A Given that, \(f(x)=\frac{x}{1+x^2}\) Let, \(y=\frac{x}{1+x^2}\) \(\mathrm{yx}^2+\mathrm{y}=\mathrm{x}\) \(\mathrm{yx}^2-\mathrm{x}+\mathrm{y}=0\) For the real value, \(\mathrm{D} \geq 0\) \(1^2-4 y \times y \geq 0\) \(1-4 y^2 \geq 0\) \((1+2 y)(1-2 y) \geq 0\) \((2 y-1)(2 y+1) \geq 0\) \(-\infty \frac{+\infty}{\frac{-1}{2}-\infty}+\infty\) \(\mathrm{y} \in\left[\frac{-1}{2}, \frac{1}{2}\right]-\{0\}\) Because, y never be zero. For, \(\mathrm{x}\) it will be zero. Hence, range of function is \(\left[\frac{-1}{2}, \frac{1}{2}\right]\).
JEE Main 11.01. 2019 Shift -I
Sets, Relation and Function
117435
Let \(\mathrm{f}:(1,3) \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\) \(\frac{x[x]}{1+x^2}\), where \([x]\) denotes the greatest integer \(\leq\) \(x\). Then the range of \(f\) is
D Given that, \(\mathrm{f}:(1,3) \rightarrow \mathrm{R}\) And, \(f(x)=\frac{x[x]}{1+x^2}\) \(f(x)=\left\{\begin{array}{l} \frac{x}{1+x^2} \cdot x \in(1,2) \\ \frac{2 x}{1+x^2}, x \in[2,3] \end{array}\right.\) On differentiating w.r.t \(\mathrm{x}\), we get- \(\begin{aligned} & f^{\prime}(x)=\left\{\begin{array}{l}\frac{\left(1+x^2\right)(1)-x(2 x)}{\left(1+x^2\right)^2}, x \in(1,2) \\ \frac{\left(1+x^2\right)(2)-2 x(2 x)}{\left(1+x^2\right)^2}, x \in[2,3]\end{array}\right. \\ & f^{\prime}(x)=\left\{\begin{array}{l}\frac{1-x^2}{1+x^2}, x \in(1,2) \\ \frac{1-2 x^2}{1+x^2}, x \in[2,3]\end{array}\right.\end{aligned}\) Therefore, \(f(x)\) is decreasing function. Hence, range of \(\mathrm{f}(\mathrm{x})\) is \(\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]\)
JEE Main 08.01.2020 Shift - II
Sets, Relation and Function
117436
If \([x]\) be the greatest integer less than or equal to \(x\), then \(\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]\) is equal to
1 0
2 4
3 -2
4 2
Explanation:
B Let, \(f(x)=\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]\) Putting the value of limits in original function . \(\mathrm{f}(\mathrm{x})=\left[\frac{8}{2}\right]+\left[\frac{-9}{2}\right]+\left[\frac{10}{2}\right]+\left[-\frac{11}{2}\right]+\) \(\ldots+\left[\frac{-99}{2}\right]+\left[\frac{100}{2}\right]\) \(=4+(-5)+5+(-6)+(6)+\ldots .(-50)+50\) \(=4-5+5-6+6+\ldots .-50+50\) \(=4\)
JEE Main 25.07. 2021 Shift-II
Sets, Relation and Function
117437
If the domain of the function \(f(x)=\frac{\cos ^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}\) is the interval \((\alpha, \beta)\), then \(\alpha+\beta\) is equal to
1 \(\frac{3}{2}\)
2 2
3 \(\frac{1}{2}\)
4 1
Explanation:
A Given that, \(f(x)=\frac{\operatorname{cox}^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}\) Defined the function \(\mathrm{f}(\mathrm{x})\) is \(0 \leq \mathrm{x}^2-\mathrm{x}+1 \leq 1\) Then, \(x^2-x+1 \leq 1\) And, \(\mathrm{x}^2-\mathrm{x}+1 \geq 0\) (it is not possible ) \(\therefore \mathrm{x}^2-\mathrm{x}+1 \leq 1\) \(\mathrm{x}^2-\mathrm{x} \leq 0\) \(\mathrm{x}(\mathrm{x}-1) \leq 0\) \(\mathrm{x} \in[0,1]\) And, \(\sin ^{-1}\left(\frac{2 x-1}{2}\right)>0\) \(\frac{2 x-1}{2}>0\) \(x>\frac{1}{2}\) \(\mathrm{x} \in\left(\frac{1}{2}, \infty\right)\) Form equation (i) and (ii), we get- \(\mathrm{x} \in\left(\frac{1}{2}, 1\right]\)Hence, \(\alpha+\beta=\frac{1}{2}+1=\frac{3}{2}\)
A Given that, \(f(x)=\frac{x}{1+x^2}\) Let, \(y=\frac{x}{1+x^2}\) \(\mathrm{yx}^2+\mathrm{y}=\mathrm{x}\) \(\mathrm{yx}^2-\mathrm{x}+\mathrm{y}=0\) For the real value, \(\mathrm{D} \geq 0\) \(1^2-4 y \times y \geq 0\) \(1-4 y^2 \geq 0\) \((1+2 y)(1-2 y) \geq 0\) \((2 y-1)(2 y+1) \geq 0\) \(-\infty \frac{+\infty}{\frac{-1}{2}-\infty}+\infty\) \(\mathrm{y} \in\left[\frac{-1}{2}, \frac{1}{2}\right]-\{0\}\) Because, y never be zero. For, \(\mathrm{x}\) it will be zero. Hence, range of function is \(\left[\frac{-1}{2}, \frac{1}{2}\right]\).
JEE Main 11.01. 2019 Shift -I
Sets, Relation and Function
117435
Let \(\mathrm{f}:(1,3) \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\) \(\frac{x[x]}{1+x^2}\), where \([x]\) denotes the greatest integer \(\leq\) \(x\). Then the range of \(f\) is
D Given that, \(\mathrm{f}:(1,3) \rightarrow \mathrm{R}\) And, \(f(x)=\frac{x[x]}{1+x^2}\) \(f(x)=\left\{\begin{array}{l} \frac{x}{1+x^2} \cdot x \in(1,2) \\ \frac{2 x}{1+x^2}, x \in[2,3] \end{array}\right.\) On differentiating w.r.t \(\mathrm{x}\), we get- \(\begin{aligned} & f^{\prime}(x)=\left\{\begin{array}{l}\frac{\left(1+x^2\right)(1)-x(2 x)}{\left(1+x^2\right)^2}, x \in(1,2) \\ \frac{\left(1+x^2\right)(2)-2 x(2 x)}{\left(1+x^2\right)^2}, x \in[2,3]\end{array}\right. \\ & f^{\prime}(x)=\left\{\begin{array}{l}\frac{1-x^2}{1+x^2}, x \in(1,2) \\ \frac{1-2 x^2}{1+x^2}, x \in[2,3]\end{array}\right.\end{aligned}\) Therefore, \(f(x)\) is decreasing function. Hence, range of \(\mathrm{f}(\mathrm{x})\) is \(\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]\)
JEE Main 08.01.2020 Shift - II
Sets, Relation and Function
117436
If \([x]\) be the greatest integer less than or equal to \(x\), then \(\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]\) is equal to
1 0
2 4
3 -2
4 2
Explanation:
B Let, \(f(x)=\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]\) Putting the value of limits in original function . \(\mathrm{f}(\mathrm{x})=\left[\frac{8}{2}\right]+\left[\frac{-9}{2}\right]+\left[\frac{10}{2}\right]+\left[-\frac{11}{2}\right]+\) \(\ldots+\left[\frac{-99}{2}\right]+\left[\frac{100}{2}\right]\) \(=4+(-5)+5+(-6)+(6)+\ldots .(-50)+50\) \(=4-5+5-6+6+\ldots .-50+50\) \(=4\)
JEE Main 25.07. 2021 Shift-II
Sets, Relation and Function
117437
If the domain of the function \(f(x)=\frac{\cos ^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}\) is the interval \((\alpha, \beta)\), then \(\alpha+\beta\) is equal to
1 \(\frac{3}{2}\)
2 2
3 \(\frac{1}{2}\)
4 1
Explanation:
A Given that, \(f(x)=\frac{\operatorname{cox}^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}\) Defined the function \(\mathrm{f}(\mathrm{x})\) is \(0 \leq \mathrm{x}^2-\mathrm{x}+1 \leq 1\) Then, \(x^2-x+1 \leq 1\) And, \(\mathrm{x}^2-\mathrm{x}+1 \geq 0\) (it is not possible ) \(\therefore \mathrm{x}^2-\mathrm{x}+1 \leq 1\) \(\mathrm{x}^2-\mathrm{x} \leq 0\) \(\mathrm{x}(\mathrm{x}-1) \leq 0\) \(\mathrm{x} \in[0,1]\) And, \(\sin ^{-1}\left(\frac{2 x-1}{2}\right)>0\) \(\frac{2 x-1}{2}>0\) \(x>\frac{1}{2}\) \(\mathrm{x} \in\left(\frac{1}{2}, \infty\right)\) Form equation (i) and (ii), we get- \(\mathrm{x} \in\left(\frac{1}{2}, 1\right]\)Hence, \(\alpha+\beta=\frac{1}{2}+1=\frac{3}{2}\)
A Given that, \(f(x)=\frac{x}{1+x^2}\) Let, \(y=\frac{x}{1+x^2}\) \(\mathrm{yx}^2+\mathrm{y}=\mathrm{x}\) \(\mathrm{yx}^2-\mathrm{x}+\mathrm{y}=0\) For the real value, \(\mathrm{D} \geq 0\) \(1^2-4 y \times y \geq 0\) \(1-4 y^2 \geq 0\) \((1+2 y)(1-2 y) \geq 0\) \((2 y-1)(2 y+1) \geq 0\) \(-\infty \frac{+\infty}{\frac{-1}{2}-\infty}+\infty\) \(\mathrm{y} \in\left[\frac{-1}{2}, \frac{1}{2}\right]-\{0\}\) Because, y never be zero. For, \(\mathrm{x}\) it will be zero. Hence, range of function is \(\left[\frac{-1}{2}, \frac{1}{2}\right]\).
JEE Main 11.01. 2019 Shift -I
Sets, Relation and Function
117435
Let \(\mathrm{f}:(1,3) \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\) \(\frac{x[x]}{1+x^2}\), where \([x]\) denotes the greatest integer \(\leq\) \(x\). Then the range of \(f\) is
D Given that, \(\mathrm{f}:(1,3) \rightarrow \mathrm{R}\) And, \(f(x)=\frac{x[x]}{1+x^2}\) \(f(x)=\left\{\begin{array}{l} \frac{x}{1+x^2} \cdot x \in(1,2) \\ \frac{2 x}{1+x^2}, x \in[2,3] \end{array}\right.\) On differentiating w.r.t \(\mathrm{x}\), we get- \(\begin{aligned} & f^{\prime}(x)=\left\{\begin{array}{l}\frac{\left(1+x^2\right)(1)-x(2 x)}{\left(1+x^2\right)^2}, x \in(1,2) \\ \frac{\left(1+x^2\right)(2)-2 x(2 x)}{\left(1+x^2\right)^2}, x \in[2,3]\end{array}\right. \\ & f^{\prime}(x)=\left\{\begin{array}{l}\frac{1-x^2}{1+x^2}, x \in(1,2) \\ \frac{1-2 x^2}{1+x^2}, x \in[2,3]\end{array}\right.\end{aligned}\) Therefore, \(f(x)\) is decreasing function. Hence, range of \(\mathrm{f}(\mathrm{x})\) is \(\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]\)
JEE Main 08.01.2020 Shift - II
Sets, Relation and Function
117436
If \([x]\) be the greatest integer less than or equal to \(x\), then \(\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]\) is equal to
1 0
2 4
3 -2
4 2
Explanation:
B Let, \(f(x)=\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]\) Putting the value of limits in original function . \(\mathrm{f}(\mathrm{x})=\left[\frac{8}{2}\right]+\left[\frac{-9}{2}\right]+\left[\frac{10}{2}\right]+\left[-\frac{11}{2}\right]+\) \(\ldots+\left[\frac{-99}{2}\right]+\left[\frac{100}{2}\right]\) \(=4+(-5)+5+(-6)+(6)+\ldots .(-50)+50\) \(=4-5+5-6+6+\ldots .-50+50\) \(=4\)
JEE Main 25.07. 2021 Shift-II
Sets, Relation and Function
117437
If the domain of the function \(f(x)=\frac{\cos ^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}\) is the interval \((\alpha, \beta)\), then \(\alpha+\beta\) is equal to
1 \(\frac{3}{2}\)
2 2
3 \(\frac{1}{2}\)
4 1
Explanation:
A Given that, \(f(x)=\frac{\operatorname{cox}^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}\) Defined the function \(\mathrm{f}(\mathrm{x})\) is \(0 \leq \mathrm{x}^2-\mathrm{x}+1 \leq 1\) Then, \(x^2-x+1 \leq 1\) And, \(\mathrm{x}^2-\mathrm{x}+1 \geq 0\) (it is not possible ) \(\therefore \mathrm{x}^2-\mathrm{x}+1 \leq 1\) \(\mathrm{x}^2-\mathrm{x} \leq 0\) \(\mathrm{x}(\mathrm{x}-1) \leq 0\) \(\mathrm{x} \in[0,1]\) And, \(\sin ^{-1}\left(\frac{2 x-1}{2}\right)>0\) \(\frac{2 x-1}{2}>0\) \(x>\frac{1}{2}\) \(\mathrm{x} \in\left(\frac{1}{2}, \infty\right)\) Form equation (i) and (ii), we get- \(\mathrm{x} \in\left(\frac{1}{2}, 1\right]\)Hence, \(\alpha+\beta=\frac{1}{2}+1=\frac{3}{2}\)
A Given that, \(f(x)=\frac{x}{1+x^2}\) Let, \(y=\frac{x}{1+x^2}\) \(\mathrm{yx}^2+\mathrm{y}=\mathrm{x}\) \(\mathrm{yx}^2-\mathrm{x}+\mathrm{y}=0\) For the real value, \(\mathrm{D} \geq 0\) \(1^2-4 y \times y \geq 0\) \(1-4 y^2 \geq 0\) \((1+2 y)(1-2 y) \geq 0\) \((2 y-1)(2 y+1) \geq 0\) \(-\infty \frac{+\infty}{\frac{-1}{2}-\infty}+\infty\) \(\mathrm{y} \in\left[\frac{-1}{2}, \frac{1}{2}\right]-\{0\}\) Because, y never be zero. For, \(\mathrm{x}\) it will be zero. Hence, range of function is \(\left[\frac{-1}{2}, \frac{1}{2}\right]\).
JEE Main 11.01. 2019 Shift -I
Sets, Relation and Function
117435
Let \(\mathrm{f}:(1,3) \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\) \(\frac{x[x]}{1+x^2}\), where \([x]\) denotes the greatest integer \(\leq\) \(x\). Then the range of \(f\) is
D Given that, \(\mathrm{f}:(1,3) \rightarrow \mathrm{R}\) And, \(f(x)=\frac{x[x]}{1+x^2}\) \(f(x)=\left\{\begin{array}{l} \frac{x}{1+x^2} \cdot x \in(1,2) \\ \frac{2 x}{1+x^2}, x \in[2,3] \end{array}\right.\) On differentiating w.r.t \(\mathrm{x}\), we get- \(\begin{aligned} & f^{\prime}(x)=\left\{\begin{array}{l}\frac{\left(1+x^2\right)(1)-x(2 x)}{\left(1+x^2\right)^2}, x \in(1,2) \\ \frac{\left(1+x^2\right)(2)-2 x(2 x)}{\left(1+x^2\right)^2}, x \in[2,3]\end{array}\right. \\ & f^{\prime}(x)=\left\{\begin{array}{l}\frac{1-x^2}{1+x^2}, x \in(1,2) \\ \frac{1-2 x^2}{1+x^2}, x \in[2,3]\end{array}\right.\end{aligned}\) Therefore, \(f(x)\) is decreasing function. Hence, range of \(\mathrm{f}(\mathrm{x})\) is \(\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]\)
JEE Main 08.01.2020 Shift - II
Sets, Relation and Function
117436
If \([x]\) be the greatest integer less than or equal to \(x\), then \(\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]\) is equal to
1 0
2 4
3 -2
4 2
Explanation:
B Let, \(f(x)=\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]\) Putting the value of limits in original function . \(\mathrm{f}(\mathrm{x})=\left[\frac{8}{2}\right]+\left[\frac{-9}{2}\right]+\left[\frac{10}{2}\right]+\left[-\frac{11}{2}\right]+\) \(\ldots+\left[\frac{-99}{2}\right]+\left[\frac{100}{2}\right]\) \(=4+(-5)+5+(-6)+(6)+\ldots .(-50)+50\) \(=4-5+5-6+6+\ldots .-50+50\) \(=4\)
JEE Main 25.07. 2021 Shift-II
Sets, Relation and Function
117437
If the domain of the function \(f(x)=\frac{\cos ^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}\) is the interval \((\alpha, \beta)\), then \(\alpha+\beta\) is equal to
1 \(\frac{3}{2}\)
2 2
3 \(\frac{1}{2}\)
4 1
Explanation:
A Given that, \(f(x)=\frac{\operatorname{cox}^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}\) Defined the function \(\mathrm{f}(\mathrm{x})\) is \(0 \leq \mathrm{x}^2-\mathrm{x}+1 \leq 1\) Then, \(x^2-x+1 \leq 1\) And, \(\mathrm{x}^2-\mathrm{x}+1 \geq 0\) (it is not possible ) \(\therefore \mathrm{x}^2-\mathrm{x}+1 \leq 1\) \(\mathrm{x}^2-\mathrm{x} \leq 0\) \(\mathrm{x}(\mathrm{x}-1) \leq 0\) \(\mathrm{x} \in[0,1]\) And, \(\sin ^{-1}\left(\frac{2 x-1}{2}\right)>0\) \(\frac{2 x-1}{2}>0\) \(x>\frac{1}{2}\) \(\mathrm{x} \in\left(\frac{1}{2}, \infty\right)\) Form equation (i) and (ii), we get- \(\mathrm{x} \in\left(\frac{1}{2}, 1\right]\)Hence, \(\alpha+\beta=\frac{1}{2}+1=\frac{3}{2}\)
