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Yukti Vault

Definite Integration - JEE Mathematics question (JEE Advanced 2015 Paper 2)

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

Let $F:\mathbb{R}\to\mathbb{R}$ be a thrice differentiable function. Suppose that $F(1)=0$, $F(3)=-4$ and $F'(x)<0$ for all $x\in(1/2,3)$. Let $f(x)=xF(x)$ for all $x\in\mathbb{R}$.
JEE Advanced 2015 Paper 2Question 1
If $\int_1^3x^2F'(x)\,dx=-12$ and $\int_1^3x^3F''(x)\,dx=40$, then the correct expression(s) is(are)
Definite Integration
Options for question 1, one or more may be correct

More than one option may be correct.

Solution

Concept. In the original problem $F(1)=0$, $F(3)=-4$ and $f(x)=xF(x)$. Integrate by parts and use these. Formulas used. (1) $\int_a^bu\,dv=\big[uv\big]_a^b-\int_a^bv\,du$(2) $f(x)=xF(x)\Rightarrow f'(x)=F(x)+xF'(x)$Step 1 - First condition. By (1), $$\int_1^3x^2F'\,dx=\Big[x^2F\Big]_1^3-2\int_1^3xF\,dx=-36-2\int_1^3f\,dx=-12.$$ So $\int_1^3f(x)\,dx=-12$. (D) is true, (B) is false. Step 2 - Second condition. By (1), $$\int_1^3x^3F''\,dx=\Big[x^3F'\Big]_1^3-3\int_1^3x^2F'\,dx=27F'(3)-F'(1)+36=40,$$ so $27F'(3)-F'(1)=4$. Step 3 - Use (2). $f'(3)=-4+3F'(3)$ and $f'(1)=F'(1)$, so $$9f'(3)-f'(1)=-36+27F'(3)-F'(1)=-32.$$ (C) is true, (A) is false. Answer: (C), (D)

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