Limits, Continuity and Differentiability
80281
If \(f^{\prime}(x)=k(\cos x-\sin x), f^{\prime}(0)=3, f\left(\frac{\pi}{2}\right)=15\) then \(\mathbf{f}(\mathbf{x})=\)
1 \(-3(\sin x+\cos x)-12\)
2 \(3(\sin x+\cos x)+12\)
3 \(12(\sin x+\cos x)+3\)
4 \(3(\sin x+\cos x)-12\)
Explanation:
(B) : Given,
\(f^{\prime}(x)=k(\cos x-\sin x), f^{\prime}(0)=3\),
\(\mathrm{f}^{\prime}\left(\frac{\pi}{2}\right)=15\)
Then, \(f^{\prime}(x)=k(\cos x-\sin x)\)
\(\mathrm{f}^{\prime}(0)=\mathrm{k}\left(\cos 0^{\circ}-\sin 0^{\circ}\right)\)
\(\mathrm{f}^{\prime}(0)=\mathrm{k}\)
\(3=\mathrm{k}\)
\(\mathrm{k}=3\)
Now, \(\int f^{\prime}(x) d x=\int k(\cos x-\sin x) \cdot d x\)
\(\int f^{\prime}(x) d x=3 \int(\cos x-\sin x) d x\)
\(f(x)=3[\sin x+\cos x]+c\)
\(\therefore \quad f\left(\frac{\pi}{2}\right)=3\left[\sin \frac{\pi}{2}+\cos \frac{\pi}{2}\right]+c\)
\(15=3[1+0]+\mathrm{c}\)
\(\mathrm{c}=15-3=12\)
Putting the value of \(\mathrm{c}\) in the equation (i) we get, Then,
\(\mathrm{f}(\mathrm{x})=3[\sin \mathrm{x}+\cos \mathrm{x}]+12\)