If f^{prime}(x)=k(cos x-sin x),f^{prime}(0)=3 | vedclass

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

Question

(A) Given $f^{\prime}(x)=k(\cos x-\sin x)$.Substitute $x=0$ into the derivative: $f^{\prime}(0)=k(\cos 0-\sin 0)=k(1-0)=k$.Since $f^{\prime}(0)=3$,we

Vedclass · Accepted Answer

$3(\sin x+\cos x)+12$

Vedclass · Answer

(A) Given $f^{\prime}(x)=k(\cos x-\sin x)$.Substitute $x=0$ into the derivative: $f^{\prime}(0)=k(\cos 0-\sin 0)=k(1-0)=k$.Since $f^{\prime}(0)=3$,we have $k=3$.Now,integrate $f^{\prime}(x)$ to find $f(x)$:$f(x)=\int 3(\cos x-\sin x) \, dx = 3(\sin x+\cos x)+C$.Use the condition $f\left(\frac{\pi}{2}\right)=15$ to find $C$:$f\left(\frac{\pi}{2}\right)=3(\sin \frac{\pi}{2}+\cos \frac{\pi}{2})+C = 3(1+0)+C = 3+C$.Setting $3+C=15$,we get $C=12$.Therefore,$f(x)=3(\sin x+\cos x)+12$.

If $f^{\prime}(x)=k(\cos x-\sin x)$,$f^{\prime}(0)=3$,and $f\left(\frac{\pi}{2}\right)=15$,then $f(x)=$

Similar Questions

If $y = \tan^{-1} \sqrt{\frac{1 + \sin 2x}{1 - \sin 2x}}$, then at $x = \frac{\pi}{6}$, the value of $\frac{dy}{dx}$ is

If $f(x) = |\cos x - \sin x|$,then $f^{\prime}\left(\frac{\pi}{6}\right)$ is equal to

If $y=\tan ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}\right)$, then $\frac{d y}{d x}=$

Differentiate the function with respect to $x$: $\sin^{3} x + \cos^{6} x$

If $y=\left(\log _{\cot x} \tan x\right)\left(\log _{\tan x} \cot x\right)+\tan ^{-1}\left(\frac{4 x}{4-x^2}\right)$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module