JEE Advance - Mathematics (2015 - Paper 2 Offline - No. 17)
$$F\left( 1 \right) = 0,F\left( 3 \right) = - 4$$ and $$F\left( x \right) < 0$$ 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)
Explanation
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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