SATHEE: Differential Equations 1 Question 3

Differential Equations 1 Question 3

3. Let $f : [0, 1] \to R$ 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

(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

$\begin{aligned} f (0) & = f (0) \cdot f (0) \\ \Rightarrow & f (0) [f (0) - 1] & = 0 \\ \Rightarrow & f (0) & = 1 as f (0) \neq 0 \end{aligned}$

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,

$\frac{d y}{d x} = f (x) \Rightarrow \frac{d y}{d x} = 1$

$\Rightarrow \int d y = \int d x$

$\Rightarrow y = x + C$

$∵ y (0) = 1$

$∴ 1 = 0 + C$

$\begin{array}{ll} \Rightarrow & C = 1 \\ ∴ & y = x + 1 \end{array}$

Now, $y \frac{1}{4} = \frac{1}{4} + 1 = \frac{5}{4}$ and $y \frac{3}{4} = \frac{3}{4} + 1 = \frac{7}{4}$

$\Rightarrow y \frac{1}{4} + y \frac{3}{4} = \frac{5}{4} + \frac{7}{4} = 3$

Differential Equations 1 Question 3

3. Let $f : [0, 1] \to R$ 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

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Differential Equations 1 Question 3

3. Let f:[0,1]→R be such that f(xy)=f(x).f(y), for all x,y∈[0,1] and f(0)≠0. If y=y(x) satisfies the differential equation, dydx=f(x) with y(0)=1, then y14+y34 is equal to

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3. Let $f : [0, 1] \to R$ 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