JEE Challenger
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MathHardAdv 2015 • Paper 2

Evaluation of Definite Integrals Involving Derivatives of a Function

Comprehension Passage

Consider a thrice differentiable function F:R→RF : \mathbb{R} \to \mathbb{R}. Assume that F(1)=0F(1) = 0, F(3)=−4F(3) = -4, and F′(x)<0F'(x) < 0 for every x∈(1/2,3)x \in (1/2, 3). Define f(x)=xF(x)f(x) = xF(x) for each x∈Rx \in \mathbb{R}.

Given that ∫13x2F′(x) dx=−12\int_1^3 x^2 F'(x)\,dx = -12 and ∫13x3F′′(x) dx=40\int_1^3 x^3 F''(x)\,dx = 40, which of the following expressions is(are) correct?

Options

A

9f′(3)+f′(1)−32=09 f'(3) + f'(1) - 32 = 0

B

∫13f(x) dx=12\int_1^3 f(x)\,dx = 12

C

9f′(3)−f′(1)+32=09 f'(3) - f'(1) + 32 = 0

Correct
D

∫13f(x) dx=−12\int_1^3 f(x)\,dx = -12

Correct

Topics & Concepts

IntegralsDefinite IntegrationProperties of Definite IntegralsDifferentiation

Step-by-Step Solution

To find the correct expressions, we evaluate the given integrals using integration by parts.

Step 1: Evaluating ∫13f(x) dx\int_1^3 f(x) \, dx

We are given the first integral: ∫13x2F′(x) dx=−12\int_1^3 x^2 F'(x) \, dx = -12

Applying integration by parts by choosing u=x2u = x^2 and dv=F′(x) dxdv = F'(x) \, dx: ∫13x2F′(x) dx=[x2F(x)]13−∫132xF(x) dx\int_1^3 x^2 F'(x) \, dx = \left[ x^2 F(x) \right]_1^3 - \int_1^3 2x F(x) \, dx

Using the given boundary values F(1)=0F(1) = 0 and F(3)=−4F(3) = -4, and the definition f(x)=xF(x)f(x) = xF(x): [x2F(x)]13=32F(3)−12F(1)=9(−4)−1(0)=−36\left[ x^2 F(x) \right]_1^3 = 3^2 F(3) - 1^2 F(1) = 9(-4) - 1(0) = -36 ∫132xF(x) dx=2∫13f(x) dx\int_1^3 2x F(x) \, dx = 2 \int_1^3 f(x) \, dx

Substituting these into the equation: −12=−36−2∫13f(x) dx-12 = -36 - 2 \int_1^3 f(x) \, dx 2∫13f(x) dx=−36+12=−242 \int_1^3 f(x) \, dx = -36 + 12 = -24 ∫13f(x) dx=−12\int_1^3 f(x) \, dx = -12

This confirms that Option (D) is correct and Option (B) is incorrect.

Step 2: Evaluating the relation between f′(3)f'(3) and f′(1)f'(1)

We are given the second integral: ∫13x3F′′(x) dx=40\int_1^3 x^3 F''(x) \, dx = 40

Applying integration by parts by choosing u=x3u = x^3 and dv=F′′(x) dxdv = F''(x) \, dx: ∫13x3F′′(x) dx=[x3F′(x)]13−∫133x2F′(x) dx\int_1^3 x^3 F''(x) \, dx = \left[ x^3 F'(x) \right]_1^3 - \int_1^3 3x^2 F'(x) \, dx

Substituting the known values ∫13x3F′′(x) dx=40\int_1^3 x^3 F''(x) \, dx = 40 and ∫13x2F′(x) dx=−12\int_1^3 x^2 F'(x) \, dx = -12: 40=(33F′(3)−13F′(1))−3(−12)40 = \left( 3^3 F'(3) - 1^3 F'(1) \right) - 3(-12) 40=27F′(3)−F′(1)+3640 = 27 F'(3) - F'(1) + 36 27F′(3)−F′(1)=4— (1)27 F'(3) - F'(1) = 4 \quad \text{--- (1)}

Now, differentiating f(x)=xF(x)f(x) = x F(x) using the product rule: f′(x)=F(x)+xF′(x)f'(x) = F(x) + x F'(x)

  • For x=1x = 1: f′(1)=F(1)+1⋅F′(1)=0+F′(1)=F′(1)  ⟹  F′(1)=f′(1)f'(1) = F(1) + 1 \cdot F'(1) = 0 + F'(1) = F'(1) \implies F'(1) = f'(1)

  • For x=3x = 3: f′(3)=F(3)+3F′(3)=−4+3F′(3)f'(3) = F(3) + 3 F'(3) = -4 + 3 F'(3) 3F′(3)=f′(3)+4  ⟹  27F′(3)=9(f′(3)+4)=9f′(3)+363 F'(3) = f'(3) + 4 \implies 27 F'(3) = 9\big(f'(3) + 4\big) = 9f'(3) + 36

Substitute F′(1)F'(1) and 27F′(3)27 F'(3) back into equation (1): (9f′(3)+36)−f′(1)=4\big(9 f'(3) + 36\big) - f'(1) = 4 9f′(3)−f′(1)+32=09 f'(3) - f'(1) + 32 = 0

This confirms that Option (C) is correct and Option (A) is incorrect.

Conclusion:

The correct options are C and D.

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