JEE Main 2024 (Online) 1st February Evening ShiftQuestion 1
If $y=\frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x\sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3\cos^{2}x-5\right)\cos^{3}x$, then $96y'\left(\frac{\pi}{6}\right)$ is equal to $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$105$
Solution
Concept. The first fraction simplifies to $x-1$; the second term differentiates to a neat product. Formulas used. (1) $s^3-1=(s-1)(s^2+s+1)$(2) $\dfrac{d}{dx}\cos^nx=-n\cos^{n-1}x\sin x$Step 1 - First term. Put $s=\sqrt x$. By (1), $$\frac{(s+1)(s^4-s)}{s^3+s^2+s}=\frac{(s+1)\,s\,(s-1)(s^2+s+1)}{s\,(s^2+s+1)}=s^2-1=x-1.$$ Step 2 - Second term. It is $\frac1{15}\left(3\cos^5x-5\cos^3x\right)$. By (2), its derivative is $$\frac1{15}\left(-15\cos^4x\sin x+15\cos^2x\sin x\right)=\cos^2x\sin x\left(1-\cos^2x\right)=\cos^2x\sin^3x.$$ Step 3 - At $x=\frac\pi6$. $y'=1+\frac34\cdot\frac18=\frac{35}{32}$. Step 4 - Value. $96\cdot\frac{35}{32}=105$. Answer: 105