The number of elements in the set [S=x R: 2 (x^2+x6)=4^x+4^-x] is :

Question

The number of elements in the set $$S=\left\{x \in \mathbb{R}: 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\}$$ is :

Accepted Answer

Given,

$$2\cos \left( {{{{x^2} + x} \over 6}} \right) = {4^x} + {4^{ - x}}$$

We know,

A.M $$\ge$$ G.M

$$\therefore$$ for 4x and 4-x

$${{{4^x} + {4^{ - x}}} \over 2} \ge \sqrt {{4^x}\,.\,{4^{ - x}}} $$

$$ \Rightarrow {{{4^x} + {4^{ - x}}} \over 2} \ge 1$$

$$ \Rightarrow {4^x} + {4^{ - x}} \ge 2$$ ...... (1)

And we know,

$$ - 1 \le \cos x \le 1$$

$$\therefore$$ $$ - 1 \le \cos \left( {{{{x^2} + x} \over 6}} \right) \le 1$$

$$ - 2 \le 2\cos \left( {{{{x^2} + x} \over 6}} \right) \le 2$$ ...... (2)

(1) and (2) both satisfies only when both equal to 2.

$$\therefore$$ $$2\cos \left( {{{{x^2} + x} \over 6}} \right) = 2$$

$$ \Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = 1$$

$$ \Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = \cos 0$$

$$ \Rightarrow {{{x^2} + x} \over 6} = 0$$

$$ \Rightarrow {x^2} + x = 0$$

$$ \Rightarrow x(x + 1) = 0$$

$$ \Rightarrow x = 0, - 1$$

When $x = 0$,

L.H.S $$ = 2\cos \left( {{{{x^2} + x} \over 6}} \right)$$

$$ = 2\cos \left( {{0 \over 6}} \right)$$

$$ = 2\cos 0$$

$$ = 2\,.\,1$$

$= 2$

R.H.S $= {4^x} + {4^{ - x}}$

$= {4^0} + {4^0}$

$= 2$

$$\therefore$$ $x = 0$ is accepted.

Now, when $x =  - 1$,

L.H.S $$ = 2\cos \left( {{{{x^2} + x} \over 6}} \right)$$

$$ = 2\cos \left( {{{1 - 1} \over 6}} \right)$$

$$ = 2\cos 0 = 2$$

R.H.S $= {4^x} + {4^{ - x}}$

$= {4^{ - 1}} + {4^1}$

$$ = {1 \over 4} + 4$$

$$ = {{17} \over 4}$$

$$\therefore$$ L.H.S $$\ne$$ R.H.S

$$\therefore$$ $x =  - 1$ is not a solution.

$$\therefore$$ Only one solution possible which is $x = 0$.

Solutions · 1