JEE Main 2021 — Mathematics Trigonometry

Given equation is

$${\mathrm{sin}}^{-1}[{x}^{2}+\frac{1}{3}]+{\mathrm{cos}}^{-1}[{x}^{2}-\frac{2}{3}]={x}^{2}$$sin−1[x2+31​]+cos−1[x2−32​]=x2

Now, ${\mathrm{sin}}^{-1}[{x}^{2}+\frac{1}{3}]$sin−1[x2+31​] is defined if $-1\leq [{x}^{2}+\frac{1}{3}]\leq 1$−1≤[x2+31​]≤1

$$\Rightarrow -1\leq {x}^{2}+\frac{1}{3}<2$$⇒−1≤x2+31​<2

$$\Rightarrow \frac{-4}{3}\leq {x}^{2}<\frac{5}{3}$$⇒3−4​≤x2<35​

$$\Rightarrow 0\leq {x}^{2}<\frac{5}{3}...(1)$$⇒0≤x2<35​...(1)

Also, and ${\mathrm{cos}}^{-1}[{x}^{2}-\frac{2}{3}]$cos−1[x2−32​] is defined if $-1\leq [{x}^{2}-\frac{2}{3}]\leq 1$−1≤[x2−32​]≤1

$$\Rightarrow -1\leq {x}^{2}-\frac{2}{3}<2$$⇒−1≤x2−32​<2

$$\Rightarrow \frac{-1}{3}\leq {x}^{2}<\frac{8}{3}$$⇒3−1​≤x2<38​

$$\Rightarrow 0\leq {x}^{2}<\frac{8}{3}...(2)$$⇒0≤x2<38​...(2)

So, from $(1)$(1) and $(2)$(2) we can conclude $0\leq {x}^{2}<\frac{5}{3}$0≤x2<35​

Case -$I:$I: If $0\leq {x}^{2}<\frac{2}{3}$0≤x2<32​

$$\Rightarrow {\mathrm{sin}}^{-1}(0)+{\mathrm{cos}}^{-1}(-1)={x}^{2}$$⇒sin−1(0)+cos−1(−1)=x2

$$\Rightarrow 0+\pi ={x}^{2}$$⇒0+π=x2

$\Rightarrow {x}^{2}=\pi$⇒x2=π but $\pi \notin [0,\frac{2}{3})$π∈/[0,32​)

$\Rightarrow$⇒ No value of $'x'$′x′

Case - $II:$II: If $\frac{2}{3}\leq {x}^{2}<\frac{5}{3}$32​≤x2<35​

$$\Rightarrow {\mathrm{sin}}^{-1}(1)+{\mathrm{cos}}^{-1}(0)={x}^{2}$$⇒sin−1(1)+cos−1(0)=x2

$$\Rightarrow \frac{\pi }{2}+\frac{\pi }{2}={x}^{2}$$⇒2π​+2π​=x2

$\Rightarrow {x}^{2}=\pi$⇒x2=π but $\pi \notin [\frac{2}{3},\frac{5}{3})$π∈/[32​,35​)

$\Rightarrow$⇒ No value of $'x'$′x′

So, number of solutions of the equation is zero.