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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 4 Inverse Trigonometric Functions Ex 4.4

Question 1.
Find the principal value of
(i) sec^-1 (\(\frac{2}{\sqrt{3}}\))
(ii) cot^-1(√3)
(iii) cosec^-1(-√2)
Solution:
(i) Let sec^-1 (\(\frac{2}{\sqrt{3}}\)) = θ
⇒ sec θ = \(\frac{2}{\sqrt{3}}\)
⇒ cos θ = \(\frac{\sqrt{3}}{2}\) = cos \(\frac{\pi}{6}\)
⇒ θ = \(\frac{\pi}{6}\)

(ii) Let cot^-1(√3) = θ
⇒ cot θ = √3
⇒ tan θ = \(\frac{1}{\sqrt{3}}\) = tan \(\frac{\pi}{6}\)
⇒ θ = \(\frac{\pi}{6}\)

(iii) Let cosec^-1 (-√2) = θ
⇒ cosec θ = -√2
⇒ sin θ = \(-\frac{1}{\sqrt{2}}\)
⇒ θ = \(-\frac{\pi}{4}\)

Question 2.
Find the value of
(i) tan^-1(√3) – sec^-1(-2)
(ii) sin^-1(-1) + cos^-1 (\(\frac{1}{2}\)) + cot^-1(2)
(iii) cot^-1(1) + sin^-1(\(-\frac{\sqrt{3}}{2}\)) – sec^-1(-√2)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.4 Additional Problems

Question 1.
Find the principal value of the following

Solution:

Question 2.
Find the value of sec²(cot^-1 3) + cosec² (tan^-1 2)
Solution:

Question 3.
Find the value of
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 4 Inverse Trigonometric Functions Ex 4.3

Question 1.
Find the domain of the following functions:
(i) \(\tan ^{-1}(\sqrt{9-x^{2}})\)
(ii) \(\frac{1}{2} \tan ^{-1}\left(1-x^{2}\right)-\frac{\pi}{4}\)
Solution:
(i) f(x) = \(\tan ^{-1}(\sqrt{9-x^{2}})\)
We know the domain of tan^-1 x is (-∞, ∞) and range is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
So, the domain of f(x) = \(\tan ^{-1}(\sqrt{9-x^{2}})\) is the set of values of x satisfying the inequality
\(-\infty \leq \sqrt{9-x^{2}} \leq \infty\)
⇒ 9 – x² ≥ 0
⇒ x² ≤ 9
⇒ |x| ≤ 3
Since tan x is an odd function and symmetric about the origin, tan^-1 x should be an increasing function in its domain.
∴ Domain is (2n + 1)\(\frac{\pi}{2}\)

The domain of y is (-∞, ∞) {x | x ∈ -1} and range is [-1, ∞) {y | y ≥ -1}
The domain for tan^-1(x² – 1) is (2n + 1)π. Since tan x is an odd function.

Question 2.
Find the value of
(i) \(\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)\)
(ii) \(\tan ^{-1}\left(\tan \left(-\frac{\pi}{6}\right)\right)\)
Solution:

Question 3.
Find the value of
(i) \(\tan \left(\tan ^{-1} \frac{7 \pi}{4}\right)\)
(ii) tan(tan^-1(1947))
(iii) tan(tan^-1(-0.2021))
Solution:
We know that tan(tan^-1 x) = x
(i) \(\tan \left(\tan ^{-1} \frac{7 \pi}{4}\right)=\frac{7 \pi}{4}\)
(ii) tan(tan^-1(1947))= 1947
(iii) tan(tan^-1 (-0.2021)) = -0.2021

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

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.3 Additional Problems

Question 1.
Find the principle value of:
Solution:

Question 2.
Find the value of
Solution:

Question 3.
Find the value of
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 11 Probability Distributions Ex 11.4

Question 1.
For the random variable X with the given probability mass function as below, find the mean and variance

Solution:

Question 2.
Two balls are drawn in succession without replacement from an urn containing four red balls and three black balls. Let X be the possible outcomes drawing red balls. Find the probability mass function and mean for X.
Solution:
Number of Red balls = 4
Number of Black balls = 3
Total number of balls = 7
Given : two balls are drawn in succession without replacement.
Let ‘X’ be the number of red balls and ‘X’ can take the values 0, 1 and 2.

Question 3.
If µ and σ² are the mean and variance of the discrete random variable X, and E (X + 3) = 10 and E(X + 3)² = 116, find µ and σ².
Solution:
Mean = µ,
Variance = σ²

Variance Var (X) = E (X²) — [E(X)]²
65 – 49 = 16 = σ²
∴ µ = 7 and σ² = 16

Question 4.
Four fair coins are tossed once. Find the probability mass function, mean and variance for number of heads occurred.
Solution:
Let ‘X’ be the number of heads occurred when four coins are tossed once. Hence ‘X’ can take the values 0, 1, 2, 3 and 4.
When 4 coins are tossed, the sample space is,
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

Question 5.
A commuter train arrives punctually at a station every half hour. Each morning, a student leaves his house to the train station. Let X denote the amount of time, in minutes, that the student waits for the train from the time he reaches the train station. It is known that the p.d.f. of X is.

Obtain and interpret the expected value of the random variable X.
Solution:

‘X’ is a continuous random variable.

Question 6.
The time to failure in thousands of hours of an electronic equipment used in a manufactured computer has the density function. Find the expected life of this electronic equipment.
Solution:

Question 7.
The probability density function of the random variable X is given by Find the mean and variance of X.
Solution:

Question 8.
A lottery with 600 tickets gives one prize of ₹ 200, four prizes of ₹ 100, and six prizes of ₹ 50. If the ticket costs is ₹ 2, find the expected winning amount of a ticket.
Solution:
Given, total number of tickets = 600
Prizes to be given : One prize of Rs. 200
Four prizes of Rs. 100
Six prizes of Rs. 50
Let ‘X’ be the random variable denotes the winning amount and it can take the values 200, 100 and 50.

Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.4 Additional Problems

Question 1.
The probability of success of an event is p and that of failure is q. Find the expected number of trials to get a first success.
Solution:
Let X be the random variable denoting ‘Number of trials to get a first success’. The success can occur in the 1^st trial. ∴ The probability of success in the 1st trial is p. The success in the 2nd trial means failure in the 1st trial. ∴ Probability is qp.
Success in the 3^rd trial means failure in the first two trials. Probability of success in the 3^rd trial is q²p. As it goes on, the success may occur in the n^th trial which mean the first (n – 1) trials are failures. probability = q^{n – 1}p.
∴ The probability distribution is as follows

Question 2.
An urn contains 4 white and 3 Red balls. Find the probability distribution of the number of red balls in three draws when a ball is drawn at random with replacement. Also find its mean and variance.
Solution:
The required probability distribution is

Question 3.
Find the mean and variance of the distribution
Solution:

Question 4.
Two cards are drawn with replacement from a well shuffled deck of 52 cards. Find the mean and variance for the number of aces.
Solution:
n (S) = 52
Number of aces = n (A) = 4

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 11 Probability Distributions Ex 11.3

Question 1.
The probability density function of X is given by Find the value of k.
Solution:

Question 2.

Solution:

Question 3.
Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function

Find (i) the value of k
(ii) the distribution function
(iii) the probability that daily sales will fall between 300 litres and 500 litres?
Solution:

Question 4.

Solution:

Question 5.
If X is a random variable with probability density function f(x) given by,

Solution:

Question 6.
If X is the random variable with probability density function F(x) given by,
then find (i) the distribution function f(x)
(ii) P(0.3 ≤ X ≤ 0.6)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.3 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

Question 3.
The probability density function of a random variable x is Find
(i) k;
(ii) P(X > 10)
Solution:

Question 4.
A continuous random variable x has the p.d.f. defined by find the value of C if a > 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 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:

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 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:

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 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:

You can Download Samacheer Kalvi 11th 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 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