JEE Main 2024 — Application Of Derivatives Question with Solution

To find the number of critical points of the function $$f(x)=(x-2{)}^{2/3}(2x+1)$$, we need to determine where its derivative $$f^{\prime }(x)$$ is equal to zero or undefined. Critical points occur where the derivative is zero or does not exist.

First, let's find the derivative of the function:

$$f(x)=(x-2{)}^{2/3}(2x+1)$$

We apply the product rule for differentiation, which states that $$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$$, where $$u=(x-2{)}^{2/3}$$ and $$v=2x+1$$.

We need the derivatives of $$u$$ and $$v$$:

For $$u=(x-2{)}^{2/3}$$, we use the chain rule:

$$u^{\prime }(x)=\frac{d}{dx}[(x-2{)}^{2/3}]=\frac{2}{3}(x-2{)}^{-1/3}\cdot 1=\frac{2}{3}(x-2{)}^{-1/3}$$

The derivative of $$v$$ is straightforward, as $$v=2x+1$$:

$$v^{\prime }(x)=2$$

Now we apply the product rule:

$$f^{\prime }(x)=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$

Substituting $$u$$, $$u^{\prime }$$, and $$v$$, we get:

$$f^{\prime }(x)=\left(\frac{2}{3}(x-2{)}^{-1/3}\right)(2x+1)+\left((x-2{)}^{2/3}\right)(2)$$

This simplifies to:

$$f^{\prime }(x)=\frac{2(2x+1)}{3(x-2{)}^{1/3}}+2(x-2{)}^{2/3}$$

For critical points, we need to solve $$f^{\prime }(x)=0$$ or where it is undefined.

1. Solve for where the derivative is zero:

$$\frac{2(2x+1)}{3(x-2{)}^{1/3}}+2(x-2{)}^{2/3}=0$$

Combining like terms in a common denominator, we get:

$$\frac{2(2x+1)+6(x-2)}{3(x-2{)}^{1/3}}=0$$

Simplifying the numerator:

$$2(2x+1)+6(x-2)=4x+2+6x-12=10x-10=10(x-1)$$

So:

$$\frac{10(x-1)}{3(x-2{)}^{1/3}}=0$$

The numerator is zero when:

$$10(x-1)=0$$

Therefore:

$$x=1$$

2. Solve for where the derivative is undefined:

The denominator, $$3(x-2{)}^{1/3}$$, is undefined when $$(x-2{)}^{1/3}=0$$, which happens at:

$$x=2$$

From the above analysis, the critical points are at $$x=1$$ and $$x=2$$. Thus, there are 2 critical points.

Therefore, the number of critical points of the function $$f(x)=(x-2{)}^{2/3}(2x+1)$$ is:

Option A