Animated Solution for Mathematics - Differentiation: Match the statements/expressions in Column I with the open intervals in Column II.

Mastering the JEE Advanced Match-the-Following

Welcome, fellow explorer of the mathematical universe! Today, we are not just solving a problem; we are embarking on a journey through four distinct landscapes of calculus.

The 'Match-the-Following' format is a classic JEE Advanced challenge. It tests not just your ability to calculate, but your ability to switch gears between differential equations, integral properties, and the nuances of function behavior. Let us break this down, one step at a time, and uncover the elegance hidden within.

The Singularity in the Differential Equation

We begin with the differential equation:

At first glance, it looks like a standard first-order equation. Our goal is to isolate the variables. By rearranging, we get:

Integrating both sides is our next logical move. The left side yields $\ln |y|$, and the right side, using the power rule for integration, gives us $\frac{1}{x-3}+C$.

Exponentiating both sides leads us to the general solution:

Now, here is the crucial part: the domain. Look at the exponent $\frac{1}{x-3}$. The function is clearly undefined at $x=3$.

Therefore, the domain of definition for any non-zero solution is all real numbers except $3$, or $x\in \mathbb{R}\setminus \{3\}$. In the context of this problem, we identify the intervals contained within this domain.

The Elegance of Symmetry in Integration

Next, we face the integral:

Expanding this polynomial would be a tedious, error-prone nightmare. Instead, let us look for symmetry. Notice the center of the interval $[1,5]$ is $3$.

Let us perform a substitution: $u=x-3$. Then $dx=du$. When $x=1$, $u=-2$; when $x=5$, $u=2$.

Our integral transforms into:

Grouping the terms, we get:

Let $f(u)=u(u^2-1)(u^2-4)$. If we replace $u$ with $-u$, we find $f(-u)=-f(u)$, confirming that $f(u)$ is an odd function.

The integral of an odd function over symmetric limits is always zero. Thus, $I=0$. We simply need to find which intervals in Column II contain the value $0$.

The Calculus of Local Maxima

Now, we turn to the function $f(x)={\cos }^{2}x+\sin x$. To find the local maxima, we must first find the critical points by setting the derivative to zero.

Using the chain rule:

Setting $f^{\prime }(x)=0$ gives us $\cos x=0$ or $\sin x=\frac{1}{2}$. This yields critical points at $x=\frac{\pi }{2}$, $x=\frac{\pi }{6}$, and $x=\frac{5\pi }{6}$.

To distinguish between maxima and minima, we use the second derivative test:

Evaluating this at our critical points, we find that $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$ result in a negative second derivative, confirming them as local maxima. We then map these points to the intervals provided.

The Increasing Nature of Inverse Trigonometry

Finally, we examine $f(x)={\tan }^{-1}(\sin x+\cos x)$. We want to know where this function is increasing, which means we need $f^{\prime }(x)>0$.

The derivative is:

Since the denominator is always positive, the sign of the derivative depends entirely on the numerator: $\cos x-\sin x>0$, or $\cos x>\sin x$.

On the unit circle, this inequality holds when $x$ is in the interval $(-\frac{3\pi }{4},\frac{\pi }{4})$. We look for the interval in Column II that is fully contained within this region, leading us to our final match.

Mathematics is not just about finding the answer; it is about the journey of discovery. By breaking down these complex expressions, we have navigated through differential equations, exploited the symmetry of integrals, mastered the second derivative test, and analyzed trigonometric inequalities.

You have successfully connected these disparate concepts into a coherent whole. Keep practicing, keep questioning, and most importantly, keep falling in love with the process!