117247
If \(g(x)=x^2+x-2\) and \(\frac{1}{2}(\operatorname{gof} f)(x)=2 x^2-5 x+2\) then one such function \(f(x)=\)
1 \(2 x-3\)
2 \(2 x+3\)
3 \(2+2 x\)
4 \(2 x^2-3 x-1\)
Explanation:
A Given, Let, \(g(x)=x^2+x-2\) and \(\frac{1}{2}(\) gof \()(x)=2 x^2-5 x+2\) \(\text { Let, } \mathrm{f}(\mathrm{x}) =(2 \mathrm{x}-3)\) \(\operatorname{gof}(\mathrm{x}) =(2 \mathrm{x}-3)^2+(2 \mathrm{x}-3)-2\) \(=4 \mathrm{x}^2+9-12 \mathrm{x}+2 \mathrm{x}-3-2\) \(=4 \mathrm{x}^2-10 \mathrm{x}+4\) \(\therefore \quad \frac{1}{2} \operatorname{gof}(\mathrm{x}) =2 \mathrm{x}^2-5 \mathrm{x}+2\) \(\therefore \quad \mathrm{f}(\mathrm{x}) =2 \mathrm{x}-3\)
Shift-II
Sets, Relation and Function
117252
Let \(\mathbf{f}: \mathbf{R}-\left\{\frac{-1}{2}\right\} \rightarrow \mathbf{R}\) be defined by \(\mathbf{f}(\mathbf{x})=\) \(\frac{\mathrm{x}-2}{2 \mathrm{x}+1}\). If \(\alpha, \beta\) satisfy the equation \(f(f(x))=-x\), then \(4\left(\alpha^2+\beta^2\right)\)
1 17
2 12
3 24
4 34
Explanation:
A Let \(f: \mathrm{R}-\left\{\frac{1}{2}\right\} \rightarrow \mathrm{R}\) and \(f(\mathrm{x})=\frac{\mathrm{x}-2}{2 \mathrm{x}+1},\) The, \(f(f(\mathrm{x}))=\frac{\frac{\mathrm{x}-2}{2 \mathrm{x}+1}-2}{2\left(\frac{\mathrm{x}-2}{2 \mathrm{x}+1}\right)+1}\) \(-x=\frac{x-2-2(2 x+1)}{2(x-2)+2 x+1}\) \(-x=\frac{x-2-4 x-2}{2 x-4+2 x+1}\) \(-x=\frac{-3 x-4}{4 x-3}\) \(\Rightarrow \quad-3 x-4=-4 x^2+3 x\) \(\text { Or } \quad 4 x^2-6 x-4=0\) \(2 x^2-3 x-2=0\) Or Given \(\alpha, \beta\) satisfy the above equation. \(\therefore \alpha+\beta=\frac{3}{2}\left(\begin{array}{l} \because \alpha+\beta=\frac{-b}{a} \\ \alpha \times \beta=\frac{c}{a} \end{array}\right)\) \(\alpha \beta=\frac{-2}{2}=-1\) \(\therefore \quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=\left(\frac{3}{2}\right)^2-2(-1)=\frac{9}{4}+2=\frac{17}{4}\) \(\therefore \quad 4\left(\alpha^2+\beta^2\right)=17\)
Shift-I
Sets, Relation and Function
117253
If \(f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)\) and \(g\left(\frac{5}{4}\right)=1\), then \(\operatorname{gof}(x)\) is equal to
1 0
2 2
3 1
4 3
Explanation:
C We have- \(f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)\) \(f(x)=\sin ^2 x+\left[\frac{1}{2} \sin x+\cos x \frac{\sqrt{3}}{2}\right]^2+\cos x\left[\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right]\) \(f(x)=\sin ^2 x+\frac{1}{4}[\sqrt{3} \cos x+\sin x]^2+\frac{1}{2}\left[\cos ^2 x-\sqrt{3} \sin x \cos x\right]\) \(f(x)=\sin ^2 x+\frac{3}{4} \cos ^2 x+\frac{\sin ^2 x}{4}+\frac{\sqrt{3}}{2} \sin x \cos x+\frac{\cos ^2 x}{2}-\frac{\sqrt{3}}{2} \sin x \cos x\) \(f(x)=\frac{5}{4} \sin ^2 x+\frac{5}{4} \cos ^2 x\) \(f(x)=\frac{5}{4}\left(\sin ^2 x+\cos ^2 x\right)\) \(f(x)=\frac{5}{4}\) Now, \(\operatorname{gof}(\mathrm{x})=\mathrm{g}(\mathrm{f}(\mathrm{x}))=\mathrm{g}\left(\frac{5}{4}\right)=1\)
Assam CEE-2017
Sets, Relation and Function
117255
If \(f(x)=\frac{x}{\sqrt{1+x^2}}\), then \(f \circ f \circ f(x)\) is equal to
117247
If \(g(x)=x^2+x-2\) and \(\frac{1}{2}(\operatorname{gof} f)(x)=2 x^2-5 x+2\) then one such function \(f(x)=\)
1 \(2 x-3\)
2 \(2 x+3\)
3 \(2+2 x\)
4 \(2 x^2-3 x-1\)
Explanation:
A Given, Let, \(g(x)=x^2+x-2\) and \(\frac{1}{2}(\) gof \()(x)=2 x^2-5 x+2\) \(\text { Let, } \mathrm{f}(\mathrm{x}) =(2 \mathrm{x}-3)\) \(\operatorname{gof}(\mathrm{x}) =(2 \mathrm{x}-3)^2+(2 \mathrm{x}-3)-2\) \(=4 \mathrm{x}^2+9-12 \mathrm{x}+2 \mathrm{x}-3-2\) \(=4 \mathrm{x}^2-10 \mathrm{x}+4\) \(\therefore \quad \frac{1}{2} \operatorname{gof}(\mathrm{x}) =2 \mathrm{x}^2-5 \mathrm{x}+2\) \(\therefore \quad \mathrm{f}(\mathrm{x}) =2 \mathrm{x}-3\)
Shift-II
Sets, Relation and Function
117252
Let \(\mathbf{f}: \mathbf{R}-\left\{\frac{-1}{2}\right\} \rightarrow \mathbf{R}\) be defined by \(\mathbf{f}(\mathbf{x})=\) \(\frac{\mathrm{x}-2}{2 \mathrm{x}+1}\). If \(\alpha, \beta\) satisfy the equation \(f(f(x))=-x\), then \(4\left(\alpha^2+\beta^2\right)\)
1 17
2 12
3 24
4 34
Explanation:
A Let \(f: \mathrm{R}-\left\{\frac{1}{2}\right\} \rightarrow \mathrm{R}\) and \(f(\mathrm{x})=\frac{\mathrm{x}-2}{2 \mathrm{x}+1},\) The, \(f(f(\mathrm{x}))=\frac{\frac{\mathrm{x}-2}{2 \mathrm{x}+1}-2}{2\left(\frac{\mathrm{x}-2}{2 \mathrm{x}+1}\right)+1}\) \(-x=\frac{x-2-2(2 x+1)}{2(x-2)+2 x+1}\) \(-x=\frac{x-2-4 x-2}{2 x-4+2 x+1}\) \(-x=\frac{-3 x-4}{4 x-3}\) \(\Rightarrow \quad-3 x-4=-4 x^2+3 x\) \(\text { Or } \quad 4 x^2-6 x-4=0\) \(2 x^2-3 x-2=0\) Or Given \(\alpha, \beta\) satisfy the above equation. \(\therefore \alpha+\beta=\frac{3}{2}\left(\begin{array}{l} \because \alpha+\beta=\frac{-b}{a} \\ \alpha \times \beta=\frac{c}{a} \end{array}\right)\) \(\alpha \beta=\frac{-2}{2}=-1\) \(\therefore \quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=\left(\frac{3}{2}\right)^2-2(-1)=\frac{9}{4}+2=\frac{17}{4}\) \(\therefore \quad 4\left(\alpha^2+\beta^2\right)=17\)
Shift-I
Sets, Relation and Function
117253
If \(f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)\) and \(g\left(\frac{5}{4}\right)=1\), then \(\operatorname{gof}(x)\) is equal to
1 0
2 2
3 1
4 3
Explanation:
C We have- \(f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)\) \(f(x)=\sin ^2 x+\left[\frac{1}{2} \sin x+\cos x \frac{\sqrt{3}}{2}\right]^2+\cos x\left[\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right]\) \(f(x)=\sin ^2 x+\frac{1}{4}[\sqrt{3} \cos x+\sin x]^2+\frac{1}{2}\left[\cos ^2 x-\sqrt{3} \sin x \cos x\right]\) \(f(x)=\sin ^2 x+\frac{3}{4} \cos ^2 x+\frac{\sin ^2 x}{4}+\frac{\sqrt{3}}{2} \sin x \cos x+\frac{\cos ^2 x}{2}-\frac{\sqrt{3}}{2} \sin x \cos x\) \(f(x)=\frac{5}{4} \sin ^2 x+\frac{5}{4} \cos ^2 x\) \(f(x)=\frac{5}{4}\left(\sin ^2 x+\cos ^2 x\right)\) \(f(x)=\frac{5}{4}\) Now, \(\operatorname{gof}(\mathrm{x})=\mathrm{g}(\mathrm{f}(\mathrm{x}))=\mathrm{g}\left(\frac{5}{4}\right)=1\)
Assam CEE-2017
Sets, Relation and Function
117255
If \(f(x)=\frac{x}{\sqrt{1+x^2}}\), then \(f \circ f \circ f(x)\) is equal to
117247
If \(g(x)=x^2+x-2\) and \(\frac{1}{2}(\operatorname{gof} f)(x)=2 x^2-5 x+2\) then one such function \(f(x)=\)
1 \(2 x-3\)
2 \(2 x+3\)
3 \(2+2 x\)
4 \(2 x^2-3 x-1\)
Explanation:
A Given, Let, \(g(x)=x^2+x-2\) and \(\frac{1}{2}(\) gof \()(x)=2 x^2-5 x+2\) \(\text { Let, } \mathrm{f}(\mathrm{x}) =(2 \mathrm{x}-3)\) \(\operatorname{gof}(\mathrm{x}) =(2 \mathrm{x}-3)^2+(2 \mathrm{x}-3)-2\) \(=4 \mathrm{x}^2+9-12 \mathrm{x}+2 \mathrm{x}-3-2\) \(=4 \mathrm{x}^2-10 \mathrm{x}+4\) \(\therefore \quad \frac{1}{2} \operatorname{gof}(\mathrm{x}) =2 \mathrm{x}^2-5 \mathrm{x}+2\) \(\therefore \quad \mathrm{f}(\mathrm{x}) =2 \mathrm{x}-3\)
Shift-II
Sets, Relation and Function
117252
Let \(\mathbf{f}: \mathbf{R}-\left\{\frac{-1}{2}\right\} \rightarrow \mathbf{R}\) be defined by \(\mathbf{f}(\mathbf{x})=\) \(\frac{\mathrm{x}-2}{2 \mathrm{x}+1}\). If \(\alpha, \beta\) satisfy the equation \(f(f(x))=-x\), then \(4\left(\alpha^2+\beta^2\right)\)
1 17
2 12
3 24
4 34
Explanation:
A Let \(f: \mathrm{R}-\left\{\frac{1}{2}\right\} \rightarrow \mathrm{R}\) and \(f(\mathrm{x})=\frac{\mathrm{x}-2}{2 \mathrm{x}+1},\) The, \(f(f(\mathrm{x}))=\frac{\frac{\mathrm{x}-2}{2 \mathrm{x}+1}-2}{2\left(\frac{\mathrm{x}-2}{2 \mathrm{x}+1}\right)+1}\) \(-x=\frac{x-2-2(2 x+1)}{2(x-2)+2 x+1}\) \(-x=\frac{x-2-4 x-2}{2 x-4+2 x+1}\) \(-x=\frac{-3 x-4}{4 x-3}\) \(\Rightarrow \quad-3 x-4=-4 x^2+3 x\) \(\text { Or } \quad 4 x^2-6 x-4=0\) \(2 x^2-3 x-2=0\) Or Given \(\alpha, \beta\) satisfy the above equation. \(\therefore \alpha+\beta=\frac{3}{2}\left(\begin{array}{l} \because \alpha+\beta=\frac{-b}{a} \\ \alpha \times \beta=\frac{c}{a} \end{array}\right)\) \(\alpha \beta=\frac{-2}{2}=-1\) \(\therefore \quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=\left(\frac{3}{2}\right)^2-2(-1)=\frac{9}{4}+2=\frac{17}{4}\) \(\therefore \quad 4\left(\alpha^2+\beta^2\right)=17\)
Shift-I
Sets, Relation and Function
117253
If \(f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)\) and \(g\left(\frac{5}{4}\right)=1\), then \(\operatorname{gof}(x)\) is equal to
1 0
2 2
3 1
4 3
Explanation:
C We have- \(f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)\) \(f(x)=\sin ^2 x+\left[\frac{1}{2} \sin x+\cos x \frac{\sqrt{3}}{2}\right]^2+\cos x\left[\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right]\) \(f(x)=\sin ^2 x+\frac{1}{4}[\sqrt{3} \cos x+\sin x]^2+\frac{1}{2}\left[\cos ^2 x-\sqrt{3} \sin x \cos x\right]\) \(f(x)=\sin ^2 x+\frac{3}{4} \cos ^2 x+\frac{\sin ^2 x}{4}+\frac{\sqrt{3}}{2} \sin x \cos x+\frac{\cos ^2 x}{2}-\frac{\sqrt{3}}{2} \sin x \cos x\) \(f(x)=\frac{5}{4} \sin ^2 x+\frac{5}{4} \cos ^2 x\) \(f(x)=\frac{5}{4}\left(\sin ^2 x+\cos ^2 x\right)\) \(f(x)=\frac{5}{4}\) Now, \(\operatorname{gof}(\mathrm{x})=\mathrm{g}(\mathrm{f}(\mathrm{x}))=\mathrm{g}\left(\frac{5}{4}\right)=1\)
Assam CEE-2017
Sets, Relation and Function
117255
If \(f(x)=\frac{x}{\sqrt{1+x^2}}\), then \(f \circ f \circ f(x)\) is equal to
117247
If \(g(x)=x^2+x-2\) and \(\frac{1}{2}(\operatorname{gof} f)(x)=2 x^2-5 x+2\) then one such function \(f(x)=\)
1 \(2 x-3\)
2 \(2 x+3\)
3 \(2+2 x\)
4 \(2 x^2-3 x-1\)
Explanation:
A Given, Let, \(g(x)=x^2+x-2\) and \(\frac{1}{2}(\) gof \()(x)=2 x^2-5 x+2\) \(\text { Let, } \mathrm{f}(\mathrm{x}) =(2 \mathrm{x}-3)\) \(\operatorname{gof}(\mathrm{x}) =(2 \mathrm{x}-3)^2+(2 \mathrm{x}-3)-2\) \(=4 \mathrm{x}^2+9-12 \mathrm{x}+2 \mathrm{x}-3-2\) \(=4 \mathrm{x}^2-10 \mathrm{x}+4\) \(\therefore \quad \frac{1}{2} \operatorname{gof}(\mathrm{x}) =2 \mathrm{x}^2-5 \mathrm{x}+2\) \(\therefore \quad \mathrm{f}(\mathrm{x}) =2 \mathrm{x}-3\)
Shift-II
Sets, Relation and Function
117252
Let \(\mathbf{f}: \mathbf{R}-\left\{\frac{-1}{2}\right\} \rightarrow \mathbf{R}\) be defined by \(\mathbf{f}(\mathbf{x})=\) \(\frac{\mathrm{x}-2}{2 \mathrm{x}+1}\). If \(\alpha, \beta\) satisfy the equation \(f(f(x))=-x\), then \(4\left(\alpha^2+\beta^2\right)\)
1 17
2 12
3 24
4 34
Explanation:
A Let \(f: \mathrm{R}-\left\{\frac{1}{2}\right\} \rightarrow \mathrm{R}\) and \(f(\mathrm{x})=\frac{\mathrm{x}-2}{2 \mathrm{x}+1},\) The, \(f(f(\mathrm{x}))=\frac{\frac{\mathrm{x}-2}{2 \mathrm{x}+1}-2}{2\left(\frac{\mathrm{x}-2}{2 \mathrm{x}+1}\right)+1}\) \(-x=\frac{x-2-2(2 x+1)}{2(x-2)+2 x+1}\) \(-x=\frac{x-2-4 x-2}{2 x-4+2 x+1}\) \(-x=\frac{-3 x-4}{4 x-3}\) \(\Rightarrow \quad-3 x-4=-4 x^2+3 x\) \(\text { Or } \quad 4 x^2-6 x-4=0\) \(2 x^2-3 x-2=0\) Or Given \(\alpha, \beta\) satisfy the above equation. \(\therefore \alpha+\beta=\frac{3}{2}\left(\begin{array}{l} \because \alpha+\beta=\frac{-b}{a} \\ \alpha \times \beta=\frac{c}{a} \end{array}\right)\) \(\alpha \beta=\frac{-2}{2}=-1\) \(\therefore \quad \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(=\left(\frac{3}{2}\right)^2-2(-1)=\frac{9}{4}+2=\frac{17}{4}\) \(\therefore \quad 4\left(\alpha^2+\beta^2\right)=17\)
Shift-I
Sets, Relation and Function
117253
If \(f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)\) and \(g\left(\frac{5}{4}\right)=1\), then \(\operatorname{gof}(x)\) is equal to
1 0
2 2
3 1
4 3
Explanation:
C We have- \(f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)\) \(f(x)=\sin ^2 x+\left[\frac{1}{2} \sin x+\cos x \frac{\sqrt{3}}{2}\right]^2+\cos x\left[\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right]\) \(f(x)=\sin ^2 x+\frac{1}{4}[\sqrt{3} \cos x+\sin x]^2+\frac{1}{2}\left[\cos ^2 x-\sqrt{3} \sin x \cos x\right]\) \(f(x)=\sin ^2 x+\frac{3}{4} \cos ^2 x+\frac{\sin ^2 x}{4}+\frac{\sqrt{3}}{2} \sin x \cos x+\frac{\cos ^2 x}{2}-\frac{\sqrt{3}}{2} \sin x \cos x\) \(f(x)=\frac{5}{4} \sin ^2 x+\frac{5}{4} \cos ^2 x\) \(f(x)=\frac{5}{4}\left(\sin ^2 x+\cos ^2 x\right)\) \(f(x)=\frac{5}{4}\) Now, \(\operatorname{gof}(\mathrm{x})=\mathrm{g}(\mathrm{f}(\mathrm{x}))=\mathrm{g}\left(\frac{5}{4}\right)=1\)
Assam CEE-2017
Sets, Relation and Function
117255
If \(f(x)=\frac{x}{\sqrt{1+x^2}}\), then \(f \circ f \circ f(x)\) is equal to
