1. Given function

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

We need the range $[a,b]$ of $f$ on $[-1,1]$, then compute $4a-b$.

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2. Rewrite using an identity

For $x\in[-1,1]$, we use

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

So,

$4\cot^{-1}x=2\pi-4\tan^{-1}x$

Hence

$f(x)=2\cos^{-1}x-4\tan^{-1}x-3x^2-2x+10+2\pi$

This form is not the easiest for extrema, so we directly differentiate the original expression.

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3. Differentiate

We know:

$\frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}}$

and

$\frac{d}{dx}(\cot^{-1}x)=-\frac{1}{1+x^2}$

(standard JEE convention).

Therefore,

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

So,

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

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4. Check the sign of $f'(x)$ on $[-1,1]$

We examine each term:

• $-\dfrac{2}{\sqrt{1-x^2}}<0$ for $-1<x<1$
• $-\dfrac{4}{1+x^2}<0$ for all $x$
• $-6x-2$

Now on $[-1,1]$, even the third part cannot make the derivative positive. Let us verify more carefully.

At $x=-1$, the third part is

$-6(-1)-2=4$

But the first two negative terms are large in magnitude; in fact for any $x\in(-1,1)$,

$\frac{2}{\sqrt{1-x^2}}>0,\qquad \frac{4}{1+x^2}\ge 2$

Thus

$f'(x)< -0-2-6x-2 = -6x-4$

This estimate alone is not enough everywhere, so let us test the largest possible value of $f'(x)$, which would occur near smaller negative contributions. For example at $x=0$,

$f'(0)=-2-4-2=-8<0$

As $x\to \pm 1$, the term $-\dfrac{2}{\sqrt{1-x^2}}\to -\infty$, so $f'(x)$ is definitely negative near endpoints.

Also for $x\in[-1,0]$,

$-6x-2\le 4$

while

$-\frac{2}{\sqrt{1-x^2}}-\frac{4}{1+x^2}\le -2-2=-4$

and in fact strict inequality holds for $-1<x<1$, so

$f'(x)<0$

For $x\in[0,1]$, clearly $-6x-2\le -2$, and the other two terms are also negative, so again $f'(x)<0$.

Hence,

$f'(x)<0 \quad \text{for all } x\in(-1,1)$

Therefore $f$ is strictly decreasing on $[-1,1]$.

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5. Find maximum and minimum values

Since $f$ is decreasing:

• maximum at $x=-1$
• minimum at $x=1$

So,

$b=f(-1), \qquad a=f(1)$

Now compute them.

At $x=1$:

$\cos^{-1}(1)=0, \qquad \cot^{-1}(1)=\frac{\pi}{4}$

Thus,

$f(1)=2(0)+4\left(\frac{\pi}{4}\right)-3(1)^2-2(1)+10$

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

So,

$a=\pi+5$

At $x=-1$:

$\cos^{-1}(-1)=\pi, \qquad \cot^{-1}(-1)=\frac{3\pi}{4}$

Thus,

$f(-1)=2\pi+4\left(\frac{3\pi}{4}\right)-3(1)-2(-1)+10$

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

$f(-1)=5\pi+9$

So,

$b=5\pi+9$

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6. Compute $4a-b$

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

$=4\pi+20-5\pi-9$

$=11-\pi$

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7. Final answer

$\boxed{11-\pi}$

So the correct option is B.