JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 10)
If $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right),$$
$$f\left( 0 \right) = 0,$$ then the value of $$f(1)$$ is :
$$f\left( 0 \right) = 0,$$ then the value of $$f(1)$$ is :
$$ - $$ $${1 \over 2}$$
$$ - $$ $${1 \over 4}$$
$${1 \over 2}$$
$${1 \over 4}$$
Explanation
$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx$$
$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$
$$f\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$
Let $${x^{ - 5}} + {x^{ - 7}} + 2 = t$$
$$\left( { - 5{x^{ - 6}} - 7{x^{ - 8}}} \right)dx = dt$$
$$\left( {5{x^{ - 6}} + 7{x^{ - 8}}} \right)dx = - dt$$
$$f(x) = \int {{{ - dt} \over {{t^2}}}} = {1 \over t} + c$$
$$f\left( x \right) = {1 \over {{x^{ - 5}} + {x^{ - 7}} + 2}} + c$$
$$f\left( x \right) = {{{x^7}} \over {{x^2} + 1 + 2{x^7}}} + c$$
$$f\left( 0 \right) = 0$$
$$ \therefore $$ $$c = 0$$
$$f\left( x \right) = {{{x^7}} \over {\left( {{x^2} + 1 + 2{x^7}} \right)}}$$
$$f(1) = {1 \over {1 + 1 + 2}} = {1 \over 4}$$
$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$
$$f\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$
Let $${x^{ - 5}} + {x^{ - 7}} + 2 = t$$
$$\left( { - 5{x^{ - 6}} - 7{x^{ - 8}}} \right)dx = dt$$
$$\left( {5{x^{ - 6}} + 7{x^{ - 8}}} \right)dx = - dt$$
$$f(x) = \int {{{ - dt} \over {{t^2}}}} = {1 \over t} + c$$
$$f\left( x \right) = {1 \over {{x^{ - 5}} + {x^{ - 7}} + 2}} + c$$
$$f\left( x \right) = {{{x^7}} \over {{x^2} + 1 + 2{x^7}}} + c$$
$$f\left( 0 \right) = 0$$
$$ \therefore $$ $$c = 0$$
$$f\left( x \right) = {{{x^7}} \over {\left( {{x^2} + 1 + 2{x^7}} \right)}}$$
$$f(1) = {1 \over {1 + 1 + 2}} = {1 \over 4}$$
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