Animated Solution for Mathematics - Differentiation: Let $f(x)=3^{(x^2-2{)}^3+4},x\in \mathbb{R}$. Then which of the following statements are true ? $P:x=0$ is a point of local minima of $f$ $Q:x=\sqrt{2}$ is a point of inflection of $f$ $R:f^{\prime }$ is increasing for $x>\sqrt{2}$

My dear student, welcome to the world of advanced calculus. Today, we are going to dissect a function that, at first glance, might make your heart skip a beat: $f(x)=3^{(x^2-2{)}^3+4}$.

It looks like a monster, doesn't it? An exponential function with a cubic polynomial in the exponent. But remember, in JEE Advanced, the most intimidating problems often hide the most elegant, simple truths. Let us peel back the layers together.

The First Derivative as Our Compass

To understand the behavior of any function—where it rises, where it falls, and where it rests—we need our primary weapon: the first derivative. We apply the chain rule here. Recall that the derivative of $a^{g(x)}$ is $a^{g(x)}\cdot \ln a\cdot g^{\prime }(x)$.

Let $g(x)=(x^2-2{)}^3+4$. Its derivative is $g^{\prime }(x)=3(x^2-2{)}^2\cdot (2x)=6x(x^2-2{)}^2$.

Putting it all together, we get:

Look at this expression. The term $3^{(x^2-2{)}^3+4}$ is always positive. The term $(x^2-2{)}^2$ is a perfect square, so it is always non-negative.

This means the sign of $f^{\prime }(x)$ is determined entirely by the term $x$. If $x<0$, $f^{\prime }(x)<0$. If $x>0$, $f^{\prime }(x)>0$.

The Valley at $x=0$

Imagine you are walking along the graph of this function. As you approach $x=0$ from the left, the slope is negative—you are walking downhill.

As you pass $x=0$, the slope becomes positive—you are walking uphill. This change from negative to positive slope is the classic signature of a local minima.

Thus, statement P is undeniably true. You have successfully identified the valley in our landscape!

The Concavity Shift at $x=\sqrt{2}$

Now, let us tackle the more complex territory: the point of inflection. Statement Q claims $x=\sqrt{2}$ is a point of inflection. To verify this, we need the second derivative $f^{\prime \prime }(x)$.

Applying the product rule to our $f^{\prime }(x)$ expression, we arrive at:

At $x=\sqrt{2}$, the term $(x^2-2)$ becomes zero, so $f^{\prime \prime }(\sqrt{2})=0$. But remember, $f^{\prime \prime }(x)=0$ is only a candidate. We must check if the concavity actually changes.

Near $x=\sqrt{2}$, the bracketed term $[5x^2-2+6x^2(x^2-2{)}^3\ln 3]$ evaluates to $8$, which is positive. Therefore, the sign of $f^{\prime \prime }(x)$ near $\sqrt{2}$ is governed by the sign of $(x^2-2)$.

For $x<\sqrt{2}$, $(x^2-2)<0$, so $f^{\prime \prime }(x)<0$ (concave down). For $x>\sqrt{2}$, $(x^2-2)>0$, so $f^{\prime \prime }(x)>0$ (concave up).

Because the concavity changes, $x=\sqrt{2}$ is indeed a point of inflection. Statement Q is true!

The Final Ascent

Finally, let us look at statement R: is $f^{\prime }(x)$ increasing for $x>\sqrt{2}$? A function increases when its derivative is positive. So, we need to check if the derivative of $f^{\prime }(x)$—which is $f^{\prime \prime }(x)$—is positive for $x>\sqrt{2}$.

We already analyzed the sign of $f^{\prime \prime }(x)$ in the previous step. For $x>\sqrt{2}$, we found that $(x^2-2)>0$ and the bracketed term is also positive.

Since all factors are positive, $f^{\prime \prime }(x)>0$ for all $x>\sqrt{2}$. This confirms that $f^{\prime }(x)$ is strictly increasing. Statement R is true!

My dear student, look at what you have achieved. You navigated through the chain rule, product rule, and sign analysis to prove that all three statements—P, Q, and R—are correct.

This is the essence of JEE Advanced: not just calculating, but visualizing the behavior of functions. Keep this curiosity alive, and no problem will ever be too intimidating again.