Animated Solution for Mathematics - Differentiation: The range of $a\in \mathbb{R}$ for which the function $f(x)=(4a-3)(x+{\log }_{e}5)+2(a-7)\cot \left(\frac{x}{2}\right){\sin }^2\left(\frac{x}{2}\right)$, $x\ne 2n\pi ,n\in \mathbb{N}$, has critical points, is

The given function is:

Many students would immediately reach for the product rule, but the true master of JEE calculus knows that simplification is the most powerful tool. Let us focus on the trigonometric term: $2\cot \left(\frac{x}{2}\right){\sin }^2\left(\frac{x}{2}\right)$.

By replacing $\cot \left(\frac{x}{2}\right)$ with $\frac{\cos (x/2)}{\sin (x/2)}$, we observe a beautiful cancellation:

Applying the double angle identity, this collapses into $\sin x$. Our function is now transformed into the elegant form:

With a clean function, finding the derivative becomes straightforward. The linear term $(4a-3)x$ differentiates to $4a-3$, the constant ${\log }_{e}5$ vanishes, and the derivative of $(a-7)\sin x$ is $(a-7)\cos x$.

Thus, the derivative is:

For the function to have critical points, the tangent must be horizontal, implying $f^{\prime }(x)=0$. This yields the equation:

We seek the range of $a$ such that this equation has a solution for $x$. We know that for any real $x$, the cosine function is bounded by:

Therefore, our expression must satisfy the same interval:

We split this into two inequalities. First, $\frac{3-4a}{a-7}\ge -1$: Using the wavy curve method, we find $a\in [-\frac{4}{3},7)$.

Second, $\frac{3-4a}{a-7}\le 1$: The wavy curve method reveals $a\in (-\infty ,2]\cup (7,\infty )$.

To find the final range, we determine the intersection of these two sets. The overlap of $[-\frac{4}{3},7)$ and $(-\infty ,2]\cup (7,\infty )$ is exactly $[-\frac{4}{3},2]$.

The final range of $a$ is $[-\frac{4}{3},2]$.