Q76 / 225 · JEE Main 2003 · Mathematics

Let $\frac{d}{d x} F(x)=\left(\frac{e^{\text {sinx }}}{x}\right) x>0$. If $\int_1^4 \frac{3}{x} e^{\text {sin }^3} d x=F(k)-F(1)$ then one of the possible values of $k$, is

  • Correct answer
    A
    64
  • B
    15
  • C
    16
  • D
    63

Solution

$$ \frac{d}{d x} F(x)=\frac{e^{\sin x}}{x} \text { or } \int_1^4 \frac{3}{x} e^{\sin x^3} d x=\int_1^4 \frac{3 x^2}{x^3} e^{\sin x^3} d x $$ Let $\mathrm{x}^3=\mathrm{t}, 3 \mathrm{x}^2 \mathrm{dx}=\mathrm{dt}$ when $x=1, t=1 \& x-4, t=64$ $$ \begin{aligned} & \mathrm{F}(\mathrm{t})=\int_1^{64} \frac{\mathrm{e}^{\sin t}}{\mathrm{t}} \mathrm{dt}=\int_1^{64} \mathrm{~F}(\mathrm{t}) \mathrm{dt}=\mathrm{F}(64)-\mathrm{F}(1) \\ & \mathrm{K}=64 \end{aligned} $$

Hence, the correct answer is option: 'A'.

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