Differential Equations 1 Question 19
19. Let $f: R \rightarrow R$ be a continuous function, which satisfies $f(x)=\int _0^{x} f(t) d t$. Then, the value of $f(\ln 5)$ is … .
(2009)
Passage Based Problems
Passage
Let $f:[0,1] \rightarrow R$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable, $f(0)=f(1)=0$ and satisfies
$$ f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^{x}, x \in[0,1] $$
(2013 Adv.)
Show Answer
Answer:
Correct Answer: 19. (c)
Solution:
- From given integral equation, $f(0)=0$.
Also, differentiating the given integral equation w.r.t. $x$
$$ \begin{array}{lc} & f^{\prime}(x)=f(x) \\ \text { If } & f(x) \neq 0 \\ \Rightarrow & \frac{f^{\prime}(x)}{f(x)}=1 \Rightarrow \log f(x)=x+c \\ \Rightarrow & f(x)=e^{c} e^{x} \\ \because & f(0)=0 \Rightarrow e^{c}=0, \text { a contradiction } \\ \therefore & f(x)=0, \forall x \in R \\ \Rightarrow & f(\ln 5)=0 \end{array} $$
Alternate Solution
Given,
$$ f(x)=\int _0^{x} f(t) d t $$
$\Rightarrow$
$$ f(0)=0 \quad \text { and } \quad f^{\prime}(x)=f(x) $$
If $f(x) \neq 0$
$$ \begin{array}{ll} \Rightarrow & \frac{f^{\prime}(x)}{f(x)}=1 \Rightarrow \ln f(x)=x+c \\ \Rightarrow & f(x)=e^{c} \cdot e^{x} \\ \because & f(0)=0 \end{array} $$
$\Rightarrow e^{c}=0, a$ contradiction
$$ \begin{array}{lc} \therefore & f(x)=0, \forall x \in R \\ \Rightarrow & f(\ln 5)=0 \end{array} $$
