Animated Solution for Mathematics - Differentiation: Comprehension Passage

Navigating Functions and Calculus

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the landscape of functions, where algebraic manipulation meets the profound beauty of calculus.

JEE Advanced problems are rarely about brute force; they are about finding the most elegant path through the woods. Let us dissect this problem together, step by step.

We begin with the function $f(x)=(1-x{)}^2{\sin }^{2}x+x^2$. Statement P asks if there exists an $x\in \mathbb{R}$ such that $f(x)+2x=2(1+x^2)$.

At first glance, this looks like a daunting algebraic equation. Instead of panicking, let us substitute the definition of $f(x)$ into the equation:

Now, let us move all the polynomial terms to the right side to see what remains:

Here is where the "Aha!" moment strikes. Look at the right-hand side: $x^2-2x+2$. It is a perfect square in disguise, which we can rewrite as $(x-1{)}^2+1$.

Since $(x-1{)}^2$ is identical to $(1-x{)}^2$, our equation becomes:

If we bring the squared term to the left, we get:

Since ${\sin }^{2}x-1=-{\cos }^{2}x$, we arrive at:

Stop right there. A square of a real number is always non-negative, and the product of two squares is also non-negative. It can never be $-1$.

We have hit a wall, but that wall is our answer: Statement P is impossible and therefore false.

Statement Q

Now, let us look at Statement Q: $2f(x)+1=2x(1+x)$. This asks for the existence of a root.

Whenever you see "exists some $x$," think of the Intermediate Value Theorem (IVT). Let us define a new function $H(x)=2f(x)+1-2x(1+x)$.

We want to know if $H(x)=0$ has a solution. Let us test the boundaries: At $x=0$, $f(0)=(1-0{)}^2{\sin }^2(0)+0^2=0$. Thus, $H(0)=2(0)+1-0=1$, which is positive.

At $x=1$, $f(1)=(1-1{)}^2{\sin }^2(1)+1^2=1$. Thus, $H(1)=2(1)+1-2(1)(2)=3-4=-1$, which is negative.

Because $H(x)$ is continuous and changes sign from positive to negative between $x=0$ and $x=1$, the IVT guarantees that there must be a root. Statement Q is true!

The Integral Function

Finally, we arrive at the integral function:

The question asks about the monotonicity of $g(x)$ for $x>1$. Many students would try to integrate this, but that is a trap. We only need the derivative.

By the Leibniz Rule:

Since $f(x)$ is a sum of squares, it is always positive. The sign of $g^{\prime }(x)$ depends entirely on $h(x)=\frac{2(x-1)}{x+1}-\ln x$.

To understand $h(x)$, let us differentiate it:

Using the quotient rule, we find:

For $x>1$, this derivative is strictly negative. This means $h(x)$ is a strictly decreasing function.

Since $h(1)=0$ and the function is decreasing, $h(x)$ must be negative for all $x>1$. Consequently, $g^{\prime }(x)<0$, which proves that $g(x)$ is strictly decreasing.