Let f(x) be a non - negative continuous function such that the area bounded by the curve y=f(x),x -axis and the ordinates x = \pi \over 4 and x = \beta \gt \pi \over 4 is ( \beta \sin \beta + \pi \over 4\cos \beta + \sqrt 2 \beta ). Then f( \pi \over 2 ) is

Area Under the Curves PYQ — JEE Main 2005 · Shift 0 · Q82 with Solution

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Correct answer: D

Interpret the given area condition

Since $f(x)$ is non-negative and continuous, the area bounded by the curve $y=f(x)$ , the $x$ -axis, and the ordinates $x=\frac{\pi}{4}$ and $x=\beta$ is

\int_{\pi/4}^{\beta} f(x)\,dx.

We are given that for $\beta>\frac{\pi}{4}$ ,

\int_{\pi/4}^{\beta} f(x)\,dx = \beta\sin\beta + \frac{\pi}{4}\cos\beta + \sqrt{2}\,\beta.

Let

F(\beta)=\int_{\pi/4}^{\beta} f(x)\,dx.

Then

F(\beta)=\beta\sin\beta + \frac{\pi}{4}\cos\beta + \sqrt{2}\beta.

Use the Fundamental Theorem of Calculus

By FTC,

F'(\beta)=f(\beta).

Differentiate the right-hand side with respect to $\beta$ :

\frac{d}{d\beta}(\beta\sin\beta)=\sin\beta+\beta\cos\beta,

\frac{d}{d\beta}\left(\frac{\pi}{4}\cos\beta\right)=-\frac{\pi}{4}\sin\beta,

\frac{d}{d\beta}(\sqrt{2}\beta)=\sqrt{2}.

Therefore,

f(\beta)=\sin\beta+\beta\cos\beta-\frac{\pi}{4}\sin\beta+\sqrt{2}.

So,

f(\beta)=\left(1-\frac{\pi}{4}\right)\sin\beta+\beta\cos\beta+\sqrt{2}.

Find $f\left(\frac{\pi}{2}\right)$

Substitute $\beta=\frac{\pi}{2}$ :

f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}\right)\sin\frac{\pi}{2}+\frac{\pi}{2}\cos\frac{\pi}{2}+\sqrt{2}.

Now,

\sin\frac{\pi}{2}=1, \qquad \cos\frac{\pi}{2}=0.

Hence,

f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}\right)(1)+0+\sqrt{2}

=1-\frac{\pi}{4}+\sqrt{2}.

Match with the options

This is

1-\frac{\pi}{4}+\sqrt{2},

which matches Option D.

More from Area Under the Curves

View written solutionFree

Correct answer: D

Interpret the given area condition

Since $f(x)$ is non-negative and continuous, the area bounded by the curve $y=f(x)$ , the $x$ -axis, and the ordinates $x=\frac{\pi}{4}$ and $x=\beta$ is

\int_{\pi/4}^{\beta} f(x)\,dx.

We are given that for $\beta>\frac{\pi}{4}$ ,

\int_{\pi/4}^{\beta} f(x)\,dx = \beta\sin\beta + \frac{\pi}{4}\cos\beta + \sqrt{2}\,\beta.

Let

F(\beta)=\int_{\pi/4}^{\beta} f(x)\,dx.

Then

F(\beta)=\beta\sin\beta + \frac{\pi}{4}\cos\beta + \sqrt{2}\beta.

Use the Fundamental Theorem of Calculus

By FTC,

F'(\beta)=f(\beta).

Differentiate the right-hand side with respect to $\beta$ :

\frac{d}{d\beta}(\beta\sin\beta)=\sin\beta+\beta\cos\beta,

\frac{d}{d\beta}\left(\frac{\pi}{4}\cos\beta\right)=-\frac{\pi}{4}\sin\beta,

\frac{d}{d\beta}(\sqrt{2}\beta)=\sqrt{2}.

Therefore,

f(\beta)=\sin\beta+\beta\cos\beta-\frac{\pi}{4}\sin\beta+\sqrt{2}.

So,

f(\beta)=\left(1-\frac{\pi}{4}\right)\sin\beta+\beta\cos\beta+\sqrt{2}.

Find $f\left(\frac{\pi}{2}\right)$

Substitute $\beta=\frac{\pi}{2}$ :

f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}\right)\sin\frac{\pi}{2}+\frac{\pi}{2}\cos\frac{\pi}{2}+\sqrt{2}.

Now,

\sin\frac{\pi}{2}=1, \qquad \cos\frac{\pi}{2}=0.

Hence,

f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}\right)(1)+0+\sqrt{2}

=1-\frac{\pi}{4}+\sqrt{2}.

Match with the options

This is

1-\frac{\pi}{4}+\sqrt{2},

which matches Option D.

Area Under the Curves question

Area Under the Curves question

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