Animated Solution for Mathematics - Definite Integration: If $f(x)=A\sin (\frac{\pi x}{2})+B$, $f^{\prime }(\frac{1}{2})=\sqrt{2}$ and ${\int }_0^{1}f(x)dx=\frac{2A}{\pi }$, then constants $A$ and $B$ are

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a function to reveal its hidden identity.

We are given a function $f(x)=A\sin (\frac{\pi x}{2})+B$, and we are tasked with finding the constants $A$ and $B$. This is a classic JEE Advanced scenario where the problem provides two distinct 'keys'—the derivative and the integral—to unlock the two 'locks'—the constants $A$ and $B$.

Our first key is the derivative condition: $f^{\prime }(\frac{1}{2})=\sqrt{2}$. In the language of geometry, the derivative $f^{\prime }(x)$ is the slope of the tangent line to the curve at any point $x$.

To use this, we must first find the general expression for the derivative. We apply the differentiation operator to our function:

Using the linearity of the derivative, we treat the terms separately. The derivative of the constant $B$ is $0$. For the sine term, we invoke the Chain Rule:

Now, we substitute $x=\frac{1}{2}$ into this expression. The argument of the cosine becomes $\frac{\pi }{2}\cdot \frac{1}{2}=\frac{\pi }{4}$. Since $\cos (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$, we equate this to our given condition:

With a bit of algebraic finesse, we multiply both sides by $2\sqrt{2}$. Since $\sqrt{2}\cdot \sqrt{2}=2$, the right side becomes $2\cdot 2=4$. We are left with $A\pi =4$, which gives us our first constant:

Now, let us turn our attention to the second key: the integral condition ${\int }_0^{1}f(x)dx=\frac{2A}{\pi }$. This integral represents the net area bounded by the curve $f(x)$ and the $x$-axis from $x=0$ to $x=1$.

We substitute our function into the integral:

Using the linearity of the integral, we split this into two parts:

Integrating the sine term, where the integral of $\sin (kx)$ is $-\frac{1}{k}\cos (kx)$, we set $k=\frac{\pi }{2}$. The integral becomes:

Let us evaluate the limits with precision. At the upper limit $x=1$, we have $-A\cdot \frac{2}{\pi }\cos (\frac{\pi }{2})+B(1)$. Since $\cos (\frac{\pi }{2})=0$, this simplifies to $B$.

At the lower limit $x=0$, we have $-A\cdot \frac{2}{\pi }\cos (0)+B(0)$. Since $\cos (0)=1$, this simplifies to $-A\cdot \frac{2}{\pi }$.

According to the Fundamental Theorem of Calculus, we subtract the lower limit from the upper limit:

We equate this to the given condition $\frac{2A}{\pi }$:

The terms $\frac{2A}{\pi }$ appear on both sides and cancel out, leaving us with $B=0$. We have successfully found our constants: $A=\frac{4}{\pi }$ and $B=0$.