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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 8 Differentials and Partial Derivatives Ex 8.3

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Here, f satisfies all the three conditions of continuity at (a, b). Hence, f’ is continuous at every point of R² as (a, b) ∈ R².

Question 7.

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 8 Differentials and Partial Derivatives Ex 8.2

Question 1.
Find differential dy for each of the following functions:

Solution:

Question 2.
Find df for f(x) = x² + 3x and evaluate it for
(i) x = 2 and dx = 0.1
(ii) x = 3 and dx = 0.02
Solution:
y = f(x) = x² + 3x
dy = (2x + 3) dx
(i) dy {when x = 2 and ate = 0.1} = [2(2) + 3] (0.1)
= 7(0.1) = 0.7

(ii) dy {when x = 3 and dx = 0.02} = [2(3) + 3] (0.0.2)
= 9(0.02) = 0.18

Question 3.
Find ∆f and df for the function f for the indicated values of x, ∆x and compare
(i) f (x) = x³ – 2x² ; x = 2, ∆x = dx = 0.5
(ii) f(x) = x² + 2x + 3; x = -0.5, ∆x = dx = 0.1
Solution:
(i) y = f(x) = x³ – 2x²
dy = (3x² – 4x) dx
dy (when x = 2 and dx = 0.5) = [3(2²) – 4(2)] (0.5)
= (12 – 8)(0.5) = 4(0.5) = 2
(i.e.,) df = 2
Now ∆f = f(x + ∆x) – f(x)
Here x = 2 and ∆x = 0.5
f(x) = x³ – 2x²
So f(x + ∆x) = f(2 + 0.5) = f(2.5) = (2.5)³ – 2 (2.5)² = (2.5)² [2.5 – 2] = 6.25 (0.5) = 3.125
f(x) = f(2) = 2³ – 2(2²) = 8 – 8 = 0
So ∆f = 3.125 – 0 = 3.125

(ii) y = f(x) = x² + 2x + 3
dy = (2x + 2) dx
dy (when x = – 0.5 and dx = 0.1)
= [2(-0.5) + 2] (0.1)
= (-1 + 2) (0.1) = (1) (0.1) = 0.1
(i.e.,) df = 0.1
Now ∆f = f(x + ∆x) – f(x)
Here x = -0.5 and ∆x = 0.1
x² + 2x + 3
f(x + ∆x) = f(-0.5 + 0.1) = f(-0.4)
= (-0.4)² + 2(-0.4) + 3
= 0.16 – 0.8 + 3 = 3.16 – 0.8 = 2.36
f(x) = f(-0.5) = (-0.5)² + 2(-0.5) + 3
= 0.25 – 1 + 3 = 3.25 – 1 = 2.25
So ∆ f = f(x + ∆x) – f(x) = 2.36 – 2.25 = 0.11

Question 4.
Assuming log₁₀ e = 0.4343, find an approximate value of log₁₀ 1003.
Solution:
To find log 1003
1003 = 1000 + 3 and
log 1000 = 3
Let y = log x

Here x = 1000 and dx = 3 = 3 log₁₀e

Question 5.
The trunk of a tree has diameter 30 cm. During the following year, the circumference grew 6 cm.
(i) Approximately, how much did the tree’s diameter grow?
(ii) What is the percentage increase in area of the tree’s cross-section?
Solution:
(i) Given r = 15 cm and rate of change of perimeter = 6 cm
To find the rate of change of diameter
Now perimeter = p = 2πr
So dp = 2πdr
Here dp = 6 cm (given)

Question 6.
An egg of a particular bird is very nearly spherical. If the radius to the inside of the shell is 5 mm and radius to the outside of the shell is 5.3 mm, find the volume of the shell approximately.
Solution:

Question 7.
Assume that the cross section of the artery of human is circular. A drug is given to a patient to dilate his arteries. If the radius of an artery is increased from 2 mm to 2.1 mm, how much is cross-sectional area increased approximately?
Solution:
Area of circle = A = πr².

Question 8.
In a newly developed city, it is estimated that the voting population (in thousands) will increase according to V (t) = 30 + 12t² – t³, 0 ≤ t ≤ 8 where t is the time in years. Find the approximate change in voters for the time change from 4 to 4\(\frac{1}{6}\) year.
Solution:

Question 9.
The relation between the number of words y a person learns in x hours is given by y = 52\(\sqrt{x}\), 0 ≤ x ≤ 9. What is the approximate number of words learned when x changes from
(i) 1 to 1.1 hour?
(ii) 4 to 4.1 hour?
Solution:

Question 10.
A circular plate expands uniformly under the influence of heat. If it’s radius increases from 10.5 cm to 10.75 cm, then find an approximate change in the area and the approximate percentage change in the area.
Solution:
Here radius is changing from 10.5 cm to 10.75 cm
⇒ r = 10.5 cm and dr = 0.25 cm
Now area = A = πr²
⇒ dA = π (2r) dr

(i) So dA
(when r = 10.5 cm and dr = 0.25 cm)
= π (2 × 10.5) (0.25)
= 5.25 π

Question 11.
A coat of paint of thickness 0.2 cm is applied to the faces of a cube whose edge is 10 cm. Use the differentials to find approximately how many cubic centimeters of paint is used to paint this cube. Also calculate the exact amount of paint used to paint this cube.
Solution:
(i) v = a³
so dv = a² da
dv (when) a = 10 cm and da = 0.20 cm
= 3(10²) (0.2)
300 × 0.2 = 60 cm³

Actual paint used = v at x + ∆x = 10.2 and x = 10 cm
= a³ at x + ∆x = 10.2 and x = 10
= (10.2)³ – (10) = 61.2 cm³

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.2 Additional Questions

Question 1.
Find the differential dy and evaluate dy for the given values of x and dx. y = x⁴ – 3x² + x – 1, x = 2, dx = 0.1
Solution:

dy = [4(8) – 6(2) + 1] (0.1) = (32 – 12 + 1) (0.1)
= (21) (0.1) = 2.1
dy = (4x³ – 6x + 1)dx; dy = 2.1

Question 2.
Find the differential dy and evaluate dy for the given values of x and dx. y = \(\sqrt{1-x}\), x = 0, dx = 0.02
Solution:

Question 3.
The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error in computing
(i) the volume of the cube and
(ii) the surface area of cube.
Solution:
The side of the cube = a = 30 cm
Error in a = da = 0.1 cm

Question 4.
The radius of a circular disc is given as 24 cm with a maximum error in measurement of 0.02 cm
(i) Use differentials to estimate the maximum error in the calculated area of the disc,
(ii) Compute the relative error.
Solution:
r = 24 cm,
dr = 0.02 cm

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.1

Question 1.
Let f(x) = \(\sqrt[3]{x}\). Find the linear approximation at x = 27. Use the linear approximation to \(\sqrt[3]{27.2}\).
Solution:

Question 2.
Use the linear approximation to find approximate values of

Solution:

Question 3.
Find a linear approximation for the following functions at the indicated points.
(i) f(x) = x³ – 5x + 12, x₀ = 2
(ii) g(x) = \(\sqrt{x^{2}+9}\) + x₀ = -4
(iii) h(x) = \(\frac{x}{x+1}\), x₀ = 1
Solution:
f(x) = x³ – 5x + 12
f'(x) = 3x² – 5
f(x₀) = f(2) = (2)³ – 5(2) + 12 = 8 – 10 + 12 = 10
f'(x₀) = f'(2) = 3(2)² – 5 = 12 – 5 = 7
The required linear approximation L(x) = f(x₀) + f'(x₀) (x – x₀)
= 10 + 7 (x – 2)
= 10 + 7x – 14
= 7x – 4

Question 4.
The radius of a circular plate is measured as 12.65 cm instead of the actual length 12.5 cm. find the following in calculating the area of the circular plate:
(i) Absolute error
(ii) Relative error
(iii) Percentage error
Solution:
We know that Area of the circular plate A(r) = πr², A'(r) = 2πr
Change in Area = A’ (12.5) (0.15) = 3.75 π cm²
Exact calculation of the change in Area = A (12.65) – A (12.5)
= 160.0225π – 156.25π
= 3.7725π cm²
(i) Absolute error = Actual value – Approximate value
= 3.7725 π – 3.75 π
= 0.0225 π cm²

Question 5.
A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9.8 cm. Find approximations for the following:
(i) change in the volume
(ii) change in the surface area
Solution:
(i) We know that Volume of sphere

∴ Volume decreases by 80π cm³

(ii) Surface area of the sphere
S(r) = 4π r²
S'(r) = 8πr
Change in surface area at r = 10 is
= S'(r) [10 – 9.8]
= 8π (10) (0.2) = 16π cm²
∴ Surface Area decreases by 16π cm²

Question 6.
The time T, taken for a complete oscillation of a single pendulum with length l, is given
by the equation T = \(2 \pi \sqrt{\frac{l}{g}}\), where g is a constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of 1.
Solution:

Question 7.
Show that the percentage error in the n^th root of a number is approximately \(\frac{1}{n}\) times the percentage error in the number.
Solution:
Let x be the number

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.1 Additional Questions Solved

Question 1.
Using differentials, find the approximate value of each of the following upto 3 places of decimal. \((255)^{\frac{1}{4}}\)
Solution:

Question 2.
Using differentials, find the approximate value of each of the following upto 3 places of decimal. \((401)^{\frac{1}{2}}\)
Solution:

Question 3.
Find approximate value of f(5.001) where f(x) = x³ – 7x² + 15.
Solution:
Here f(x) = x³ – 7x² + 15
f'(x) = 3x² – 14x
Let x = 5 and x + ∆x = 5.001
∴ ∆x (x + ∆x) – x = 5.001 – 5
= 0.001
f(5.001) = f(x + ∆x)
Now ∆y = f(x + ∆x) – f(x)
f(x + ∆x) = f(x) + ∆y
= f(x) + f'(x).∆x [∴ ∆y = f'(x).∆x ]
= (x³ – 7x² + 15) + (3x² – 14x)(0.001)
= (5³ – 7 × 5² + 15) + (3 × 5² – 14 × 5) [0.001]
= [125 – 175 + 15] + [75 – 70][0.001]
= -35 + 0.005 = – 34.995
Thus approximate value of f(5.001) is – 34.995

Question 4.
If the radius of a sphere is measured as 7m with an error of 0.02 m then find the approximate error in calculating its volume.
Solution:
Le r be the radius of the sphere and ∆r be the error in measuring the radius.
Then r = 7 m and ∆r = 0.02 m
Now volume of a sphere is given by

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 4.6

Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
The value of sin^-1(cos x), 0 ≤ x ≤ π is …………
(a) π – x
(b) x – \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{2}\) – x
(d) π – x
Solution:
(c) \(\frac{\pi}{2}\) – x
Hint:

Question 2.
If sin^-1 + sin^-1 y = \(\frac{2 \pi}{3}\); then cos^-1x + cos^-1y is equal to ………….
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{6}\)
(d) π
Solution:
(b) \(\frac{\pi}{3}\)
Hint:

Question 3.
……………
(a) 2π
(b) π
(c) 0
(d) tan^-1\(\frac{12}{65}\)
Solution:
(c) 0
Hint:

Question 4.
If sin^-1x = 2 sin^-1 α has a solution, then ……………

Solution:
(a) |α| ≤ \(\frac{1}{\sqrt{2}}\)
Hint:

Question 5.
sin^-1 (cos x) = \(\frac{\pi}{2}\) – x is valid for ……………..
(a) -π ≤ x ≤ 0
(b) 0 ≤ x ≤ π
(c) \(-\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\)
(d) \(-\frac{\pi}{4}\) ≤ x ≤ \(\frac{3 \pi}{4}\)
Solution:
(b) 0 ≤ x ≤ π

Question 6.
If sin^-1 x + sin^-1 y + sin^-1 z = \(\frac{3 \pi}{2}\), the value of x²⁰¹⁷ + y²⁰¹⁸ + z²⁰¹⁹ – \(\frac{9}{x^{101}+y^{101}+z^{101}}\) is ……………….
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
The maximum value of sin^-1 x is \(\frac{\pi}{2}\) and sin^-1 1 = \(\frac{\pi}{2}\)
Here it is given that
sin^-1 x + sin^-1 y + sin^-1 z = \(\frac{3 \pi}{2}\)
⇒ x = y = z = 1
and so 1 + 1 + 1 – \(\frac{9}{1+1+1}\) = 3 – 3 = 0

Question 7.
If cot^-1 x = \(\frac{2 \pi}{5}\) for some x ∈ R, the value of tan^-1 x is …………
(a) \(-\frac{\pi}{10}\)
(b) \(\frac{\pi}{5}\)
(c) \(\frac{\pi}{10}\)
(d) \(-\frac{\pi}{5}\)
Solution:
(c) \(\frac{\pi}{10}\)
Hint:

Question 8.
The domain of the function defined by f(x) = sin^-1 \(\sqrt{x-1}\) is …………….
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Solution:
(a) [1, 2]
Hint:
The domain for sin^-1 x is [0, 1]
So \(\sqrt{x-1}\) = 0 ⇒ x – 1 = 0 ⇒ x = 1
\(\sqrt{x-1}\) = 1 ⇒ x – 1 = 0 ⇒ x = 2
∴ The domain is [1, 2]

Question 9.
If x = \(\frac{1}{5}\), the value of cos(cos^-1 x + 2 sin^-1 x) is …………….

Solution:
(d) \(-\frac{1}{5}\)
Hint:

Question 10.
\(\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\) is equal to …………

Solution:
(d) tan^-1\(\frac{1}{2}\)
Hint:

Question 11.
If the function f(x) = sin^-1(x² – 3), then x belongs to …………..
(a) [1, -1]
(b) [\(\sqrt{2}\), 2]
(c) \([-2,-\sqrt{2}] \cup[\sqrt{2}, 2]\)
(d) \([-2,-\sqrt{2}] \cap[\sqrt{2}, 2]\)
Solution:
(c) \([-2,-\sqrt{2}] \cup[\sqrt{2}, 2]\)
Hint:
f(x) = sin^-1(x² – 3)
Domain of sin^-1 (x) is [-1, 1]
⇒ -1 ≤ x² – 3 ≤ 1 ⇒ 2 ≤ x² ≤ 4
⇒ \(\sqrt{2}\) ≤ x ≤ 2 ⇒ \(\sqrt{2}\) ≤ |x| ≤ 2
x ∈ \([-2,-\sqrt{2}] \cup[\sqrt{2}, 2]\)

Question 12.
If cot^-1 2 and cot^-1 3 are two angles of a triangle, then the third angle is …………..

Solution:
(b) \(\frac{3 \pi}{4}\)
Hint:

Question 13.
\(\sin ^{-1}\left(\tan \frac{\pi}{4}\right)-\sin ^{-1}(\sqrt{\frac{3}{x}})=\frac{\pi}{6}\). Then x is a root of the equation …………..
(a) x² – x – 6 = 0
(b) x² – x – 12 = 0
(c) x² + x – 12 = 0
(d) x² + x – 6 = 0
Solution:
(b) x² – x – 12 = 0
Hint:

Question 14.
sin^-1(2 cos² x – 1) + cos^-1(1 – 2 sin² x) = ……………

Solution:
(a) \(\frac{\pi}{2}\)
Hint:
2 cos²x – 1 = cos 2x
1 – 2 sin² x = cos 2x
∴ sin^-1 x(cos 2x) + cos^-1(cos 2x) = \(\frac{\pi}{2}\) (∵ sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\))

Question 15.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})\) = u, then cos 2u is equal to …………..
(a) tan² α
(b) 0
(c) -1
(d) tan 2α
Solution:
(c) -1
Hint:
cot^-1 x + tan^-1 x = \(\frac{\pi}{2}\) ⇒ u = \(\frac{\pi}{2}\) so 2u = π
∴ cos 2u = cos π = -1

Question 16.
If |x| ≤ 1, then 2tan^-1 x – sin^-1 \(\frac{2 x}{1+x^{2}}\) is equal to ………….
(a) tan^-1 x
(b) sin^-1 x
(c) 0
(d) π
Solution:
(c) 0
Hint:
Let x = tan θ so \(\frac{2 x}{1+x^{2}}\) = sin 2θ.
Now 2 tan^-1(tanθ) – sin^-1(sin 2θ) = 2θ – 2θ = 0

Question 17.
The equation tan^-1 x – cot^-1 x = tan^-1 \(\left(\frac{1}{\sqrt{3}}\right)\) has …………..
(a) no solution
(b) unique solution
(c) two solutions
(d) infinite number of solutions
Solution:
(b) unique solution
Hint:

Question 18.
If sin^-1 x + cot^-1 \(\left(\frac{1}{2}\right)=\frac{\pi}{2}\), then x is equal to …………

Solution:
(b) \(\frac{1}{\sqrt{5}}\)
Hint:
sin^-1 x + cot^-1 \(\left(\frac{1}{2}\right)=\frac{\pi}{2}\)

Question 19.
If sin^-1\(\frac{x}{5}\) + cosec^-1\(\frac{5}{4}\) = \(\frac{\pi}{2}\), then the value of x is ………
(a) 4
(b) 5
(c) 2
(d) 3
Solution:
(d) 3
Hint:

Question 20.
sin(tan^-1), |x| < 1 is equal to ………….

Solution:
(d) \(\frac{x}{\sqrt{1+x^{2}}}\)
Hint:

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.6 Additional Questions

Question 1.
Find the principal value of
Solution:

Question 2.
Find the principal value of
Solution:

Question 3.

Solution:

Question 4.
Evaluate

Solution:

Question 5.
Evaluate

Solution:

Question 6.

Question 7.

Solution:

Question 8.

Solution:
x = \(\frac{1}{6}\)

Question 9.
Find the values of each of the following:

Solution:

Question 10.
Solve for x:

Solution:

Question 11.
Prove:

Question 12.
Evaluate: sin(tan^-1 x + cot^-1 x)

Question 13.
The value of sin^-1(1) + sin^-1(0) is …….

Solution:
(a) \(\frac{\pi}{2}\)
Hint:

Question 14.


Solution:
(d) \(\frac{\pi}{2}\)
Hint:

Question 15.
tan^-1x + cot^-1x = ……..
(a) 1
(b) – π
(c) \(\frac{\pi}{2}\)
(d) π
Solution:
(c) \(\frac{\pi}{2}\)
Hint:

Question 16.


Solution:
(a) \(\frac{-\pi}{2}\)
Hint:

Question 17.


Solution:
(b) \(\frac{\pi}{2}\)
Hint:

Question 18.


Solution:
(a) \(\sin ^{-1} \frac{1}{\sqrt{2}}\)
Hint:

Question 19.


Solution:
(d) 2π
Hint:

Question 20.


Solution:
(c) 2π
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 11 Probability Distributions Ex 11.6

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

Question 1.
Let X be random variable with probability density function

Which of the following statement is correct
(a) both mean and variance exist
(b) mean exists but variance does not exist
(c) both mean and variance do not exist
(d) variance exists but Mean does not exist
Solution:
(b) mean exists but variance does not exist
Hint:

Question 2.
A rod of length 21 is broken into two pieces at random. The probability density function of the shorter of the two pieces is The mean and variance of the shorter of the two pieces are respectively ……..

Solution:
(d) \(\frac{l}{2}, \frac{l^{2}}{12}\)
Hint:

Question 3.
Consider a game where the player tosses a six-sided fair die. If the face that comes up is 6, the player wins ₹ 36, otherwise he loses ₹ k², where k is the face that comes up k = {1, 2, 3, 4, 5}. The expected amount to win at this game in ₹ is ……..

Solution:
(b) \(-\frac{19}{6}\)
Hint:

Question 4.
A pair of dice numbered 1, 2, 3, 4, 5, 6 of a six-sided die and 1, 2, 3, 4 of a four-sided die is rolled and the sum is determined. Let the random variable X denote this sum. Then the number of elements in the inverse image of 7 is ………
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4
Hint:
Sample points to get the sum ‘7’ are (3, 4), (4, 3), (5, 2), (6, 1)
Number of elements in inverse image of 7 = 4

Question 5.
A random variable X has binomial distribution with n = 25 and p = 0.8 then standard deviation of X is ……
(a) 6
(b) 4
(c) 3
(d) 2
Solution:
(d) 2
Hint:
n = 25, p = 0.8, ∴ q = 1 – 0.8 = 0.2

Question 6.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. Then the possible values of X are
(a) i + 2n, i = 0, 1, 2 … n
(b) 2i – n, i = 0, 1, 2 … n
(c) n – i, i = 0, 1, 2 … n
(d) 2i + 2n, i = 0, 1, 2 … n
Solution:
(b) 2i – n, i = 0, 1, 2 … n

Question 7.
If the function represents a probability density function of a continuous random variable X, then which of the following cannot be the value of a and b?
(a) 0 and 12
(b) 5 and 17
(c) 7 and 19
(d) 16 and 24
Solution:
(d) 16 and 24
Hint:

Question 8.
Four buses carrying 160 students from the same school arrive at a football stadium. The buses carry, respectively, 42, 36, 34, and 48 students. One of the students is randomly selected. Let X denote the number of students that were on the bus carrying the randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on that bus. Then E[X] and E[Y] respectively are
(a) 50, 40
(b) 40, 50
(c) 40.75, 40
(d) 41,41
Solution:
(c) 40.75, 40
Hint:
E[X] > E[Y], Since the students selected is more likely to come from the bus carrying maximum number of students whereas the drivers are equally likely to be selected.

Question 9.
Two coins are to be flipped. The first coin will land on heads with probability 0.6, the second with Probability 0.5. Assume that the results of the flips are independent, and let X equal the total number of heads that result. The value of E[X] is …..
(a) 0.11
(b) 1.1
(c) 11
(d) 1
Solution:
(b) 1.1
Hint:
P (X = 0) = P (TT)
= P (T) P (T) = (0.4) (0.5) = 0.20
P (X = 1) = P (HT) + P(TH) = (0.6) (0.5) + (0.4) (0.5) = 0.5
P (X = 2) = P (HH)
= P (H). P (H) = (0.6) (0.5) = 0.30

= (0).(0.20) + (1) (0.5) + (2) (0.30)
= 0.5 + 0.6 = 1.1

Question 10.
On a multiple-choice exam with 3 possible destructives for each of the 5 questions, the probability that a student will get 4 or more correct answers just by guessing is …….

Solution:
(a) \(\frac{11}{243}\)
Hint:

Question 11.
If P{X = 0} = 1 – P{X = 1}. If E[X] = 3Var(X), then P{X = 0}.

Solution:
(d) \(\frac{1}{3}\)
Hint:
Given P (X = 0) = 1 – P (X = 1) ⇒ P (X = 1) = 1 – P (X = 0)
Let n = P(X = 0), ∴ P(X = 1) = 1 -n
E(X) = (0) (n) + (1) (1 – n) = 1 – n
E(X²) = (0)² (n) + (1)² (1 – n) = 1 – n

Question 12.
If X is a binomial random variable with expected value 6 and variance 2.4, Then P{X = 5} is ……….

Solution:
(d) \(\left(\begin{array}{c}{10} \\ {5}\end{array}\right)\left(\frac{3}{5}\right)^{5}\left(\frac{2}{5}\right)^{5}\)
Hint:

Question 13.

(a) 1 and \(\frac{1}{2}\)
(b) \(\frac{1}{2}\) and 1
(c) 2 and 1
(d) 1 and 2
Solution:
(a) 1 and \(\frac{1}{2}\)
Hint:

Question 14.
Suppose that X takes on one of the values 0, 1, and 2. If for some constant k,
P(X = i) = k P(X = i – 1) for i = 1, 2 P(X = 0) = \(\frac{1}{7}\). Then the value of k is …..
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Hint:
P(X = i) = k P(X = i – 1) for i = 1, 2 and P(X = 0) = \(\frac{1}{7}\)
Solving the equation, we get k = 2

Question 15.
Which of the following is a discrete random variable?
I. The number of cars crossing a particular signal in a day.
II. The number of customers in a queue to buy train tickets at a moment.
III. The time taken to complete a telephone call.
(a) I and II
(b) II only
(c) III only
(d) II and III
Solution:
(a) I and II

Question 16.
If is a probability density function of a random variable, then the value of a is ……..
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(a) 1
Hint:

Question 17.
The probability function of a random variable is defined as ……..

Then E(X) is equal to …….

Solution:
(d) \(\frac{2}{3}\)
Hint:

Question 18.
Let X have a Bernoulli distribution with mean 0.4, then the variance of (2X – 3) is ……
(a) 0.24
(b) 0.48
(c) 0.6
(d) 0.96
Solution:
(d) 0.96
Hint:
Mean = 0.4 ⇒ µ = p = 0.4
Var (2X – 3) = 2² Var X
= 4(pq) = 4(0.4)(0.6) = 0.96

Question 19.
If in 6 trials, X is a binomial variate which follows the relation 9P(X = 4) = P(X = 2), then the probability of success is …..
(a) 0.125
(b) 0.25
(c) 0.375
(d) 0.75
Solution:
(b) 0.25
Hint:

9p² = q² [∵ 6C₄ = 6C₂]
9p² = (1 – p)²
9p² = 1 – 2p+ p²
8p² + 2p – 1 = 0
(4p – 1)(2p + 1) = 0

Question 20.
A computer salesperson knows from his past experience that he sells computers to one in every twenty customers who enter the showroom. What is the probability that he will sell a computer to exactly two of the next three customers?

Solution:
(a) [ltaex]\frac{57}{20^{3}}[/latex]
Hint:

Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.6 Additional Questions Solved

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

Question 1.
is a probability density function then the value of k is ……..

Solution:
(c) \(\frac{1}{9}\)
Hint:
Given that f(x) is a probability density function

Question 2.
is a probability density function of a continuous random variable X, then the value of A is …………
(a) 16
(b) 8
(c) 4
(d) 1
Solution:
(c) 4
Hint:

Question 3.
A random variable X has the following probability mass function as follows:

Then the value of λ is ……..
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Hint:

Question 4.


Solution:
(b) \(\frac{24}{169}\)

Question 5.
Given E(x + c) = 8 and E(x – c) = 12, then the value of c is …….
(a) -2
(b) 4
(c) -4
(d) 2
Solution:
(a) -2
Hint:
E(X + c) = 8
E(X) + E(c) = 8
E(X) + c = 8 ….. (1)
E(X – c) = 12
E(X) – E(c) = 12 ⇒ E(X) – c = 12 ……… (2)
(1) – (2) ⇒ 2c = -4 = c = -2

Question 6.
X is random variable taking the values 3, 4 and 12 with probabilities \(\frac{1}{3}\), \(\frac{1}{4}\) and \(\frac{5}{12}\). Then E(X) is ……..
(a) 5
(b) 7
(c) 6
(d) 3
Solution:
(b) 7
Hint:

Question 7.
Variance of the random variable X is 4. Its mean is 2. Then E(X²) is ……..
(a) 2
(b) 4
(c) 6
(d) 8
Solution:
(d) 8
Hint:
Given, Var (X) = 4
E(X²) – [E(X)]² = 4
E(X²) – 2² = 4 ⇒ E(X²) = 4 + 4 = 8

Question 8.
In 5 throws of a die, getting 1 or 2 is a success. The mean number of success is ……

Hint:
Given, n = 5

Question 9.
The mean of a binomial distribution is 5 and its standard deviation is 2. Then the value of n and p are ……..

Solution:
(d) \(\left(25, \frac{1}{5}\right)\)
Hint:

Question 10.
If the mean and standard deviation of a binomial distribution are 12 and 2 respectively. Then the value of its parameter p is ……

Solution:
(c) \(\frac{2}{3}\)
Hint:

Question 11.
A box contains 6 red and 4 white balls. If 3 balls are drawn at random, the probability of getting 2 white balls without replacement is ………

Solution:
(d) \(\frac{3}{10}\)
Hint:
Total number of balls = 10
Number of ways of choosing 3 balls from 10 balls = 10C₃
Number of ways of choosing 2 white balls from 4 white balls = 4C₃
Number of ways of choosing 1 red ball from 6 red balls = 6C₁

Question 12.
If 2 cards are drawn from a well shuffled pack of 52 cards, the probability that they are of the same colours without replacement is ………

Solution:
(c) \(\frac{25}{51}\)
Hint:

Question 13.
The distribution function F(X) of a random variable X is ……..
(a) a decreasing function
(b) a non-decreasing function
(c) a constant function
(d) increasing first and then decreasing
Solution:
(b) a non-decreasing function

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.8

Question 1.
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours?
Solution:
Let ‘P’ be the number of bacteria present at time ‘t’

log P = kt + c
P = e^{kt + c}
P = e^kt c …(1)
Initially, when t = 0, let P = P₀
P₀ = e° c ⇒ c = P₀
Given t = 5, P = 3P₀
Substituting in (1)
3 P₀ = e^5k. P₀ ⇒ e^5k = 3
Again, when t = 10
P = e^10k. c
P = (e^5k)² . P₀ = (3)² P₀
P = 9P₀
After 10 hours, the number of bacteria is 9 times the original bacteria present.

Question 2.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 3,00,000 to 4,00,000.
Solution:
Let the population of a city be ’P’ at time ‘t’

Question 3.
The equation of electromotive force for an electric circuit containing resistance and self inductance is E = Ri + L \(\frac{d i}{d t}\), where E is the electromotive force is given to the circuit, R the resistance and L, the coefficient of induction. Find the current i at time t when E = 0.
Solution:
Given equation of electromotive force is

This is a Linear differential equation

Substituting in (1)

Question 4.
The engine of a motor boat moving at 10 m/s is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equal to the velocity at that time. Find the velocity after 2 seconds of switching off the engine.
Solution:
Given, the retardation experienced by a moving motor boat after its engine is shut off is given by

Again, given t = 2 ⇒ v = 10 e²

Question 5.
Suppose a person deposits 10,000 Indian rupees in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later?
Solution:
Let P be the principal
Rate of interest 5 %

Given,when t = 0,P = 10000
P = 10000 e^{0.05(1.5) = 10000 e0.075}

Question 6.
Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. In a certain sample 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. What percentage of the original radioactive nuclei will remain after 1000 years?
Solution:
Let x (t) denote the number of nuclei present in a sample at anytime ‘t’

Let x_o denote the number of nuclei present in the given sample at time t = 0
∴ from (1) ⇒ c = log k₀
∴ x = x₀e^{k t}

∴ Required percentage after 1000 years

of radioactive element will remain after 1000 years.

Question 7.
Water at temperature 100° C cools in 10 minutes to 80° C in a room temperature of 25° C. Find
(i) The temperature of water after 20 minutes
(ii) The time when the temperature is 40° C

Solution:

Question 8.
At 10.00 A.M. a woman took a cup of hot instant coffee from her microwave oven placed it on a nearby Kitchen counter to cool. At this instant the temperature of the coffee was 180° F, and 10 minutes later it was 160° F . Assume that constant temperature of the kitchen was 70° F.
(i) What was the temperature of the coffee at 10.15 A.M.?
(ii) The woman likes to drink coffee when its temperature is between 130° F and 140° F. between what times should she have drunk the coffee?
Solution:
Change in Temperature is proportional to the difference in temperature (Newton’s Cooling
Law)

(ii) Substituting T = 130 and T = 140
We get t = 22.523

Question 9.
A pot of boiling water at 100° C is removed from a stove at time t = 0 and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to 80° C, and another 5 minutes later it has dropped to 65° C. Determine the temperature of the kitchen.
Solution:
Let A be the temperature of the kitchen

log (A – T) = kt + c
⇒ T = A + ce^-kt …. (1)
Given, T(0) = 100,
T(5) = 80,
T (10) = 65
Substituting in (1), we get
100 = A+c
80 = A + ce^-5k
65 = A + ce ^-10k
Solving these linear equations,
We get A = 11° (Kitchen Temperature)

Question 10.
A tank initially contains 50 liters of pure water. Starting at time t = 0 a brine containing with 2 grams of dissolved salt per litre flows into the tank at the rate of 3 liters per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time t > 0 .
Solution:
Let ‘x’ be the amount of salt present


Since the tank initially contains only pure water, the initial amount of salt is zero. i.e., x (0) = 0

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.8 Additional Problems

Question 1.
In a certain chemical reaction the rate of conversion of a substance at time t is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour, 60 grams remain and at the end of 4 hours 21 grams. How many grams of the substance was there initially?
Solution:
Let A be the substance at time t

Initially i.e., when t = 0, A = c = 85.15 gms (app.)
Hence initially there was 85.15 gms (approximately) of the substance.

Question 2.
The temperature T of a cooling object drops at a rate proportional to the difference T – S, where S is constant temperature of surrounding medium. If initially T = 150° C, find the temperature of the cooling object at any time t.
Solution:
Let T be the temperature of the cooling object at any time t

when t = 0, T = 150 ⇒ 150 = S + c ⇒ c = 150 – S
∴ The temperature of the cooling object at any time T = S + (150 – S)e^kt
Note:
Since k is negative, as t increases T decreases.
It is a decay problem. Instead of k one may take – k where k > 0. Then the answer is
T = S + (150 – S) e^-kt.
Again, as t increases T decreases.

Question 3.
The sum of Rs. 1000 is compounded continuously, the nominal rate of interest being four per cent per annum. In how many years, will the amount be twice the original principal? [log_e 2 = 0.6931]
Solution:
Rate of interest = 4 %

Question 4.
A cup of coffee at temperature 100° C is placed in a room whose temperature is 15° C and it cools to 60° C in 5 minutes. Find its temperature after a further interval of 5 minutes.
Solution:

Question 5.
The rate at which the population of a city increases at any time is proportional to the population at that time. If there were 1,30,000 people in the city in 1960 and 1,60,000 in 1990, what population may be anticipated in 2020? [log_e\(\left(\frac{16}{3}\right)\) = 0.2070; e^-0.42 = 1.52]
Solution:

∴ P = 1,30,000; e^0.007 is the population at any time ‘t’
In the year 2020, (i.e.,) when t = 60
x = 130000 × e^0.42 = 1,30,000 × 1.52 = 1,97,600
So, the anticipated population in 2020 will be 1, 97, 600

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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:

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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:

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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

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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: