Search any question & find its solution
Question:
Answered & Verified by Expert
$\quad$ If $f(x)=A \sin \left(\frac{\pi x}{2}\right)+B$ and $f^{\prime}\left(\frac{1}{2}\right)=\sqrt{2}$ and
$\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{2 \mathrm{~A}}{\pi}$, then what is the value of $\mathrm{B}$ ?
Options:
$\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{2 \mathrm{~A}}{\pi}$, then what is the value of $\mathrm{B}$ ?
Solution:
1010 Upvotes
Verified Answer
The correct answer is:
0
Given function $\mathrm{f}(\mathrm{x})=\mathrm{A} \sin \left(\frac{\pi \mathrm{x}}{2}\right)+\mathrm{B}$
Differentiating w. r. t. x $f^{\prime}(x)=A \cos \left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}$
$\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\sqrt{2}=\mathrm{A}\left(\cos \frac{\pi}{4}\right) \frac{\pi}{2}=\mathrm{A} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2}$
$\Rightarrow \quad A=\frac{(\sqrt{2} \times \sqrt{2}) \times 2}{\pi}=\frac{4}{\pi}$
Now, $\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{2 \mathrm{~A}}{\pi}$
$\Rightarrow \quad \int_{0}^{1}\left\{\mathrm{~A} \sin \left(\frac{\pi \mathrm{x}}{2}\right)+\mathrm{B}\right\} \mathrm{d} \mathrm{x}=\frac{2 \times 4}{\pi^{2}}$
$\Rightarrow\left[-\mathrm{A} \cos \frac{\pi \mathrm{x}}{2} \cdot \frac{2}{\pi}+\mathrm{Bx}\right]_{0}^{1}=\frac{8}{\pi^{2}}$
$\Rightarrow \quad-\frac{4}{\pi} \cdot \frac{2}{\pi} \cos \frac{\pi}{2}+\mathrm{B}+\frac{4}{\pi} \cdot \frac{2}{\pi} \cos 0=\frac{8}{\pi^{2}}$
$\Rightarrow \quad B+\frac{8}{\pi^{2}}=\frac{8}{\pi^{2}} \Rightarrow B=0$
Differentiating w. r. t. x $f^{\prime}(x)=A \cos \left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}$
$\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\sqrt{2}=\mathrm{A}\left(\cos \frac{\pi}{4}\right) \frac{\pi}{2}=\mathrm{A} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2}$
$\Rightarrow \quad A=\frac{(\sqrt{2} \times \sqrt{2}) \times 2}{\pi}=\frac{4}{\pi}$
Now, $\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{2 \mathrm{~A}}{\pi}$
$\Rightarrow \quad \int_{0}^{1}\left\{\mathrm{~A} \sin \left(\frac{\pi \mathrm{x}}{2}\right)+\mathrm{B}\right\} \mathrm{d} \mathrm{x}=\frac{2 \times 4}{\pi^{2}}$
$\Rightarrow\left[-\mathrm{A} \cos \frac{\pi \mathrm{x}}{2} \cdot \frac{2}{\pi}+\mathrm{Bx}\right]_{0}^{1}=\frac{8}{\pi^{2}}$
$\Rightarrow \quad-\frac{4}{\pi} \cdot \frac{2}{\pi} \cos \frac{\pi}{2}+\mathrm{B}+\frac{4}{\pi} \cdot \frac{2}{\pi} \cos 0=\frac{8}{\pi^{2}}$
$\Rightarrow \quad B+\frac{8}{\pi^{2}}=\frac{8}{\pi^{2}} \Rightarrow B=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.