JEE Main 2021 — Mathematics Algebra

The domain of the function

$$f(x)={\mathrm{sin}}^{-1}(\frac{3{x}^{2}+x-1}{(x-1{)}^{2}})+{\mathrm{cos}}^{-1}(\frac{x-1}{x+1})$$f(x)=sin−1((x−1)23x2+x−1​)+cos−1(x+1x−1​)

For, ${\mathrm{cos}}^{-1}(\frac{x-1}{x+1})$cos−1(x+1x−1​)

$$-1\leq \frac{x-1}{x+1}\leq 1$$−1≤x+1x−1​≤1

$$\Rightarrow -1\leq 1-\frac{2}{x+1}\leq 1$$⇒−1≤1−x+12​≤1

$$\Rightarrow -2\leq \frac{-2}{x+1}\leq 0$$⇒−2≤x+1−2​≤0

$$\Rightarrow 0\leq \frac{1}{x+1}\leq 1$$⇒0≤x+11​≤1

$$\Rightarrow x+1\in [1,\infty )$$⇒x+1∈[1,∞)

$$\Rightarrow x\in [0,\infty )....(i)$$⇒x∈[0,∞)....(i)

For, ${\mathrm{sin}}^{-1}(\frac{3{x}^{2}+x-1}{(x-1{)}^{2}})$sin−1((x−1)23x2+x−1​)

$$-1\leq \frac{3{x}^{2}+x-1}{{(x-1)}^{2}}\leq 1$$−1≤(x−1)23x2+x−1​≤1

$$\Rightarrow -{(x-1)}^{2}\leq 3{x}^{2}+x-1\leq {(x-1)}^{2},x\neq 1$$⇒−(x−1)2≤3x2+x−1≤(x−1)2,x=1

Now, $-{(x-1)}^{2}\leq 3{x}^{2}+x-1,x\neq 1$−(x−1)2≤3x2+x−1,x=1

$$\Rightarrow 4{x}^{2}-x\geq 0,x\neq 1$$⇒4x2−x≥0,x=1

$$\Rightarrow x(4x-1)\geq 0,x\neq 1$$⇒x(4x−1)≥0,x=1

$$\Rightarrow x\in (-\infty ,0]\cup [\frac{1}{4},\infty )-{1}....(ii)$$⇒x∈(−∞,0]∪[41​,∞)−1....(ii)

And $3{x}^{2}+x-1\leq {(x-1)}^{2},x\neq 1$3x2+x−1≤(x−1)2,x=1

$$\Rightarrow 2{x}^{2}+3x-2\leq 0,x\neq 1$$⇒2x2+3x−2≤0,x=1

$$\Rightarrow (x+2)(2x-1)\leq 0,x\neq 1$$⇒(x+2)(2x−1)≤0,x=1

$$\Rightarrow x\in [-2,\frac{1}{2}]....(iii)$$⇒x∈[−2,21​]....(iii)

Domain of the function ${\mathrm{sin}}^{-1}(\frac{3{x}^{2}+x-1}{(x-1{)}^{2}})$sin−1((x−1)23x2+x−1​) from the equations (ii)&(iii) is

$$\Rightarrow x\in [-2,0]\cup [\frac{1}{4},\frac{1}{2}]....(iv)$$⇒x∈[−2,0]∪[41​,21​]....(iv)

Now the domain of the given function will be the intersection of the equation (i)&(iv)

Hence, domain is $x\in [\frac{1}{4},\frac{1}{2}]\cup {0}$x∈[41​,21​]∪0