Application of Derivative for Error Determination - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Application of Derivative for Error Determination”.

1. What will be the increment of the differentiable function f(x) = 2x² – 3x + 2 when x changes from 3.02 to 3?
a) 0.18
b) 0.018
c) 0.16
d) 0.016
View Answer

Answer: a
Explanation: Let, y = f(x) = 2x² – 3x + 2
So, f’(x) = 4x – 3
Clearly, as x changes from 3 to 3.02, the increment of x is:
3.02 – 3 = 0.02
So, increment in f(x) = f(3.02) – f(3)
= 2(3.02)² – 3(3.02) + 2 – 2(3)² – 3(3) + 2
= 0.1808
Now, differential of y is dy
We get, f’(3) * 0.02 = (4.3 – 3)*0.02
= 0.18

2. If log₁₀3 = 0.4771 and log₁₀e = 0.4343, then what is the value of log₁₀30.5?
a) 1.43
b) 1.5
c) 1.484
d) 1.4
View Answer

Answer: c
Explanation: Let, y = log₁₀x
Then f’(x) = d/dx[log₁₀x]
= d/dx[log_ex * log₁₀e]
= 1/x(log₁₀e)
Now, f(x + δx) = f(x) + f’(x) δx
Putting x = 30 and δx = 0.5 in the above equation we get,
f(30 + 0.5) = f(30) + f’(30) δx
=> f(30.5) = log₁₀30 + (1/30) log₁₀e * 0.5
Putting the values we get,
log₁₀30.5 = 1.4843 = 1.484

3. If 1° = 0.01745 then, what is the value of cos62°?
a) 0.4588
b) 0.4788
c) 0.4688
d) 0.3688
View Answer

Answer: c
Explanation: Let, y = f(x) = cosx
And, x = 60° = π/3, δx = 2° = 2 * 0.01745 = 0.03490
Since f(x) = cosx, hence
f’(x) = -sinx
Now, we have f(x + δx) = f(x) + f’(x) δx
Or, f(60° + 2°) = f(60°) + f’(60°) * 0.0349
Or, f(62°) = cos(60°) – sin60° * 0.0349
Or, cos62° = 0.5 – 0.866 * 0.0349 = 0.4688

4. The length of a side of a cube is 10cm; if an error of 0.05cm is made in measuring the side, then what is the value of approximate error in calculating its volume?
a) 16 cu cm
b) 15 cu cm
c) 15.5 cu cm
d) 14 cu cm
View Answer

Answer: b
Explanation: Let, v cubic cm be the volume of the cube of the cube of side x cm.
Then, v = x³
Thus, dv/dx = d/dx(x³) = 3x²
Now, dv = dv/dx * δx = 3x² * δx
By problem, x = 10 and δx = 0.05
Therefore, approximate error = dv = 3 * 10² * 0.05 = 15 cu cm.

5. The length of a side of a cube is 10cm; if an error of 0.05cm is made in measuring the side, then what is the value of relative error in calculating its volume?
a) 0.0016
b) 0.014
c) 0.015
d) 0.0015
View Answer

Answer: c
Explanation: Let, v cubic cm be the volume of the cube of the cube of side x cm.
Then, v = x³
Thus, dv/dx = d/dx(x³) = 3x²
Now, dv = dv/dx * δx = 3x² * δx
By problem, x = 10 and δx = 0.05
Therefore, approximate error = dv = 3 * 10² * 0.05 = 15 cc cm.
Hence, relative error = dv/v = 15/10³ = 0.015

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6. The length of a side of a cube is 10cm; if an error of 0.05cm is made in measuring the side, then what is the value of percentage error in calculating its volume?
a) 1.65
b) 1.45
c) 0.015
d) 1.5
View Answer

Answer: d
Explanation: Let, v cubic cm be the volume of the cube of the cube of side x cm.
Then, v = x³
Thus, dv/dx = d/dx(x³) = 3x²
Now, dv = dv/dx * δx = 3x² * δx
By problem, x = 10 and δx = 0.05
Therefore, approximate error = dv = 3 * 10² * 0.05 = 15 cc cm.
Hence, relative error = dv/v = 15/10³ = 0.015
Thus, percentage error = relative error * 100 = 0.015 * 100 = 1.5

7. What will be the estimate error made in calculating the area of the triangle ABC in which the sides a and b are measured accurately as 25 cm and 16 cm, while the angle C is measured as 60° but (1/2)° in error?
a) 55/63 sq cm
b) 53/63 sq cm
c) 55/67 sq cm
d) Data not sufficient
View Answer

Answer: a
Explanation: Let, y be the area of the triangle ABC.
Then, y = ½(ab)(sinC) where a and b are constants.
Now, dy = dy/dC * dC = ½(ab)(sinC) dC
Now, by problem, a = 25, b = 16, C = 60° and δC = (1/2)° = π/180 * ½ radian
Therefore, dy = 1/2 * 25 * 16 * cos60° * π/180 * ½ sq cm
=> dy = 25 * 8 * ½ * 22/7 * 1/180 * ½ sq cm
= 55/63 sq cm

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