Finding the range of a mixed inverse trigonometric function

Expert intro: This question tests how to combine inverse trigonometric ranges with a quadratic expression on a closed interval.

Detailed walkthrough

Step 1: Split the function into manageable parts

We are given

$f(x)=2\cos^{-1}x+4\cot^{-1}x-3x^2-2x+10, \qquad x\in[-1,1].$

To find the range, it is natural to inspect the behavior of each term. The inverse trig part depends on standard identities, while the polynomial part is smooth on the closed interval. Since the interval is compact, the absolute maximum and minimum must occur at endpoints or where the derivative is zero.

Now use the identity valid for $x\in[-1,1]$:

$\cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x,$

but an even cleaner route is to differentiate directly and observe monotonicity.

Step 2: Differentiate and simplify

Differentiate term by term:

$f'(x)=2\cdot\frac{-1}{\sqrt{1-x^2}}+4\left(-\frac{1}{1+x^2}\right)-6x-2.$

So

$f'(x)=-\frac{2}{\sqrt{1-x^2}}-\frac{4}{1+x^2}-6x-2.$

Every term here is non-positive on $[-1,1]$, and in fact the first two are strictly negative in the interior. Also, $-6x-2\le 4$ on the interval, but the inverse-trig derivatives dominate enough to keep the function decreasing throughout the interval. Hence the maximum occurs at the left endpoint and the minimum at the right endpoint.

Evaluate the endpoints:

At $x=-1$, $f(-1)=2\cos^{-1}(-1)+4\cot^{-1}(-1)-3(1)-2(-1)+10.$ Using $\cos^{-1}(-1)=\pi$ and the standard principal value $\cot^{-1}(-1)=\frac{3\pi}{4}$,

$f(-1)=2\pi+3\pi-3+2+10=11+5\pi.$

At $x=1$, $f(1)=2\cos^{-1}(1)+4\cot^{-1}(1)-3-2+10,$ with $\cos^{-1}(1)=0$ and $\cot^{-1}(1)=\frac{\pi}{4}$, so

$f(1)=\pi+5.$

Thus the range is $[\pi+5,\,11+5\pi]$. Therefore $a=\pi+5$ and $b=11+5\pi$, giving

$4a-b=4(\pi+5)-(11+5\pi)=9-\pi.$

Key idea to remember

For range problems on a closed interval, the safest method is endpoint checking after confirming monotonicity. Inverse trig derivatives often make the sign analysis straightforward. When the function includes standard principal inverse trig values, endpoint evaluation can produce the exact interval endpoints quickly.

💡 Pitfall guide

A common mistake is to treat $\cot^{-1}x$ as if it were undefined on negative inputs or to use the wrong principal-value convention. In JEE-style problems, the principal range matters, and $\cot^{-1}(-1)$ is not $-\frac{\pi}{4}$ under the standard convention. Another frequent error is to guess the range from the polynomial part alone and ignore how the inverse trig terms shift the endpoints. Always check the interval, the derivative sign, and the exact principal values at $x=-1$ and $x=1$.

🔄 Real-world variant

If the question changed to $f(x)=2\cos^{-1}x+4\cot^{-1}x-3x^2-2x+c$ on $[-1,1]$, the same monotonicity argument would still work. Only the final numerical endpoints would shift by the constant $c-10$. If the interval were instead $( -1,1 )$, the function would still have the same supremum and infimum values, but they would no longer be attained at endpoints because the interval is open. That variant tests whether you know the difference between range and attained extrema.

🔍 Related terms

inverse trigonometric functions, principal value branch, absolute extrema on a closed interval