Differential Equations 1 Question 3
3. Let $f:[0,1] \rightarrow R$ be such that $f(x y)=f(x) . f(y)$, for all $x, y \in[0,1]$ and $f(0) \neq 0$. If $y=y(x)$ satisfies the differential equation, $\frac{d y}{d x}=f(x)$ with $y(0)=1$, then $y \frac{1}{4}+y \frac{3}{4}$ is equal to
(2019 Main, 9 Jan II)
(a) 5
(b) 3
(c) 2
(d) 4
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Answer:
Correct Answer: 3. (b)
Solution:
- Given, $f(x y)=f(x) \cdot f(y), \forall x, y \in[0,1]$
Putting $x=y=0$ in Eq. (i), we get
$$ \begin{aligned} & & f(0) & =f(0) \cdot f(0) \\ \Rightarrow & & f(0)[f(0)-1] & =0 \\ \Rightarrow & & f(0) & =1 \text { as } f(0) \neq 0 \end{aligned} $$
Now, put $y=0$ in Eq. (i), we get
$$ \begin{array}{ll} & f(0)=f(x) \cdot f(0) \\ \Rightarrow & f(x)=1 \end{array} $$
So,
$$ \frac{d y}{d x}=f(x) \Rightarrow \frac{d y}{d x}=1 $$
$\Rightarrow \quad \int d y=\int d x$
$$ \Rightarrow \quad y=x+C $$
$\because \quad y(0)=1$
$$ \therefore \quad 1=0+C $$
$$ \begin{array}{ll} \Rightarrow & C=1 \\ \therefore & y=x+1 \end{array} $$
Now, $\quad y \frac{1}{4}=\frac{1}{4}+1=\frac{5}{4}$ and $y \frac{3}{4}=\frac{3}{4}+1=\frac{7}{4}$
$\Rightarrow \quad y \frac{1}{4}+y \frac{3}{4}=\frac{5}{4}+\frac{7}{4}=3$
