Question

If $A\le2\sin^{-1}x+\cos^{-1}x\le B$, then A and B respectively, are

• A. $\frac{\pi}{2}$ and $\pi$
• B. $0$ and $2\pi$
• C. $\frac{\pi}{4}$ and $\frac{\pi}{2}$
• D. $\frac{\pi}{4}$ and $\pi$
• E. $0$ and $\pi$

Hint:

Math Tip: Whenever you see a linear combination of inverse sine and inverse cosine, always extract the pair $(\sin^{-1}x + \cos^{-1}x)$ to swap it out for the constant $\frac{\pi}{2}$. This instantly reduces a two-variable problem into a single-variable limit evaluation.

Answer 1

Answer: (E)

Concept:
Inverse Trigonometric Functions - Complementary Angle Identity.
Recall the standard complementary identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
Step 1: Rewrite the function to utilize the identity.
Let $f(x) = 2\sin^{-1}x + \cos^{-1}x$. Split the $2\sin^{-1}x$ term into two separate terms: $$ f(x) = \sin^{-1}x + \sin^{-1}x + \cos^{-1}x $$
Step 2: Apply the complementary identity.
Substitute $\frac{\pi}{2}$ for the $(\sin^{-1}x + \cos^{-1}x)$ portion of the expression: $$ f(x) = \sin^{-1}x + \left( \sin^{-1}x + \cos^{-1}x \right) $$ $$ f(x) = \sin^{-1}x + \frac{\pi}{2} $$
Step 3: Determine the bounded range of the base function.
To find the minimum and maximum values of $f(x)$, we must look at the range of the remaining variable term, $\sin^{-1}x$. The principal range of $\sin^{-1}x$ is: $$ -\frac{\pi}{2} \le \sin^{-1}x \le \frac{\pi}{2} $$
Step 4: Calculate the minimum value (A).
Substitute the minimum possible value of $\sin^{-1}x$ into our equation for $f(x)$: $$ A = \left(-\frac{\pi}{2}\right) + \frac{\pi}{2} $$ $$ A = 0 $$
Step 5: Calculate the maximum value (B).
Substitute the maximum possible value of $\sin^{-1}x$ into our equation for $f(x)$: $$ B = \left(\frac{\pi}{2}\right) + \frac{\pi}{2} $$ $$ B = \pi $$ Therefore, the values are $A = 0$ and $B = \pi$.