Let y=f(x)=sin ^3left(frac{pi}{3}cos left(fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $y = \sin^3\left(\frac{\pi}{3} \cos(g(x))\right)$ where $g(x) = \frac{\pi}{3\sqrt{2}}(-4x^3 + 5x^2 + 1)^{3/2}$.At $x=1$,$g(1) = \frac{\pi}{3

Vedclass · Accepted Answer

$2 y^{\prime}+3 \pi^2 y=0$

Vedclass · Answer

(B) Given $y = \sin^3\left(\frac{\pi}{3} \cos(g(x))\right)$ where $g(x) = \frac{\pi}{3\sqrt{2}}(-4x^3 + 5x^2 + 1)^{3/2}$.At $x=1$,$g(1) = \frac{\pi}{3\sqrt{2}}(-4+5+1)^{3/2} = \frac{\pi}{3\sqrt{2}}(2)^{3/2} = \frac{\pi}{3\sqrt{2}}(2\sqrt{2}) = \frac{2\pi}{3}$.$y(1) = \sin^3\left(\frac{\pi}{3} \cos\left(\frac{2\pi}{3}\right)\right) = \sin^3\left(\frac{\pi}{3} \cdot \left(-\frac{1}{2}\right)\right) = \sin^3\left(-\frac{\pi}{6}\right) = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8}$.Now,$y' = 3\sin^2\left(\frac{\pi}{3}\cos(g(x))\right) \cdot \cos\left(\frac{\pi}{3}\cos(g(x))\right) \cdot \left(-\frac{\pi}{3}\sin(g(x))\right) \cdot g'(x)$.$g'(x) = \frac{\pi}{3\sqrt{2}} \cdot \frac{3}{2}(-4x^3 + 5x^2 + 1)^{1/2} \cdot (-12x^2 + 10x) = \frac{\pi}{2\sqrt{2}} \sqrt{-4x^3 + 5x^2 + 1} (-12x^2 + 10x)$.$g'(1) = \frac{\pi}{2\sqrt{2}} \cdot \sqrt{2} \cdot (-2) = -\pi$.Substituting $x=1$ into $y'$:$y'(1) = 3\sin^2(-\pi/6) \cdot \cos(-\pi/6) \cdot \left(-\frac{\pi}{3}\sin(2\pi/3)\right) \cdot (-\pi) = 3 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \cdot \left(-\frac{\pi}{3} \cdot \frac{\sqrt{3}}{2}\right) \cdot (-\pi) = \frac{3\pi^2}{16}$.Checking option $B$: $2y'(1) + 3\pi^2 y(1) = 2\left(\frac{3\pi^2}{16}\right) + 3\pi^2\left(-\frac{1}{8}\right) = \frac{3\pi^2}{8} - \frac{3\pi^2}{8} = 0$.

Let $y=f(x)=\sin ^3\left(\frac{\pi}{3}\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^3+5 x^2+1\right)^{\frac{3}{2}}\right)\right)$. Then,at $x =1$,

Similar Questions

Let $f : R \rightarrow R$ be a differentiable function and $f(1) = 4$. Then the value of $\lim_{x \rightarrow 1} \int_{4}^{f(x)} \frac{2t \, dt}{x - 1}$ is:

$\begin{aligned} & \text{If } y = \tan^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\} \\ & + \sin \left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\} \text{, then } \frac{dy}{dx} = \end{aligned}$

The derivative of $\sec^{-1}\left( \frac{1}{2x^2 - 1} \right)$ with respect to $\sqrt{1 - x^2}$ at $x = \frac{1}{2}$ is:

If $y = \operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}$,then $\frac{dy}{dx} = $

$\frac{d}{dx} \tan^{-1} \left[ \frac{\cos x - \sin x}{\cos x + \sin x} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module