JEE Main 2022 — Mathematics Trigonometry

Given,

$$2\mathrm{cos}(\frac{{x}^{2}+x}{6})={4}^{x}+{4}^{-x}$$2cos(6x2+x​)=4x+4−x

Now we know that maximum value of $\mathrm{cos}$cos function is $1$1, so $2\mathrm{cos}(\frac{{x}^{2}+x}{6})\leq 2$2cos(6x2+x​)≤2 and using $A.M\geq G.M$A.M≥G.M we get ${4}^{x}+{4}^{-x}\geq 2$4x+4−x≥2

So, L.H.S.\leq 2.&R.H.s\geq 2

Hence L.H.S.=2&R.H.S.=2

Now equating both to $2$2 we get,

$2\mathrm{cos}(\frac{{x}^{2}+x}{6})=2$2cos(6x2+x​)=2 and ${4}^{x}+{4}^{-x}=2$4x+4−x=2

Now solving $2\mathrm{cos}(\frac{{x}^{2}+x}{6})=2$2cos(6x2+x​)=2

$$\Rightarrow \mathrm{cos}(\frac{{x}^{2}+x}{6})=1$$⇒cos(6x2+x​)=1

$$\Rightarrow \frac{{x}^{2}+x}{6}=0$$⇒6x2+x​=0

$$\Rightarrow {x}^{2}+x=0$$⇒x2+x=0

$$\Rightarrow x=0\text{or}-1...........(1)$$⇒x=0or−1...........(1)

Now solving ${4}^{x}+{4}^{-x}=2$4x+4−x=2 we get $x=0.....(2)$x=0.....(2)

Now from equation (1)&(2) we can see common solution is $x=0$x=0

Hence, possible solution is only one.