JEE Main 2023 (Online) 24th January Evening ShiftQuestion 1
Let $f$ be a differentiable function defined on $\left[0,\frac\pi2\right]$ such that $f(x)>0$ and $f(x)+\int_0^{x}f(t)\sqrt{1-\left(\log_e f(t)\right)^{2}}\,dt=e$ for all $x\in\left[0,\frac\pi2\right]$. Then $\left\{6\log_e\left(f\left(\frac\pi6\right)\right)\right\}^{2}$ is equal to
Differential Equation
Answer$27$
Solution
Concept. Differentiate; with $u=\ln f$ the equation is $u'=-\sqrt{1-u^2}$. Formulas used. (1) $\int\dfrac{du}{\sqrt{1-u^2}}=\sin^{-1}u$Step 1 - Initial value. At $x=0$: $f(0)=e$, so $u(0)=1$. Step 2 - Differentiate. $f'+f\sqrt{1-(\ln f)^2}=0$. Divide by $f$: $u'=-\sqrt{1-u^2}$. Step 3 - Solve. By (1), $\sin^{-1}u=c-x$, and $u(0)=1$ gives $c=\frac\pi2$. So $u=\sin\left(\frac\pi2-x\right)=\cos x$. Step 4 - Value. $6\ln f\left(\frac\pi6\right)=6\cos\frac\pi6=3\sqrt3$, and its square is $27$. Answer: 27