JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 1)
If $$\int\limits_0^x \, $$f(t) dt = x2 + $$\int\limits_x^1 \, $$ t2f(t) dt then f '$$\left( {{1 \over 2}} \right)$$ is -
$${{18} \over {25}}$$
$${{6} \over {25}}$$
$${{24} \over {25}}$$
$${{4} \over {5}}$$
Explanation
$$\int\limits_0^x \, $$f(t) dt = x2 + $$\int\limits_x^1 \, $$ t2f(t) dt f '$$\left( {{1 \over 2}} \right)$$ = ?
Differentiate w.r.t. 'x'
f(x) = 2x + 0 $$-$$ x2 f(x)
f(x) = $${{2x} \over {1 + {x^2}}}$$ $$ \Rightarrow $$ f '(x) = $${{\left( {1 + {x^2}} \right)2 - 2x\left( {2x} \right)} \over {{{\left( {1 + {x^2}} \right)}^2}}}$$
f '(x) = $${{2{x^2} - 4{x^2} + 2} \over {{{\left( {1 + {x^2}} \right)}^2}}}$$
f '$$\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}$$
Differentiate w.r.t. 'x'
f(x) = 2x + 0 $$-$$ x2 f(x)
f(x) = $${{2x} \over {1 + {x^2}}}$$ $$ \Rightarrow $$ f '(x) = $${{\left( {1 + {x^2}} \right)2 - 2x\left( {2x} \right)} \over {{{\left( {1 + {x^2}} \right)}^2}}}$$
f '(x) = $${{2{x^2} - 4{x^2} + 2} \over {{{\left( {1 + {x^2}} \right)}^2}}}$$
f '$$\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}$$
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