Let f be a differentiable function satisfying f(x)=\frac2\sqrt3 \int\limits_0^\sqrt3 f(\frac\lambda^2 x3) d \lambda, x\gt 0 and f(1)=\sqrt3 . If y=f(x) passes through the point (\alpha, 6) , then \alpha is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2022 · 27 Jul · Shift 2 · Q39 with Solution

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Correct answer: 12

f(x)=\frac{2}{\sqrt{3}}\int_{0}^{\sqrt{3}} f\left(\frac{\lambda^2 x}{3}\right)\,d\lambda, \qquad x>0

and also $f(1)=\sqrt{3}.$ We need the point $(\alpha,6)$ on the curve $y=f(x)$ , i.e. solve $f(\alpha)=6.$

Simplify the integral equation by substitution. Let $t=\frac{\lambda^2 x}{3}.$ Then $\lambda=\sqrt{\frac{3t}{x}},$ so $d\lambda=\frac{1}{2}\sqrt{\frac{3}{x}}\frac{1}{\sqrt{t}}dt.$ When $\lambda=0$ , $t=0$ ; and when $\lambda=\sqrt{3}$ , $t=\frac{(\sqrt{3})^2x}{3}=x.$ Thus

\int_0^{\sqrt{3}} f\left(\frac{\lambda^2x}{3}\right)d\lambda =\int_0^x f(t)\cdot \frac{1}{2}\sqrt{\frac{3}{x}}\frac{1}{\sqrt{t}}dt.

Multiplying by $\frac{2}{\sqrt{3}}$ , we get

f(x)=\frac{1}{\sqrt{x}}\int_0^x \frac{f(t)}{\sqrt{t}}dt.

So the equation becomes

\boxed{f(x)=x^{-1/2}\int_0^x f(t)t^{-1/2}\,dt.}

\sqrt{x}\,f(x)=\int_0^x \frac{f(t)}{\sqrt{t}}dt.

Now differentiate both sides with respect to $x$ .

Left side:

\frac{d}{dx}[\sqrt{x}\,f(x)] = \frac{f(x)}{2\sqrt{x}}+\sqrt{x}f'(x).

Right side, by Fundamental Theorem of Calculus:

\frac{d}{dx}\int_0^x \frac{f(t)}{\sqrt{t}}dt=\frac{f(x)}{\sqrt{x}}.

Therefore,

\frac{f(x)}{2\sqrt{x}}+\sqrt{x}f'(x)=\frac{f(x)}{\sqrt{x}}.

Multiply by $\sqrt{x}$ :

\frac{f(x)}{2}+xf'(x)=f(x).

Hence

xf'(x)=\frac{f(x)}{2}.

\frac{f'(x)}{f(x)}=\frac{1}{2x}.

Integrating,

\ln f(x)=\frac{1}{2}\ln x + C.

Thus

f(x)=C\sqrt{x}.

\sqrt{3}=C\sqrt{1}=C.

\boxed{f(x)=\sqrt{3x}}.

f(\alpha)=6.

\sqrt{3\alpha}=6.

Squaring:

3\alpha=36 \quad\Rightarrow\quad \alpha=12.

Therefore,

\boxed{\alpha=12}.

View written solutionFree

Correct answer: 12

f(x)=\frac{2}{\sqrt{3}}\int_{0}^{\sqrt{3}} f\left(\frac{\lambda^2 x}{3}\right)\,d\lambda, \qquad x>0

and also $f(1)=\sqrt{3}.$ We need the point $(\alpha,6)$ on the curve $y=f(x)$ , i.e. solve $f(\alpha)=6.$

Simplify the integral equation by substitution. Let $t=\frac{\lambda^2 x}{3}.$ Then $\lambda=\sqrt{\frac{3t}{x}},$ so $d\lambda=\frac{1}{2}\sqrt{\frac{3}{x}}\frac{1}{\sqrt{t}}dt.$ When $\lambda=0$ , $t=0$ ; and when $\lambda=\sqrt{3}$ , $t=\frac{(\sqrt{3})^2x}{3}=x.$ Thus

\int_0^{\sqrt{3}} f\left(\frac{\lambda^2x}{3}\right)d\lambda =\int_0^x f(t)\cdot \frac{1}{2}\sqrt{\frac{3}{x}}\frac{1}{\sqrt{t}}dt.

Multiplying by $\frac{2}{\sqrt{3}}$ , we get

f(x)=\frac{1}{\sqrt{x}}\int_0^x \frac{f(t)}{\sqrt{t}}dt.

So the equation becomes

\boxed{f(x)=x^{-1/2}\int_0^x f(t)t^{-1/2}\,dt.}

\sqrt{x}\,f(x)=\int_0^x \frac{f(t)}{\sqrt{t}}dt.

Now differentiate both sides with respect to $x$ .

Left side:

\frac{d}{dx}[\sqrt{x}\,f(x)] = \frac{f(x)}{2\sqrt{x}}+\sqrt{x}f'(x).

Right side, by Fundamental Theorem of Calculus:

\frac{d}{dx}\int_0^x \frac{f(t)}{\sqrt{t}}dt=\frac{f(x)}{\sqrt{x}}.

Therefore,

\frac{f(x)}{2\sqrt{x}}+\sqrt{x}f'(x)=\frac{f(x)}{\sqrt{x}}.

Multiply by $\sqrt{x}$ :

\frac{f(x)}{2}+xf'(x)=f(x).

Hence

xf'(x)=\frac{f(x)}{2}.

\frac{f'(x)}{f(x)}=\frac{1}{2x}.

Integrating,

\ln f(x)=\frac{1}{2}\ln x + C.

Thus

f(x)=C\sqrt{x}.

\sqrt{3}=C\sqrt{1}=C.

\boxed{f(x)=\sqrt{3x}}.

f(\alpha)=6.

\sqrt{3\alpha}=6.

Squaring:

3\alpha=36 \quad\Rightarrow\quad \alpha=12.

Therefore,

\boxed{\alpha=12}.