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JEE Advance - Mathematics (2024 - Paper 2 Online - No. 16)

Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$.
Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$.
Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$.
The value of $2 \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x-\int\limits_0^{\frac{\pi}{2}} g(x) d x$ is ____________.
Answer
0

Explanation

$$\mathrm{I}=2 \int_\limits0^{\frac{\pi}{2}} \underbrace{\sin ^2 \mathrm{x} \cdot \sqrt{\frac{\pi x}{2}-\mathrm{x}^2}}_{\mathrm{i}_1}-\int_\limits0^{\frac{\pi}{2}} \mathrm{~g}(\mathrm{x}) \mathrm{dx}$$

$$\text { Let } I_1=\int_\limits0^{\frac{\pi}{2}} \sin ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} \quad \text { (making perfect square) }$$

apply kings

$$I_1=\int_\limits0^{\frac{\pi}{2}} \cos ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(\frac{\pi}{2}-x\right)^2}$$

add both

$$2 I_1=\int_\limits0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$$

i.e. $$2 I_1=\int_\limits0^{\frac{\pi}{2}} g(x)$$

Now $$I=2 I_1-\int_\limits0^{\frac{\pi}{2}} g(x)=0$$

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