Integral Calculus Question 173
Question: If $ \int{f(x)\cos xdx=\frac{1}{2}f^{2}(x)+c,} $ then $ f(x) $ can be
Options:
A) x
B) 1
C) $ \cos x $
D) $ sinx $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Since $ \int{f{{[x]}^{n}}f’(x)dx=\frac{{{[f(x)]}^{n+1}}}{n+1}+c} $
$ \therefore \int{f(x)\cos xdx=\frac{f^{2}(x)}{2}+c} $
$ \Rightarrow f’(x)=\cos x\Rightarrow f(x)=\sin x $
BETA
