QJEEAdv15Ma0fb8: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Ma0fb8 Solution: Let be a thrice differentiable function. Suppose that and for…

Question

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Let

F:R \to R

be a thrice differentiable function. Suppose that

F\left( 1 \right) = 0,F\left( 3 \right) = - 4

and

F\left( x \right) &lt; 0

for all

x \in \left( {{1 \over 2},3} \right).

Let

f\left( x \right) = xF\left( x \right)

for all

x \in R.

If $\int_1^3 {{x^2}F'\left( x \right)dx = - 12}$ and $\int_1^3 {{x^3}F''\left( x \right)dx = 40,}$ then the correct expression(s) is (are)

Accepted Answer

Given, $\int_1^3 {{x^2}F'(x)dx = - 12}$

$\Rightarrow [{x^2}F(x)]_1^3 - \int_1^3 {2x\,.\,F(x)dx = - 12}$

$\Rightarrow 9F(3) - F(1) - 2\int_1^3 {f(x)dx = - 12}$

[

\because

xF(x) = f(x)

given]

$\Rightarrow - 36 - 0 - 2\int_1^3 {f(x)dx = - 12}$

$\therefore$ $\int_1^3 {f(x)dx = - 12}$

and $\int_1^3 {{x^3}F''(x)dx = 40}$

$\Rightarrow [{x^3}F'(x)]_1^3 - \int_1^3 {3{x^2}F'(x)dx = 40}$

$\Rightarrow [{x^2}(xF'(x)]_1^3 - 3 \times ( - 12) = 40$

$\Rightarrow \{ {x^2}\,.\,[f'(x) - F(x)]\} _1^3 = 4$

$\Rightarrow 9[f'(3) - F(3)] - [f'(1) - F(1)] = 4$

$\Rightarrow 9[f'(3) + 4] - [f'(1) - 0] = 4$

$\Rightarrow 9f'(3) - f'(1) = - 32$

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