Download Samacheer Kalvi 12th 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 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.7

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
A zero of x³ + 64 is _______
(a) 0
(b) 4
(c) 4i
(d) -4
Answer:
(d) -4
Hint: x³ + 64 = 0
⇒ x³ = -64
⇒ x³ = (-4)³
⇒ x = -4

Question 2.
If f and g are polynomials of degrees m and n respectively, and if h(x) = (f 0 g) (x), then the degree of h is ______
(a) mn
(b) m + n
(c) mⁿ
(d) n^m
Answer:
(a) mn

Question 3.
A polynomial equation in x of degree n always has _______
(a) n distinct roots
(b) n real roots
(c) n imaginary roots
(d) at most one root.
Answer:
(c) n imaginary roots (Every real number is also imaginary)

Question 4.
If α, β and γ are the zeros of x³ + px² + qx + r, then \(\sum \frac{1}{\alpha}\) is ______
(a) \(-\frac{q}{r}\)
(b) \(\frac{q}{p}\)
(c) \(\frac{q}{r}\)
(d) \(-\frac{q}{p}\)
Answer:
(a) \(-\frac{q}{r}\)
Hint:

Question 5.
According to the rational root theorem, which number is not possible rational zero of 4x⁷ + 2x⁴ – 10x³ – 5?
(a) -1
(b) \(\frac{5}{4}\)
(c) \(\frac{4}{5}\)
(d) 5
Answer:
(c) \(\frac{4}{5}\)
Hint:
a_n = 4; a₀ = 5
Let \(\frac{p}{q}\) be the root of P (x). P must divide 5, possible values of P are ±1, ±5
q must divide 4, possible values of q are ±1, ±2, ±4
Possible roots are \(\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}\)

Question 6.
The polynomial x³ – kx² + 9x has three real zeros if and only if, k satisfies.
(a) |k| ≤ 6
(b) k = 0
(c) |k| > 6
(d) |k| ≥ 6
Answer:
(d) |k| ≥ 6
Hint:
x³ – kx² + 9x = 0
⇒ x (x² – kx + 9) = 0
x = 0 is one real root. If the remaining roots to be real if the
b² – 4ac ≥ 0
⇒ k² – 36 ≥ 0
⇒ k² ≥ 36
⇒ |k| ≥ 6

Question 7.
The number of real numbers in [0, 2π] satisfying sin⁴ x – 2sin² x + 1 is ______
(a) 2
(b) 4
(c) 1
(d) ∞
Answer:
(c) 1
Hint:
sin⁴ x – 2sin² x + 1 = 0
⇒ t² – 2t + 1 = 0
⇒ (t – 1)² = 0
⇒ t – 1 = 0
⇒ t = 1
⇒ sin² x = 1
⇒ \(\frac{1-\cos 2 x}{2}=1\)
⇒ 1 – cos 2x = 2
⇒ cos 2x = cos 0
⇒ 2x = 2nπ
⇒ x = nπ
n = 0, x = 0
n = 1, x = π
n = 2, x = 2π

Question 8.
If x³ + 12x² + 10ax + 1999 definitely has a positive zero, if and only if _______
(a) a ≥ 0
(b) a > 0
(c) a < 0
(d) a ≤ 0
Answer:
(c) a < 0
Hint:
If a < 0, then P(x) = x³ + 12x² + 10ax + 1999 has 2 changes of sign.
∴ P (x) has atmost two positive roots. So a < 0

Question 9.
The polynomial x³ + 2x + 3 has _______
(a) one negative and two imaginary zeros
(b) one positive and two imaginary zeros
(c) three real zeros
(d) no zeros
Answer:
(a) one negative and two imaginary zeros
Hint:
P(x) = x³ + 2x + 3; No positive root.
P(-x) = -x³ – 2x + 3; Only one change in the sign.
∴ One negative root.

Question 10.
The number of positive zeros of the polynomial is ______
(a) 0
(b) n
(c) <n
(d) r
Answer:
(b) n

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6

Question 1.
Discuss the maximum possible number of positive and negative roots of the polynomial equation 9x⁹ – 4x⁸ + 4x⁷ – 3x⁶ + 2x⁵ + x³ + 7x² + 7x + 2 = 0.
Solution:
P(x) = 9x⁹ – 4x⁸ + 4x⁷ – 3x⁶ + 2x⁵ + x³ + 7x² + 7x + 2
The number of sign changes in P(x) is 4.
∴P(x) has atmost 4 positive roots.
P(-x) = -9x⁹ – 4x⁸ – 4x⁷ – 3x⁶ – 2x⁵ – x³ + 7x² – 7x + 2
The number of sign changes in P(-x) is 3.
P(x) has almost 3 negative roots. Since the difference between the number of sign changes in co-efficient P(-x) and the number of negative roots of the polynomial P(x) is even.
The number of negative roots = at most 2.

Question 2.
Discuss the maximum possible number of positive and negative zeros of the polynomials x² – 5x + 6 and x² – 5x + 16. Also, draw a rough sketch of the graphs.
Solution:
P(x) = x² – 5x + 6
The number of sign changes in P(x) is 2.
P(x) has atmost 2 positive roots. P(-x) = x² + 5x + 6.
The number of sign changes in P(-x) is 0.
∴ P (x) has no negative roots. P (x) = x² – 5x + 16

Question 3.
Show that the equation x⁹ – 5x⁵ + 4x⁴ + 2x² +1 = 0 has atleast 6 imaginary solutions.
Solution:
P(x) = x⁹ – 5x⁵ + 4x⁴ + 2x² + 1
(i) The number of sign changes in P(x) is 2. The number of positive roots is atmost 2.
(ii) P(-x) = -x⁹ + 5x⁵ + 4x⁴ + 2x² + 1. The number of sign changes in P(-x) is 1. The number of negative roots of P (x) is atmost 1. Since the difference of number of sign changes in P(-x) and number of negative zeros is even.
P(x) has one negative root.
(iii) 0 is not the zero of the polynomial P(x). So the number of real roots is almost 3.
∴ The number of imaginary roots at least 6.

Question 4.
Determine the number of positive and negative roots of the equation x⁹ – 5x⁸ – 14x⁷ = 0.
Solution:
x⁹ – 5x⁸ – 14x⁷ = 0
P (x) = x⁹ – 5x⁸ – 14x⁷. The number of sign changes is P(x) is 1.
The number of positive roots is 1. P (-x) = -x⁹ – 5x⁸ + 14x⁷
The number of sign changes is P(-x) is one. The number of negative zero of P(-x) is 1. It is clear that 0 is a root of the equation.
∴ The number of the imaginary roots is at least 6.

Question 5.
Find the exact number of real zeros and imaginary of the polynomial x⁹ + 9x⁷ + 7x⁵ + 5x³ + 3x.
Solution:
P(x) = x⁹ + 9x⁷ + 7x⁵ + 5x³ + 3x.
There is no change in the sign of P(x) and P(-x), P(x) has no positive and no negative real roots, but 0 is the root of the polynomial equation P(x).

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6 Additional Problems

Question 1.
Find the maximum possible number of real roots of the equation, x⁵ – 6.x² – 4x + 5 = 0.
Solution:
Let f(x) = x⁵ – 6x² – 4x + 5
Check the terms when it changes sign
Number of changes = 2
∴ Maximum number of positive real roots = 2
f(-x) = (- x)⁵ – 6 (- x)² – 4 (-x) + 5
= -x⁵ – 6x² + 4x + 5
Check the terms when it changes signs.
Number of changes = 1
∴ Maximum number of negative real roots = 1
∴ Total maximum number of real roots = 2 + 1 = 3

Question 2.
Find the number of real roots of the equation, x² + 5 |x| + 6 = 0.
Solution:
|x²| – 5|x| + 6 = 0
(|x| -2) (|x| -3) = 0

It has four real roots. The real roots are 2, -2, 3, -3

Question 3.
Find the real roots of the equation, x² + 5 |x| + 6 = 0.
Solution:
x² + 5 |x| + 6 = 0
Case (i) If x ≥ 0
x² + 5x + 6 = 0
(x + 2) (x + 3) = 0
x = -2 and x = -3
Case (ii) If x < 0
x² – 5x + 6 = 0
(x – 2) (x – 3) = 0
x = 2 and x = 3

Question 4.
Solve x⁴ – 4x² + 8x + 35 = 0. Given (2 + \(i \sqrt{3}\)) is a root.
Solution:

⇒ x² – 4x + 7 is a factor of P (x). Dividing the polynomial P(x) = 0 by x² – 4x + 7.
We get x² + 4x + 5 = 0 is a other factor. The roots of x² + 4x + 5 = 0 are

Question 5.
Solve x⁴ – 5x³ + 4x² + 8x – 8 = 0. Given (1 – \(\sqrt{5}\)) is a root of the polynomial equation.
Solution:
Since (1 – \(\sqrt{5}\)) is a root of the polynomial P(x) = 0
(1 + \(\sqrt{5}\)) is also a root of P (x) = 0
⇒ x² – [(1 + \(\sqrt{5}\)) + (1 – \(\sqrt{5}\))] x + (1 + \(\sqrt{5}\) ) (1 – \(\sqrt{5}\) ) = 0 is a factor of P(x) = 0 ⇒ x² – 2x – 4 = 0 is a factor of P(x) = 0.
Dividing the polynomial by x² – 2x – 4 = 0
We get the other factor x² – 3x + 2 = 0
The roots of x² – 3x + 2 = 0
(x – 2) (x – 1) = 0
x = 1, 2
The roots are 1, 2, 1 ± \(\sqrt{5}\)

Question 6.
Find a polynomial equation of the lowest degree with rational co-efficient having \(\sqrt{3}\), (1 – 2i) as two of its roots.
Solution:
When \(\sqrt{3}\) is a root, – \(\sqrt{3}\) will also be a root.
Now the quadratic equation with \(\sqrt{3}\) , – \(\sqrt{3}\) are roots is x² – (\(\sqrt{3}\) – \(\sqrt{3}\))x + (\(\sqrt{3}\))(-\(\sqrt{3}\)) = 0
(i.e) x² – 3 = 0
When 1 – 2i is a root, 1 + 2i will be another root.
Now the quadratic equation with roots 1 – 2i and 1 + 2i is
x² – (1 – 2i + 1 + 2i)x + (1 – 2i)(1 + 2i) = 0
(i.e) x² – 2x + 5 = 0
∴ The equation with roots ± \(\sqrt{3}\) and 1 ± 2i is (x² – 3) (x² – 2x + 5) = 0
(i.e) x⁴ – 2x³ + 5x² – 3x² + 6x – 15 = 0 ,
(i.e) x⁴ – 2x³ + 2x² + 6x – 15 = 0

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5

Question 1.
Solve the following equations:
(i) sin²x – 5sinx + 4 = 0
(ii) 12x³ + 8x = 29x² – 4
Solution:
(i) sin²x – 5sinx + 4 = 0
Let y = sin x
(y² – 5y + 4 = 0
(y – 1) (y – 4) = 0
(y – 1) = o or (y – 4) = 0
y = 1 or y = 4
sin x = 1 or sin x = 4 [not possible since sin x ≤ 1]
sin x = sin \(\frac{\pi}{2}\)
x = nπ + (-1)ⁿ α, n ∈ z.
x = nπ + (-1)ⁿ \(\frac{\pi}{2}\)
(ii) 12x³ + 8x = 29x² – 4
12x³ – 29x² + 8x + 4 = 0 ….. (1)
By Trail and error method, (x – 2) is a factor of (1)
The other factor is 12x² – 5x – 2
The roots is 12x² – 5x – 2 = 0
(3x – 2) (4x + 1) = 0
x = \(\frac{2}{3}\), x = \(-\frac{1}{4}\)
The roots are 2, \(\frac{2}{3}\), \(-\frac{1}{4}\)

[Here a_n = 12, a₀ = 4; Let \(\frac{p}{q}\) be the root of the equation (1)
The factors of a₀ : ±1, ±2, ±4 (P must divisible by 4)
The factor of a_n : ±1, ±2, ±3, ±4, ±6, ±12
q must divide as (12)
Using these p and q we can form \(\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{2}{3}, \pm \frac{1}{6}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm 3\) are the possible roots of equation. (1)]

Question 2.
Examine for the rational roots of:
(i) 2x³ – x² – 1 = 0
(ii) x⁸ – 3x + 1 = 0
Solution:
(i) 2x³ – x² – 1 = 0
Sum of the co-efficients = 0
∴ (x – 1) is a factor
The other factor is 2x² + x + 1.
The root is (2x² + x + 1) = 0
Here ∆ = b² – 4ac = (1)² – 4(2) (1) = 1 – 8 = -7 < 0
The remaining roots are imaginary.
The only rational root is x = 1

(ii) x⁸ – 3x + 1 = 0 …. (1)
Here a_n = 1, a₀ = 1
If \(\frac{p}{q}\) is a rational root of (1)
Then q is a factor a_n, p is a factor of a₀
The possible values of p and q are ± 1.
Among the possible values 1, -1, [(p, q) = 1]
None of them satisfies the equation (1)
The above equation has no rational roots.

Question 3.
Solve: \(8 x^{\frac{3}{2 n}}-8 x^{\frac{-3}{2 n}}=63\)
Solution:

Squaring on both sides

only possible solution is x = 4ⁿ

Question 4.
Solve: \(2 \sqrt{\frac{x}{a}}+3 \sqrt{\frac{a}{x}}=\frac{b}{a}+\frac{6 a}{b}\)
Solution:

Question 5.
Solve the equations:
(i) 6x⁴ – 35x³ + 62x² – 35x + 6 = 0
(ii) x⁴ + 3x³ – 3x – 1 = 0
Solution:
(i) 6x⁴ – 35x³ + 62x² – 35x + 6 = 0 ….. (1)
It is a even degree reciprocal equation as p(x) = \(x^{n} p\left(\frac{1}{x}\right)\)
Dividing equation (1) by x²,


(ii) x⁴ + 3x³ – 3x – 1 = 0 …… (1)
It is an even degree reciprocal function of type II.
1, -1 are the solution of equation (1)
(x – 1), (x + 1) are the factor of (1)
(x² – 1) is a factor of (1)
Dividing (1) by (x² – 1)
we get, x² + 3x + 1 = 0 is the other factor.

Question 6.
Find all real numbers satisfying 4^x – 3(2^x+2) + 2⁵ = 0
Solution:
4^x – 3 (2^x+2) + 2⁵ = 0
⇒ (2²)^x – 3(2^x . 2²) + 2⁵ = 0
(2²)^x – 12 . 2^x+ 32 = 0
Let y = 2^x
y² – 12y + 32 = 0
⇒ (y – 4) (y – 8) = 0
y – 4 = 0 or y – 8 = 0
Case (i): 2^x = 4
⇒ 2^x = (2)²
⇒ x = 2
Case (ii): 2x = 8
⇒ 2^x = (2)³
⇒ x = 3
∴ The roots are 2, 3

Question 7.
Solve the equation 6x⁴ – 5x³ – 38x² – 5x + 6 = 0 if it is known that \(\frac{1}{3}\) is a solution.
Solution:
6x⁴ – 5x³ – 38x² – 5x + 6 = 0 …… (1)
x = \(\frac{1}{3}\) is a Solution
∴ (3x – 1) is a factor of (1)
(1) is a Reciprocal equation even degree divide (1) by x².

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5 Additional Problems

Question 1.
Solve: 3^{2x + 4} + 1 = 2.3^{x + 2}
Solution:
3^{2x + 4} = 3^{x + 2} + 3^{x + 2} – 1
3^{2x + 4} – 3^{x + 2} = 3^{x + 2} – 1 ⇒ 3^{x + 2}[3^{x + 2} – 1] = [3^{x + 2} – 1]
3^{x + 2}= 1 ⇒ 3^{x + 2} = 3⁰
x + 2 = 0 ⇒ x = -2

Question 2.
Solve: 2^x – 2^{x + 3} + 2⁴ = 0.
Solution:
2^2x – (2^x.2³) + 2⁴ = 0 since 2³ = 8 = 4 + 4 = 2² + 2²
2^2x – (4 + 4)2^x + 2⁴ = 0 ⇒ 2^2x – (2² + 2²)2^x + 2⁴ = 0
(2^x – 2²)(2^x – 2²) = 0 ⇒ (2^x – 2²)² = 0
2^x = 2² ⇒ x = 2

Question 3.
Solve: (x – 4) (x + 2) (x + 3) (x – 3) + 8 = 0.
Solution:
(x – 4) (x + 3) (x + 2) (x – 3) + 8 = 0
(x² – x – 12) (x² – x – 6) + 8 = 0
Let y = x² – x
(y – 12) (y – 6) + 8 = 0 ⇒ y² – 18y + 72 + 8 = 0
y² – 18y + 80 = 0 ⇒ (y – 10)( y – 8) = 0
Case (i) y – 10 = 0
x² – x – 8 = 0

Question 4.

Solution:

Question 5.
5^{x – 1} + 5^{1 – x} = 26
Solution:

Question 6.
Solve: 12x⁴ – 56x³ + 89x² – 56x + 12 = 0
Solution:
Since the co-efficients of the equations are equal from both ends.
Divide the equation by x²

Question 7.

Solution:

Question 8.
Solve: x⁴ + 4x³ + 5x² + 4x + 1 = 0
Solution:

You can Download Samacheer Kalvi 12th 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 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4

Question 1.
Show that each of the following expressions is a solution of the corresponding given differential equation.

(i) y = 2x² ; xy’ = 2y
Solution:
v = 2x² …(1)
Differential equation: xy’ = 2y
Differentiate with respect to ‘x’

On simplifying, 2y = xy’
∴ (1) is solution of the given differential equation.

(ii) y = ae^x + be^-x ; y” – y = 0
Solution:
y = ae^x + be^-x …(1) Differential equation: y” – y = 0
Differentiate with respect to ‘x’
y’ = ae^x – be^-x
Again differentiate with respect to ‘x’
y” = ae^x + be^-x
y” = y ⇒ y” – y = 0
∴ (1) is the solution of the given differential equation.

Question 2.
Find value of m so that the function y = e^mx is a solution of the given differential equation.
(i) y + 2y = 0
Solution:
Given solution y = e^mx
Differentiate with respect to ‘x’

(ii) y” – 5y’ + 6y = 0

Given differential equation is y” – 5y’ + 6y = 0
Substitute (1), (2) and (3) in this

Question 3.
The slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through (2, 5). Find the equation of the curve.
Solution:
Slope of the tangent is the reciprocal of four times the ordinate

Question 4.

Solution:

Question 5.
Show that y = ax + \(\frac{b}{x}\), x ≠ 0, is a solution of the differential equation x²y”+ xy’ – y = 0.
Solution:

Here ‘a’ and ‘b’ are arbitrary constants

Differentiate with respect to ‘x’
xy’ + y . 1 = a (2x) = 2ax ……….. (2)
Differentiate again with respect to ‘x’ .
xy” + y’ . 1 + y = 2a ⇒ xy” + 2y’ = 2a …….. (3)
Substitute (3) in (2)
xy’ + y = (xy” + 2y’)x
xy’ + y = x²y” + 2xy’ ⇒ x²y” + xy’ – y = 0
Hence proved.

Question 6.
Show that y = ae^-3x + b, where a and b are arbitrary constants, is a solution of the differential equation

Solution:

Question 7.
Show that the differential equation representing the family of curves where a is a positive parameter, is
Solution:

Question 8.
Show that, y = a cos bx is a solution of the differential equation .
Solution:
y = a cos bx …(1) (a is an arbitrary constant)
Differentiating with respect to ‘x’

Again, differentiating with respect to ‘x’

Hence proved

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 Additional Problems

Question 1.
Verify that the function y = a cos x + b sin x is a solution of the differential equation cos \(\frac{d y}{d x}\) + y sin x = b. dx
Solution:
The given function is y = a cos x + b sin x
Differentiating both sides with respect to x, we have

Putting values of \(\frac{d y}{d x}\) and y in the given differential equation, we have
L.H.S. = cos x (- a sin x + b cos x) + {a cos x + b sin x) sin x
= – a sin x cos x + b cos2 x + a sin x cos x + b sin2 x = b (cos2 x + sin2 x)
= b × 1 = b = R.H.S
Thus, y = a cos x + b sin x is a solution of differential equation

Question 2.
Verify that the function y = 4 sin 3x is a solution of the differential equation
Solution:
The given function is y = 4 sin 3x
Differentiating both sides with respect to x, we have

Again, differentiating both sides with respect to x, we have

Putting values of \(\frac{d^{2} y}{d x^{2}}\) and y in the given differential equation, we have
L.H.S. = – 36 sin 3x + 9 (4 sin 3x) = – 36 sin 3x + 36 sin 3x = 0 = R.H.S.
Thus, y = 4 sin 3x is a solution of differential equation

Question 3.
Verify that the function y = ax² + bx + c is a solution of the differential equation   .
Solution:
The given function is y = ax² + bx + c
Differentiating both sides with respect to x, we have

Again differentiating both sides with respect to x, we have

Which is the given differential equation.
Thus, y = ax² + bx + c is a solution of differential equation \(\frac{d^{2} y}{d x^{2}}\)= 2a.

Question 4.
Verify that the function y = e^-3x is a solution of the differential equation
Solution:
The given function is y = e^-3x
Differentiating both sides with respect to x, we have

Again, differentiating both sides with respect to x, we have

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

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Question 1.
The equation of the locus of the point whose distance from y-axis is half the distance from origin is ……..
(a) x² + 3y² = 0
(b) x² – 3y² = 0
(c) 3x² + y² = 0
(d) 3x² – y² = 0
Solution:
(c) 3x² + y² = 0
Hint:
Given that PA = \([\frac{1}{2}/latex]OP
2PA = OP

4PA² = OP²
4(x)² = x² + y² ⇒ 3x² – y² = 0

Question 2.
Which of the following equation is the locus of (at², 2at) ……

Solution:
(d) y² = 4ax
Hint:
y² = 4ax ⇒ Equation that satisfies the given point (at², 2at)

Question 3.
Which of the following point lie on the locus of 3x² + 3y² – 8x – 12y + 17 = 0?
(a) (0, 0)
(b) (-2, 3)
(c) (1, 2)
(d) (0, -1)
Solution:
(c) (1, 2)
Hint:
The point that satisfies the given equations (0, 0) ⇒ 17 ≠ 0
(-2, 3) ⇒ 3 (4) + 3 (9) + 16 – 36 + 17 ≠ 0
(1, 2) ⇒ 3 + 3 (4) – 8 (1) – 12 (2) + 17
32 – 32 = 0, 0 = 0

Question 4.

(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3

Question 5.
Straight line joining the points (2, 3) and (-1, 4) passes through the point (α, β) if
(a) α + 2β = 7
(b) 3α + β = 9
(c) α + 3β = 11
(d) 3α + 3β = 11
Solution:
(c) α + 3β = 11
Hint:
Equation joining (2, 3), (-1, 4)

3y – 12 = – x -1 ⇒ x + 3y – 11 = 0, (α, β) lies on it ⇒ α + 3β – 11 = 0.

Question 6.
The slope of the line which makes an angle 45° with the line 3x – y = – 5 are
(a) 1, -1
(b) [latex]\frac{1}{2},-2\)
(c) \(1, \frac{1}{2}\)
(d) \(2,-\frac{1}{2}\)
Solution:
(c) \(1, \frac{1}{2}\)
Hint:
Equation of line 3x – y = -5, y = 3x + 5, m₁ = 3

Question 7.
Equation of the straight line that forms an isosceles triangle with coordinate axes in the I-quadrant with perimeter \(4+2 \sqrt{2}\) is
(a)x + y + 2 = 0
(b) x + y – 2 = 0
(c) x + y – \(\sqrt{2}\) = 0
(d) x + y + \(\sqrt{2}\) = 0
Solution:
(b) x + y – 2 = 0
Hint.
Let the sides be x, x

Question 8.
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
Solution:
(b) x + y = 1
Hint:
This equation passes through (-1, 2)
-1 + 2 = 1 ⇒ 1 = 1

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
Solution:
(b) 5, 5

Question 10.
The equation of the line with slope 2 and the length of the perpendicular from the origin equal to \(\sqrt{5}\) is ……
(a) x + 2y = \(\sqrt{5}\)
(b) 2x + y = \(\sqrt{5}\)
(c) 2x + y = 5
(d) x + 2y – 5 = 0
Solution:
(c) 2x + y = 5

The required line is y = 2x + 5 ⇒ 2x – y + 5 = 0

Question 11.
A line perpendicular to the line 5x – y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5 sq. units, then its equation is …….

Solution:
(a) x + 5y ± 5\(\sqrt{2}\) = 0
Hint:
Equation of a line perpendicular to 5x – y = 0 is

Question 12.
Equation of the straight line perpendicular to the line x – y + 5 = o, through the point of intersection the y-axis and the given line …….
(a) x – y – 5 = 0
(b) x + y – 5 = 0
(c) x + y + 5 = 0
(d) x + y + 10 = 0
Solution:
(b) x + y – 5 = 0
Hint:
x – y + 5 = 0 ⇒ put x = 0, y = 5
The point is (0, 5)
Equation of a line perpendicular to x – y + 5 = 0 is x + y + k = 0
This passes through (0, 5)
k = -5
x + 7 – 5 = 0

Question 13.
If the equation of the base opposite to the vertex (2, 3) of an equilateral triangle is x + y = 2, then the length of a side is ………

Solution:
\(\sqrt{6}\)
Hint:
In an equilateral ∆ the perpendicular will bisects the base in to two equal parts. Length of the perpendicular drawn from (2, 3) to the line x + 7 – 2 = 0

Question 14.
The line (p + 2q) x + (p – 3q)y = p – q for different values of p and q passes through the point ……

Solution:
(d) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)
Hint:
(p + 2 q)x + (p – 3q)y = p – q
px + 2qx + py – 3qy = p – q
P(x + y) + q (2x – 3y) = p – q
The fourth option x = 2/5, y = 3/5

= p – q = RHS

Question 15.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …
(a) (7, 3)
(b) (4, 1)
(c) (1, -1)
(d) (-2, 3)
Solution:
(b) (4, 1)
Hint:
Let (a, b) be on 2x – 3y = 5 ⇒ 2a – 3b = 5
It is equidistance from (1, 2) and (3, 4)

(a – 1)² + (b – 2)² = (a – 3)² + (6 – 4)²
a² – 2a + 1 + b² – 4b + 4 = a² – 6a + 9 + b² – 8b + 16
4a + 4b = 20
2a+ 2b = 10
2a – 3b = 5
5b = 5
b = 1 ∴ a = 4
∴ The point is (4, 1)

Question 16.
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)
Solution:
(a) (-3, -2)

x – 2 = -5, y – 3 = -5
x = -3, y = -2
(-3,-2)

Question 17.
The length of ⊥ from the origin to the line \(\frac{x}{3}-\frac{y}{4}=1\) is ……

Solution:
(c) \(\frac{12}{5}\)
Hint:
4x – 3y = 12 ⇒ 4x – 3y – 12 = 0

Question 18.
The y-intercept of the straight line passing through (1, 3) and perpendicular to 2x – 3y + 1 = 0 is ……..

Solution:
(b) \(\frac{9}{2}\)
Hint:
Equation of a line perpendicular to 2x – 3y + 1 = 0 is 3x + 2y = k. It passes through (1, 3).
3 + 6 = k ⇒ k = 9, 3x + 2y = 9
To find y-intercept x = 0, 2y = 9, y = 9/2

Question 19.
If the two straight lines x + (2k – 7)y + 3 = 0 and 3kx + 9y – 5 = 0 are perpendicular then the value of k is ……

Solution:
(a) k = 3
Hint.

Since the lines are perpendicular m₁m₂ = – 1

Question 20.
If a vertex of a square is at the origin and its one side lies along the line 4x + 3y – 20 = 0, then the area of the square is ……..
(a) 20 sq. units
(b) 16 sq. units
(c) 25 sq. units
(d) 4 sq.units
Solution:
(b) 16 sq. units
Hint:
One side of a square = Length of the perpendicular from (0, 0) to the line.

Question 21.
If the lines represented by the equation 6x² + 41xy – 7y² = 0 make angles α and β with
x – axis, then tan α tan β =

Solution:
(a) \(-\frac{6}{7}\)
Hint.
6x² + 41xy – 7y² = 0 ⇒ 6x² – xy + 42xy – 7y² = 0 ⇒ x (6x – y) + 7y (6x – y) = 0
(x + 7y) (6x – y) = 0 ⇒ x + 7y = 0, 6x – y = 0

Question 22.
The area of the triangle formed by the lines x² – 4y² = 0 and x = a is …….

Solution:
(c) \(\frac{1}{2} a^{2}\)
Hint:
x² – 4y² = 0 , (x – 2y) (x + 2y) = 0 ⇒ x – 2y = 0, x + 2y = 0

Question 23.
If one of the lines given by 6x² – x + 4x² = 0 is 3x + 4y = 0, then c equals to ……
(a) -3
(b) -1
(c) 3
(d) 1
Solution:
(a) -3
Hint.
6x² – xy + 4cy² = 0, 3x + 4y = 0
The other line may be (2x + by)
(3x + 4y) (2x + by) = 6x² – xy + 4cy²
6x² + 3xby + 8xy + 4by² = 6x² – xy + 4cy²
6x² + xy (3b + 8) + 4by² = 6x² – xy + 4cy²
paring, 3b + 8 = -1
3b = -9 ⇒ b = -3
4b = 4c ⇒ 4(-3) = 4c
-12 = 4c ⇒ c = -3

Question 24.


Solution:
(c) \(\frac{5}{9}\)
Hint:

Question 25.
The equation of one the line represented by the equation x² + 2xy cot θ – y² = 0 is ………
(a) x – y cotθ = 0
(b) x + y tan θ = 0
(e) x cos θ + y(sin θ + 1) = 0
(d) x sin θ + y(cos θ + 1) = 0
Solution:
(d) x sin θ + y(cos θ + 1)=0