The number of values \(x\) satisfying the equation \(\left.\left.\cot ^{-1}[\sqrt{\\{x}(x+1)\\}\right]+\cos ^{-1}\left[\sqrt{(} x^{2}+x+1\right)\right]=(\pi / 2)\) is (a) 0 (b) 1 (c) 2 (d) 3

Short Answer

Expert verified
The number of values \(x\) satisfying the equation \(\cot ^{-1}[\sqrt{x(x+1)}]+\cos ^{-1}[\sqrt{x^{2}+x+1}]=\frac{\pi}{2}\) is (c) 2, which are \(x = 0\) and \(x = -1\).

Step by step solution

01

Examine the given equation and properties of inverse trigonometric functions

The given equation is: \(\cot ^{-1}[\sqrt{x(x+1)}]+\cos ^{-1}[\sqrt{x^{2}+x+1}]=\frac{\pi}{2}\) We should recall that the sum of the angles whose cotangent and cosine are complementary, i.e. \(\cot ^{-1}(A) + \cos ^{-1}(B) = \frac{\pi}{2}\) if and only if \(A*B=1\). So, we need to find x such that: \(\sqrt{x(x+1)}\cdot\sqrt{x^{2}+x+1} = 1\)
02

Square both sides of the equation

To remove the square roots, we will square both sides of the equation: \((\sqrt{x(x+1)}\cdot\sqrt{x^{2}+x+1})^{2} = 1^{2}\) This simplifies to: \(x(x+1)(x^{2}+x+1) = 1\)
03

Solve the resulting polynomial equation

Now we need to solve the polynomial equation: \(x(x+1)(x^{2}+x+1) = 1\) This simplifies to a quartic polynomial equation: \(x^{4}+2x^{3}+3x^{2}+2x = 0\)
04

Factor the polynomial equation

We observe that the quartic polynomial can be factored as: \((x^{2}+x)(x^{2}+x+1) = 0\) Now we have two equations to solve: 1. \(x^{2}+x=0\) 2. \(x^{2}+x+1=0\)
05

Solve the two quadratic equations

First, solve equation 1: \(x^{2}+x=0 \Rightarrow x(x+1) = 0\) This equation has two solutions: \(x = 0\) and \(x = -1\) Next, solve equation 2: \(x^{2}+x+1=0\) Using the quadratic formula, we find that the solutions of this equation are \(x = \frac{-1\pm i\sqrt{3}}{2}\), which are complex numbers. Since we are looking for real values of x, we discard the solutions from equation 2.
06

Count the number of real solutions

As there are two real solutions, \(x = 0\) and \(x = -1\), the correct answer is: (c) 2

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