NIMCET 2009 — Mathematics PYQ

To find the number of solutions for the given equation, we must first determine the domain where both inverse trigonometric functions are defined.

1. Analyzing the Domain of ${\tan }^{-1}\sqrt{x(x+1)}$

For the square root $\sqrt{x(x+1)}$ to be a real number, the expression inside must be non-negative:

This inequality holds when:

• $x\ge 0$ or $x\le -1$

2. Analyzing the Domain of ${\sin }^{-1}\sqrt{x^2+x+1}$

For ${\sin }^{-1}(f(x))$ to be defined, the argument must satisfy $-1\le f(x)\le 1$. Since we have a square root, it must be $0\le \sqrt{x^2+x+1}\le 1$.

Squaring the inequality:

Solving $x^2+x+1\le 1$:

This inequality holds when:

• $-1\le x\le 0$

3. Finding the Intersection of Domains

We have two conditions for $x$:

1. From the first term: $x\in (-\infty ,-1]\cup [0,\infty )$

2. From the second term: $x\in [-1,0]$

The only values of $x$ that satisfy both conditions are the boundary points:

4. Verifying the Solutions

Now, we substitute these values back into the original equation to see if they satisfy it.

• Case 1: $x=0$

  $${\tan }^{-1}\sqrt{0(1)}+{\sin }^{-1}\sqrt{0^2+0+1}$$
  $${\tan }^{-1}(0)+{\sin }^{-1}(1)=0+\frac{\pi }{2}=\frac{\pi }{2}$$

  (This is a valid solution)

• Case 2: $x=-1$

  $${\tan }^{-1}\sqrt{-1(0)}+{\sin }^{-1}\sqrt{(-1{)}^2+(-1)+1}$$
  $${\tan }^{-1}(0)+{\sin }^{-1}\sqrt{1-1+1}=0+{\sin }^{-1}(1)=\frac{\pi }{2}$$

  (This is also a valid solution)

Conclusion

There are exactly two values of $x$ ($0$ and $-1$) that satisfy the equation.

Correct Option:

(c) two