JEE Main 2021 — Mathematics Algebra

Given function is $f(x)=\frac{{\mathrm{cos}}^{-1}\sqrt{{x}^{2}-x+1}}{\sqrt{{\mathrm{sin}}^{-1}(\frac{2x-1}{2})}}$f(x)=sin−1(22x−1​)​cos−1x2−x+1​​

For, {\mathrm{sin}}^{-1}x&{\mathrm{cos}}^{1}x to be defined $-1\leq x\leq 1$−1≤x≤1 and for $\sqrt{f(x)}$f(x)​ to be defined $f(x)\geq 0$f(x)≥0 and the denominator of any expression can never be zero.

Thus, to define the domain of the given function, we have

$-1\leq \sqrt{{x}^{2}-x+1}\leq 1,{x}^{2}-x+1\geq 0$−1≤x2−x+1​≤1,x2−x+1≥0 and ${\mathrm{sin}}^{-1}(\frac{2x-1}{2})>0$sin−1(22x−1​)>0

Taking the common values and using the range of ${\mathrm{sin}}^{-1}x$sin−1x i.e. for positive values $0<{\mathrm{sin}}^{-1}x\leq \frac{\pi }{2},$0<sin−1x≤2π​, we get

$0\leq {x}^{2}-x+1\leq 1$0≤x2−x+1≤1 and $0<{\mathrm{sin}}^{-1}(\frac{2x-1}{2})\leq \frac{\pi }{2}$0<sin−1(22x−1​)≤2π​

$\Rightarrow {x}^{2}-x\leq 0$⇒x2−x≤0 and $0<\frac{2x-1}{2}\leq 1$0<22x−1​≤1

$\Rightarrow x(x-1)\leq 0$⇒x(x−1)≤0 and $0<2x-1\leq 2$0<2x−1≤2

$\Rightarrow x\in [0,1]$⇒x∈[0,1] and $1<2x\leq 3$1<2x≤3

$\Rightarrow x\in [0,1]$⇒x∈[0,1] and $\frac{1}{2}<x\leq \frac{3}{2}$21​<x≤23​

Now, taking intersection, we get $x\in (\frac{1}{2},1]$x∈(21​,1]

Given, domain of the function is $(\alpha ,\beta ]=(\frac{1}{2},1]$(α,β]=(21​,1]

$$\Rightarrow \alpha =\frac{1}{2},\beta =1$$⇒α=21​,β=1

$$\Rightarrow \alpha +\beta =\frac{3}{2}.$$⇒α+β=23​.