# Mathematics Questions and Answers – Application of Derivative for Error Determination

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This set of Mathematics Objective Questions and Answers for Class 12 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

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

Explanation: Let, y = log10x
Then f’(x) = d/dx[log10x]
= d/dx[logex * 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

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
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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

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

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

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

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

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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