Question

Assertion(A): Let $f\left(x\right)$ be twice differentiable function such that $f^{{}^{\prime \prime }}\left(x\right)=-f\left(x\right)$ and $f^{{}^{\prime }}\left(x\right)=g\left(x\right)$. lf $h\left(x\right)=\left[f\right(x){]}^2+[g\left(x\right){]}^2$ and $h\left(1\right)=8$, then $h\left(2\right)=8$

Reason (R): Derivative of a constant function is zero.

• A. Both A and R are true R is correct reason of A
• B. A is true but R is false
• C. Both A and R are true R is not correct reason of A
• D. A is false but R is true

Answer 1

Answer: (A)

Let $f\left(x\right)=c$
$f^{\prime }\left(x\right)=0$ So Reason is True

Now, $h\left(x\right)=\left[f\right(x){]}^2+[g\left(x\right){]}^2$
$h^{\prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)+2g\left(x\right)g^{\prime }\left(x\right)$
$=2f\left(x\right)g\left(x\right)+2g\left(x\right)f^{\prime \prime }\left(x\right)$ ...since $f^{\prime }\left(x\right)=g\left(x\right)\Rightarrow f^{\prime \prime }\left(x\right)=g^{\prime }\left(x\right)$
$=2g\left(x\right)\left[f\right(x)+f^{\prime \prime }(x\left)\right]$

$=2g\left(x\right)\left(f\right(x)-f(x\left)\right)$
$\Rightarrow h^{\prime }\left(x\right)=0$
means $h\left(x\right)=C$ (a constant function)
Given $h\left(1\right)=8$
$\Rightarrow h\left(x\right)=8$
Hence, $f\left(2\right)=8$

So, both A and R are correct and R is the correct explanation of A.