JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 3)
Let f : R $$ \to $$ R be defined as f(x) = e$$-$$xsinx. If F : [0, 1] $$ \to $$ R is a differentiable function with that F(x) = $$\int_0^x {f(t)dt} $$, then the value of $$\int_0^1 {(F'(x) + f(x)){e^x}dx} $$ lies in the interval
$$\left[ {{{331} \over {360}},{{334} \over {360}}} \right]$$
$$\left[ {{{330} \over {360}},{{331} \over {360}}} \right]$$
$$\left[ {{{335} \over {360}},{{336} \over {360}}} \right]$$
$$\left[ {{{327} \over {360}},{{329} \over {360}}} \right]$$
Explanation
F(x) = $$\int_0^x {f(t)dt} $$
$$ \Rightarrow $$ F'(x) = f(x) by Leibnitz theorem
I = $$\int\limits_0^1 {(F'(x) + f(x)){e^x}dx = \int\limits_0^1 {2f(x){e^x}dx} } $$
$$I = \int\limits_0^1 {2\sin x\,dx} $$
$$I = 2(1 - \cos 1)$$
$$ = 2\left\{ {1 - \left( {1 - {{{1^2}} \over {2!}} + {{{1^4}} \over {4!}} - {1 \over {6!}} + ...} \right)} \right\}$$
$$ = 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}}} \right)} \right\} < 2(1 - \cos 1) < 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}} - {1 \over {720}}} \right)} \right\}$$
$${{330} \over {360}} < 2(1 - \cos 1) < {{331} \over {360}}$$
$${{330} \over {360}} < 1 < {{331} \over {360}}$$
$$ \Rightarrow $$ F'(x) = f(x) by Leibnitz theorem
I = $$\int\limits_0^1 {(F'(x) + f(x)){e^x}dx = \int\limits_0^1 {2f(x){e^x}dx} } $$
$$I = \int\limits_0^1 {2\sin x\,dx} $$
$$I = 2(1 - \cos 1)$$
$$ = 2\left\{ {1 - \left( {1 - {{{1^2}} \over {2!}} + {{{1^4}} \over {4!}} - {1 \over {6!}} + ...} \right)} \right\}$$
$$ = 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}}} \right)} \right\} < 2(1 - \cos 1) < 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}} - {1 \over {720}}} \right)} \right\}$$
$${{330} \over {360}} < 2(1 - \cos 1) < {{331} \over {360}}$$
$${{330} \over {360}} < 1 < {{331} \over {360}}$$
Comments (0)
