Limit Continuity And Differentiability Ques 7
- If $f^{\circ}(x)=-f(x)$, where $f(x)$ is a continuous double differentiable function and $g(x)=f^{\prime}(x)$.
If $F(x)=\left\{f\left(\frac{x}{2}\right)\right\}^2+\left\{g\left(\frac{x}{2}\right)\right\}^2$ and $F(5)=5$, then $F(10)$ is
(2006, 3M)
(a) $0$
(b) $5$
(c) $10$
(d) $25$
Show Answer
Answer:
Correct Answer: 7.(b)
Solution: (b) Since, $f^{\prime \prime}(x)=-f(x)$
$\Rightarrow \quad \frac{d}{d x}\left\{f^{\prime}(x)\right\}=-f(x)$
$\Rightarrow \quad g^{\prime}(x)=-f(x) \quad\left[\because g(x)=f^{\prime}(x)\right.$, given $]$ $\quad$ ……..(i)
Also, $\quad F(x)=\left\{f\left(\frac{x}{2}\right)\right\}^2+\left\{g\left(\frac{x}{2}\right)\right\}^2$
[from Eq.(i)]
$\Rightarrow \quad F^v(x)=2\left(f\left(\frac{x}{2}\right)\right) \cdot f^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2}$ $+2\left(g\left(\frac{x}{2}\right)\right) \cdot g\left(\frac{x}{2}\right) \cdot \frac{1}{2}=0$
$\therefore \quad F(x)$ is constant
$\Rightarrow \quad F(10)=F(5)=5$
BETA
