Let f be a differentiable function satisfying f ( x )=2 3 _ 0 ^ 3… - Vidyadip

MCQ

Let $f$ be a differentiable function satisfying $f ( x )=\frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left(\frac{\lambda^{2} x }{3}\right) d \lambda, x >0$ and $f (1)=\sqrt{3}$. If $y=f(x)$ passes through the point $(\alpha, 6)$, then $\alpha$ is equal to $.........$

A
$6$
✓
$12$
C
$4$
D
$3$

✓

Answer

Correct option: B.

$12$

b
Let, $\frac{\lambda^{2} x }{3}= t$

$\frac{2 \lambda x }{3} d \lambda= dt$

$d \lambda=\frac{3}{2} \cdot \frac{1 \sqrt{ x }}{ x \cdot \sqrt{3} \sqrt{ t }} dt$

$d \lambda=\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{ x }} \cdot \frac{ dt }{\sqrt{ t }}$

So, $f(x)=\frac{1}{\sqrt{x}} \int_{0}^{x} \frac{f(t)}{\sqrt{t}} d t$

$\sqrt{x} \cdot f^{\prime}(x)+\frac{f(x)}{2 \sqrt{x}}=\frac{f(x)}{\sqrt{x}}$

$\sqrt{ x } \cdot f ^{\prime}( x )=\frac{ f ( x )}{2 \sqrt{ x }}$

$\frac{ dy }{ y }=\frac{ dx }{2 x }$

$\ln y=\frac{1}{2} \ln x+c \Rightarrow f(x)=\sqrt{x}$

$y =\sqrt{3 x } \quad$ $\{as$ $f (1)=\sqrt{3}\}$

So, $f(x)=\sqrt{3 x}$

Now, $f(\alpha)=6 \Rightarrow 36=3 \alpha$

$\alpha=12$

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