TS Inter 1st Year Maths 1B Errors and Approximations Important Questions

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Errors and Approximations Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1B Errors and Approximations Important Questions

Question 1.
If y = x2 + 3x + 6, find ∆y and dy when x = 10, ∆x = 0.01. [Mar. ’15 (TS), ’14, ’11, ’05; May ’15 (AP)]
Solution:
Let y = f(x) = x2 + 3x + 6, x = 10, ∆x = 0.01
∆y = f(x + ∆x) – f(x)
= (x + ∆x)2 + 3(x + ∆x) + 6 – (x2 + 3x + 6)
= x2 + (∆x)2 + 2 . x . ∆x + 3x + 3(∆x) + 6 – x2 – 3x – 6
= (∆x)2 + 2 . x . ∆x + 3 . ∆x
= (2x + 3) ∆x + (∆x)2
= (2.10 + 3) (0.01) + (0. 01)2
= (23)(0.01) + 0.0001
= 0.23 + 0.0001
= 0.2301
dy = \(\frac{d y}{d x}\) × ∆x
= (2x + 3) ∆x
= (2.10 + 3)(0.01)
= 23(0.01)
= 0.23

TS Inter First Year Maths 1B Errors and Approximations Important Questions

Question 2.
Find ∆y and dy for the function y = ex + x when x = 5, ∆x = 0.02. [May ’13]
Solution:
Let y = f(x) = ex + x, x = 5, ∆x = 0.02
∆y = f(x + ∆x) – f(x)
= e(x+∆x) + (x + ∆x) – (ex + x)
= ex+∆x + x + ∆x – ex – x
= ex+∆x + ∆x – ex
= e5+0.02 + 0.02 – e5
dy = \(\frac{d y}{d x}\) × ∆x
= (ex + 1) × ∆x
= (e5 + 1) 0.02

Question 3.
Find the approximate value of √82. [Mar. ’13; May ’09]
Solution:
f(x) = √x, x = 81, ∆x = 1
∆f = df = \(\frac{d f}{d x}\) . ∆x
= \(\frac{1}{2 \sqrt{81}} \cdot 1\)
= \(\frac{1}{18}\)
= 0.05556
Now, √82 = f(x + ∆x) = ∆f + f(x)
= 0.05556 + 9
= 9.05556

Question 4.
Find the approximate value of \(\sqrt[3]{65}\).
Solution:
Let f(x) = \(\sqrt[3]{65}\), x = 64, ∆x = 1
TS Inter First Year Maths 1B Errors and Approximations Important Questions Q4
f(x) = \(\sqrt[3]{x}\) = \(\sqrt[3]{64\) = 4
Now, \(\sqrt[3]{65}\) = f(x + ∆x) = ∆f + f(x)
= 4 + 0.02083
= 4.02083

TS Inter First Year Maths 1B Errors and Approximations Important Questions

Question 5.
If the increase in the side of a square is 2% then find the approximate percentage of increase in its area. [Mar. ’18 (TS); Mar. ’12, ’08; May ’05]
Solution:
TS Inter First Year Maths 1B Errors and Approximations Important Questions Q5
Let x be the side and A be the area of a square.
Given, that % error in x = 2
\(\frac{\Delta \mathrm{x}}{\mathrm{x}}\) × 100 = 2
The area of a square, A = x2
TS Inter First Year Maths 1B Errors and Approximations Important Questions Q5.1
∴ % error in area = 4.

Question 6.
The diameter of a sphere is measured to be 40 cm. If an error of 0.02 cm is made in it, then find approximate errors in the volume and surface area of the sphere. [Mar. ’13(old), ’09; May ’03]
Solution:
TS Inter First Year Maths 1B Errors and Approximations Important Questions Q6
Let r, d, s, v be the radius, diameter, surface area, and volume of a sphere.
Given, diameter, d = 40 cm,
radius, r = 20 cm
error in diameter, ∆d = 0.02 cm
error in radius, ∆r = 0.01 cm
(i) Volume of a sphere, V = \(\frac{4}{3}\)πr3
error in V = ∆V = \(\frac{\mathrm{dV}}{\mathrm{dr}}\) . ∆r
= \(\frac{4}{3}\)π(3r2) . ∆r
= 4πr2 . ∆r
= 4π(20)2 (0.01)
= 16π cu. cm
(ii) Surface area of a sphere, s = 4πr2
error in surface area of a sphere ∆s = \(\frac{\mathrm{ds}}{\mathrm{dr}}\) ∆dr
= 4π(2r) . ∆r
= 8π(20)(0.01)
= 1.6π sq. cm
∴ Approximate error in the volume of sphere = 16π cu. cm.
∴ Approximate error in the area of sphere = 1.6π sq. cm.

Some More Maths 1B Errors and Approximations Important Questions

Question 7.
If y = 5x2 + 6x + 6, find ∆y and dy when x = 2, ∆x = 0.001.
Solution:
Let y = f(x) = 5x2 + 6x + 6, x = 2, ∆x = 0.001
∆y = f(x + ∆x) – f(x)
= 5(x + ∆x)2 + 6(x + ∆x) + 6 – (5x2 + 6x + 6)
= 5(x2 + (∆x)2 + 2 . x . ∆x) + 6x + 6(∆x) + 6 – 5x2 – 6x – 6
= 5x2 + 5(∆x)2 + 10 . x . ∆x + 6(∆x) – 5x2
= [5(∆x) + 10x + 6)] ∆x
= [5(0.001) + 10(2) + 6] (0.001)
= [0.005 + 20 + 6] 0.001
= 0.026005
dy = \(\frac{\mathrm{dy}}{\mathrm{dx}}\) × ∆x
= (5(2x) + 6) ∆x
= (10x + 6) ∆x
= (10 . 2 + 6) 0.001
= (26) 0.001
= 0.026

TS Inter First Year Maths 1B Errors and Approximations Important Questions

Question 8.
Find the approximate value of \(\sqrt{\mathbf{25.001}}\).
Solution:
Let f(x) = √x, x = 25, ∆x = 0.001
TS Inter First Year Maths 1B Errors and Approximations Important Questions DTP Q2
f(x) = √x = √25 = 5
Now, \(\sqrt{\mathbf{25.001}}\) = f(x + ∆x)
= ∆f + f(x)
= 0.0001 + 5
= 5.0001

Question 9.
If the increase in the side of a square is 4% then find the approximate percentage of increase in the area of the square.
Solution:
Let x be the side and A be the area of a square.
Given, that % error in the side = 4
∴ \(\frac{\Delta x}{x}\) × 100 = 4
Area of a square, A = x2
∆A = 2x . ∆x (∵ ∆A = \(\frac{\mathrm{dA}}{\mathrm{dx}}\) . ∆x)
Now, % error in area = \(\frac{\Delta \mathrm{A}}{\mathrm{A}}\) × 100
= \(\frac{2 \mathrm{x} \cdot \Delta \mathrm{x}}{\mathrm{x}^2}\) × 100
= \(\frac{2 \cdot \Delta x}{x}\) × 100
= 2 × 4
= 8
∴ % error in area = 8

Question 10.
Find dy and ∆y of y = f(x) = x2 + x, at x = 10 when ∆x = 0.1. [Mar. ’16 (TS), ’15 (AP); May ’15 (TS); Mar. ’17 (AP & TS)]
Solution:
Given, y = f(x) = x2 + x, x = 10, ∆x = 0.1
∆y = f(x + ∆x) – f(x)
= (x + ∆x)2 + x + ∆x – (x2 + x)
= x2 + (∆x)2 + 2 . x . ∆x + x + ∆x – x2 – x
= (∆x)2 + 2 . x . ∆x + ∆x
= (2x + 1) ∆x + (∆x)2
= (2 . 10 + 1) 0.1 + (0.1)2
= (21) (0.1) + 0.01
= 2.1 + 0.01
= 2.11
dy = \(\frac{\mathrm{dy}}{\mathrm{dx}}\) × ∆x
= (2x + 1) ∆x
= (2 . 10 + 1) 0.1
= (21) (0.1)
= 2.1

TS Inter First Year Maths 1B Errors and Approximations Important Questions

Question 11.
Find ∆y and dy of y = \(\frac{1}{x+2}\), x = 8 and ∆x = 0.02.
Solution:
Let y = f(x) = \(\frac{1}{x+2}\), x = 8, ∆x = 0.02
∆y = f(x + ∆x) – f(x)
TS Inter First Year Maths 1B Errors and Approximations Important Questions Some More Q2
TS Inter First Year Maths 1B Errors and Approximations Important Questions Some More Q2.1

Question 12.
Find ∆y and dy of y = cos x, x = 60°, and ∆x = 1°. [Mar. ’19 (TS)]
Solution:
Let y = f(x) = cos x, x = 60°,
∆x = 1° = 1(0.01745) = 0.01745
∆y = f(x + ∆x) – f(x)
= cos(x + ∆x) – cos x
= cos (60° + 1°) – cos 60°
= cos 61° – cos 60°
= 0.4848 – \(\frac{1}{2}\)
= 0.4848 – 0.5
= -0.0152
dy = \(\frac{\mathrm{dy}}{\mathrm{dx}}\) . ∆x
= -sin x . ∆x
= -sin 60° . (0.01745)
= \(\frac{-\sqrt{3}}{2}\) . (0.01745)
= -(0.8660)(0.01745)
= -0.015

Question 13.
Find the approximate value of \(\sqrt[3]{999}\). [Mar. ’19 (AP)]
Solution:
TS Inter First Year Maths 1B Errors and Approximations Important Questions Some More Q4

Question 14.
Find the approximate value of \(\sqrt[3]{7.8}\).
Solution:
TS Inter First Year Maths 1B Errors and Approximations Important Questions Some More Q5

Question 15.
Find the approximate value of sin 62°.
Solution:
Let y = f(x) = sin x, x = 60°,
∆x = 2° = 2(0.01745) = 0.03490
∆f = df = \(\frac{d f}{d x}\) . ∆x
= cos x . ∆x
= cos 60° (0.03490)
= \(\frac{0.03490}{2}\)
= 0.01745
f(x) = sin x = sin 60° = 0.8660
Now, sin 62° = f(x + ∆x) = f(x) + ∆f
= 0.8660 + 0.01745
= 0.88345

TS Inter First Year Maths 1B Errors and Approximations Important Questions

Question 16.
Find the approximate value of cos(60°5′).
Solution:
Let y = f(x) = cos x, x = 60°,
∆x = 5′ = \(\left(\frac{5}{60}\right)^{\circ}=\left(\frac{1}{12}\right)^{\circ}\)
= \(\frac{1}{12}\)(0.01745)
= 0.001454
∆f = df = \(\frac{\mathrm{df}}{\mathrm{dx}}\) . ∆x
= -sin x . ∆x
= -sin 60° . (0.001454)
= -0.8660 . (0.001454)
= -0.001259
f(x) = cos x = cos 60° = 0.5
Now, cos(60° 5′) = f(x + ∆x) = f(x) + ∆x
= 0.5 – 0.001259
= 0.498741

Question 17.
Find the approximate value of \(\sqrt[4]{17}\).
Solution:
TS Inter First Year Maths 1B Errors and Approximations Important Questions Some More Q8

Question 18.
The radius of a sphere is measured as 14 cm. Later it was found that there is an error of 0.02 cm in measuring the radius. Find the approximate error in the surface area of the sphere.
Solution:
Let r be the radius and s be the surface area of the sphere.
Given, radius, r = 14 cm
error in radius, ∆r = 0.02 cm
the surface area of a sphere, s = 4πr2
error in surface area of sphere = ∆s
= \(\frac{\mathrm{ds}}{\mathrm{dr}}\) . ∆r
= 4π(2r) . ∆r
= 8π(14)(0.02)
= 2.24π
= 2.24(3.14)
= 7.03356 or 7.04 sq.cm.

Question 19.
The side of a square is increased from 3 cm to 3.01 cm. Find the approximate increase in the area of the square.
Solution:
Let x be the side and A be the area of a square.
Given that x = 3, ∆x = 0.01
Area of a square, A = x2
error in area = ∆A = \(\frac{\mathrm{dA}}{\mathrm{dx}}\) . ∆x
= 2x . ∆x
= 2.3(0.01)
= 6(0.01)
= 0.06 sq. cm

TS Inter First Year Maths 1B Errors and Approximations Important Questions

Question 20.
If the radius of a sphere is increased from 7 cm to 7.02 cm then find the approximate increase in the volume of the sphere.
Solution:
Ler r be the radius of a sphere and V be its volume.
Given that, radius r = 7 cm, ∆r = 0.02 cm
Volume of the sphere, V = \(\frac{4}{3} \pi r^3\)
error in volume = ∆V = \(\frac{\mathrm{dV}}{\mathrm{dr}}\) . ∆r
TS Inter First Year Maths 1B Errors and Approximations Important Questions Some More Q11

Question 21.
If y = f(x) = kxn then show that the approximate relative error (or increase) in y is n times the relative error (or increase) in x where n and k are constants.
Solution:
Given that y = f(x) = k . xn
error in y = ∆y = \(\frac{d y}{d x}\) . ∆x
= k . n . xn-1 . ∆x
relative error in y = \(\frac{\Delta \mathrm{y}}{\mathrm{y}}\)
TS Inter First Year Maths 1B Errors and Approximations Important Questions Some More Q12
∴ The relative error in y is n times the relative error in x.

Question 22.
The time t, of a complete oscillation of a simple pendulum of length l, is given by t = \(2 \pi \sqrt{\frac{l}{g}}\), where g is gravitational constant. Find the approximate percentage of error in ‘t’ when the percentage of error in l is 1%.
Solution:
TS Inter First Year Maths 1B Errors and Approximations Important Questions Some More Q13

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