116980
If \(f(x)= \begin{cases}\frac{\sin [x]}{[x]}, & {[x] \neq 0} \\ 0, & {[x]=0}\end{cases}\) Where \([x]\) denotes the greatest integer less than or equal to \(x\), then \(\lim _{x \rightarrow 0} f(x)\) equals
116981
For which of the following values of ' \(x\) ', does the function \(f(x)=\log\) \(\left[\left\{\sqrt{\left(25-\mathbf{x}^2\right)}\right\} /(2-x)\right]\) have the real values?
1 \(-5\lt x\lt 5\)
2 \(-5\lt x\lt 2\)
3 \(x>-2\)
4 \(x\lt 2\)
Explanation:
B Given function, \(f(x)=\log \left[\frac{\left\{\sqrt{25-x^2}\right\}}{2-x}\right]\) \(f(x)=\log \left(\sqrt{25-x^2}\right)-\log (2-x)\) \(\sqrt{25-x^2}>0 \text { and } 2-x>0\) \(=25-x^2>0,-x>-2\) \(=x^2-25\lt 0, x\lt 2\) \(=(x-5)(x+5)\lt 0 \quad x \in(-\infty, 2)\) \(=x \in(-5,5), x \in(-5,2)\) \(=-5\lt x\lt 2\)
J and K CET-2019
Sets, Relation and Function
116982
Let function \(f(x)=(x-1)^2(x+1)^3\). Then which of the following is false?
1 There exists a point where \(f(x)\) has a maximum value
2 There exists a point where \(f(x)\) has a minimum value
3 There exists a point where \(f(x)\) has neither maximum nor minimum value
4 All of the above
Explanation:
D We have, \(f(x)=(x-1)^2(x+1)^3\) Now, differentiating we get - \(f^{\prime}(x)=(x-1)^2 3(x+1)^2+(x+1)^3 2(x-1)\) For maxima or minima- \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3(\mathrm{x}-1)^2(\mathrm{x}+1)^2+2(\mathrm{x}+1)^3(\mathrm{x}-1)=0\) \((\mathrm{x}-1)(\mathrm{x}+1)^2(5 \mathrm{x}-1)=0\) \(\mathrm{x}=-1,1, \frac{1}{5}\) Again differentiating \(w\). r. \(t x\) we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(5 \mathrm{x}^2-6 \mathrm{x}+1\right) 2(\mathrm{x}+1)+(\mathrm{x}+1)^2(10 \mathrm{x}-6)\) \(=2\left[5 \mathrm{x}^3-\mathrm{x}^2-5 \mathrm{x}+1\right]+2\left[5 \mathrm{x}^3+7 \mathrm{x}^2-\mathrm{x}-3\right]\) \(f^{\prime \prime}(\mathrm{x})=2\left[10 \mathrm{x}^3+6 \mathrm{x}^2-6 \mathrm{x}-2\right]\) \(\text { at } \quad \mathrm{x}=-1\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})=0\) \(\text { at } \mathrm{x}=\frac{1}{5}\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})\lt 0\) \(\text { at } \quad \mathrm{x}=1\) \(f^{\prime \prime}(x)>0\) At, \(\mathrm{x}=-1 \mathrm{f}(\mathrm{x})\) has neither maximum nor minimum at \(x=-1 \mathrm{f}(\mathrm{x})\) has minimum value. \(x=\frac{1}{5} f(x)\) has maximum value.
J and K CET-2015
Sets, Relation and Function
116983
The number of the solutions of the equation \(\mathbf{5}^{2 \mathrm{x}-1}+\mathbf{5}^{\mathrm{x}+1}=\mathbf{2 5 0}\) is/are
1 0
2 1
3 2
4 infinitely many
Explanation:
B We have given equation. \(5^{2 \mathrm{x}-1}+5^{\mathrm{x}+1}=250\) \(\frac{5^{2 \mathrm{x}}}{5}+5^{\mathrm{x}} \cdot 5=250\) \(5^{2 \mathrm{x}}+25 \cdot 5^{\mathrm{x}}=1250\) Let, \(5^{\mathrm{x}}=\mathrm{t}\) \(\mathrm{t}^2+25 \mathrm{t}-1250=0\) \(\mathrm{t}=\frac{-25 \pm \sqrt{(25)^2-4 \times 1 \times(-1250)}}{2 \times 1}\) \(\mathrm{t}=\frac{-25 \pm \sqrt{625+5000}}{2}\) \(\mathrm{t}=\frac{-25 \pm 75}{2}\) \(\mathrm{t}=\frac{-25+75}{2} \text { or } \mathrm{t}=\frac{-25-75}{2}\) \(\mathrm{t}=25 \text { and } \mathrm{t}=-50\) \(5^{\mathrm{x}}=5^2 \quad 5^{\mathrm{x}}=-50\) \(\mathrm{x}=2\) \(5^{\mathrm{x}}=-50 \text { can not express in } 5^{\mathrm{x}}\) Hence, the number of solution is one.
116980
If \(f(x)= \begin{cases}\frac{\sin [x]}{[x]}, & {[x] \neq 0} \\ 0, & {[x]=0}\end{cases}\) Where \([x]\) denotes the greatest integer less than or equal to \(x\), then \(\lim _{x \rightarrow 0} f(x)\) equals
116981
For which of the following values of ' \(x\) ', does the function \(f(x)=\log\) \(\left[\left\{\sqrt{\left(25-\mathbf{x}^2\right)}\right\} /(2-x)\right]\) have the real values?
1 \(-5\lt x\lt 5\)
2 \(-5\lt x\lt 2\)
3 \(x>-2\)
4 \(x\lt 2\)
Explanation:
B Given function, \(f(x)=\log \left[\frac{\left\{\sqrt{25-x^2}\right\}}{2-x}\right]\) \(f(x)=\log \left(\sqrt{25-x^2}\right)-\log (2-x)\) \(\sqrt{25-x^2}>0 \text { and } 2-x>0\) \(=25-x^2>0,-x>-2\) \(=x^2-25\lt 0, x\lt 2\) \(=(x-5)(x+5)\lt 0 \quad x \in(-\infty, 2)\) \(=x \in(-5,5), x \in(-5,2)\) \(=-5\lt x\lt 2\)
J and K CET-2019
Sets, Relation and Function
116982
Let function \(f(x)=(x-1)^2(x+1)^3\). Then which of the following is false?
1 There exists a point where \(f(x)\) has a maximum value
2 There exists a point where \(f(x)\) has a minimum value
3 There exists a point where \(f(x)\) has neither maximum nor minimum value
4 All of the above
Explanation:
D We have, \(f(x)=(x-1)^2(x+1)^3\) Now, differentiating we get - \(f^{\prime}(x)=(x-1)^2 3(x+1)^2+(x+1)^3 2(x-1)\) For maxima or minima- \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3(\mathrm{x}-1)^2(\mathrm{x}+1)^2+2(\mathrm{x}+1)^3(\mathrm{x}-1)=0\) \((\mathrm{x}-1)(\mathrm{x}+1)^2(5 \mathrm{x}-1)=0\) \(\mathrm{x}=-1,1, \frac{1}{5}\) Again differentiating \(w\). r. \(t x\) we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(5 \mathrm{x}^2-6 \mathrm{x}+1\right) 2(\mathrm{x}+1)+(\mathrm{x}+1)^2(10 \mathrm{x}-6)\) \(=2\left[5 \mathrm{x}^3-\mathrm{x}^2-5 \mathrm{x}+1\right]+2\left[5 \mathrm{x}^3+7 \mathrm{x}^2-\mathrm{x}-3\right]\) \(f^{\prime \prime}(\mathrm{x})=2\left[10 \mathrm{x}^3+6 \mathrm{x}^2-6 \mathrm{x}-2\right]\) \(\text { at } \quad \mathrm{x}=-1\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})=0\) \(\text { at } \mathrm{x}=\frac{1}{5}\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})\lt 0\) \(\text { at } \quad \mathrm{x}=1\) \(f^{\prime \prime}(x)>0\) At, \(\mathrm{x}=-1 \mathrm{f}(\mathrm{x})\) has neither maximum nor minimum at \(x=-1 \mathrm{f}(\mathrm{x})\) has minimum value. \(x=\frac{1}{5} f(x)\) has maximum value.
J and K CET-2015
Sets, Relation and Function
116983
The number of the solutions of the equation \(\mathbf{5}^{2 \mathrm{x}-1}+\mathbf{5}^{\mathrm{x}+1}=\mathbf{2 5 0}\) is/are
1 0
2 1
3 2
4 infinitely many
Explanation:
B We have given equation. \(5^{2 \mathrm{x}-1}+5^{\mathrm{x}+1}=250\) \(\frac{5^{2 \mathrm{x}}}{5}+5^{\mathrm{x}} \cdot 5=250\) \(5^{2 \mathrm{x}}+25 \cdot 5^{\mathrm{x}}=1250\) Let, \(5^{\mathrm{x}}=\mathrm{t}\) \(\mathrm{t}^2+25 \mathrm{t}-1250=0\) \(\mathrm{t}=\frac{-25 \pm \sqrt{(25)^2-4 \times 1 \times(-1250)}}{2 \times 1}\) \(\mathrm{t}=\frac{-25 \pm \sqrt{625+5000}}{2}\) \(\mathrm{t}=\frac{-25 \pm 75}{2}\) \(\mathrm{t}=\frac{-25+75}{2} \text { or } \mathrm{t}=\frac{-25-75}{2}\) \(\mathrm{t}=25 \text { and } \mathrm{t}=-50\) \(5^{\mathrm{x}}=5^2 \quad 5^{\mathrm{x}}=-50\) \(\mathrm{x}=2\) \(5^{\mathrm{x}}=-50 \text { can not express in } 5^{\mathrm{x}}\) Hence, the number of solution is one.
116980
If \(f(x)= \begin{cases}\frac{\sin [x]}{[x]}, & {[x] \neq 0} \\ 0, & {[x]=0}\end{cases}\) Where \([x]\) denotes the greatest integer less than or equal to \(x\), then \(\lim _{x \rightarrow 0} f(x)\) equals
116981
For which of the following values of ' \(x\) ', does the function \(f(x)=\log\) \(\left[\left\{\sqrt{\left(25-\mathbf{x}^2\right)}\right\} /(2-x)\right]\) have the real values?
1 \(-5\lt x\lt 5\)
2 \(-5\lt x\lt 2\)
3 \(x>-2\)
4 \(x\lt 2\)
Explanation:
B Given function, \(f(x)=\log \left[\frac{\left\{\sqrt{25-x^2}\right\}}{2-x}\right]\) \(f(x)=\log \left(\sqrt{25-x^2}\right)-\log (2-x)\) \(\sqrt{25-x^2}>0 \text { and } 2-x>0\) \(=25-x^2>0,-x>-2\) \(=x^2-25\lt 0, x\lt 2\) \(=(x-5)(x+5)\lt 0 \quad x \in(-\infty, 2)\) \(=x \in(-5,5), x \in(-5,2)\) \(=-5\lt x\lt 2\)
J and K CET-2019
Sets, Relation and Function
116982
Let function \(f(x)=(x-1)^2(x+1)^3\). Then which of the following is false?
1 There exists a point where \(f(x)\) has a maximum value
2 There exists a point where \(f(x)\) has a minimum value
3 There exists a point where \(f(x)\) has neither maximum nor minimum value
4 All of the above
Explanation:
D We have, \(f(x)=(x-1)^2(x+1)^3\) Now, differentiating we get - \(f^{\prime}(x)=(x-1)^2 3(x+1)^2+(x+1)^3 2(x-1)\) For maxima or minima- \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3(\mathrm{x}-1)^2(\mathrm{x}+1)^2+2(\mathrm{x}+1)^3(\mathrm{x}-1)=0\) \((\mathrm{x}-1)(\mathrm{x}+1)^2(5 \mathrm{x}-1)=0\) \(\mathrm{x}=-1,1, \frac{1}{5}\) Again differentiating \(w\). r. \(t x\) we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(5 \mathrm{x}^2-6 \mathrm{x}+1\right) 2(\mathrm{x}+1)+(\mathrm{x}+1)^2(10 \mathrm{x}-6)\) \(=2\left[5 \mathrm{x}^3-\mathrm{x}^2-5 \mathrm{x}+1\right]+2\left[5 \mathrm{x}^3+7 \mathrm{x}^2-\mathrm{x}-3\right]\) \(f^{\prime \prime}(\mathrm{x})=2\left[10 \mathrm{x}^3+6 \mathrm{x}^2-6 \mathrm{x}-2\right]\) \(\text { at } \quad \mathrm{x}=-1\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})=0\) \(\text { at } \mathrm{x}=\frac{1}{5}\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})\lt 0\) \(\text { at } \quad \mathrm{x}=1\) \(f^{\prime \prime}(x)>0\) At, \(\mathrm{x}=-1 \mathrm{f}(\mathrm{x})\) has neither maximum nor minimum at \(x=-1 \mathrm{f}(\mathrm{x})\) has minimum value. \(x=\frac{1}{5} f(x)\) has maximum value.
J and K CET-2015
Sets, Relation and Function
116983
The number of the solutions of the equation \(\mathbf{5}^{2 \mathrm{x}-1}+\mathbf{5}^{\mathrm{x}+1}=\mathbf{2 5 0}\) is/are
1 0
2 1
3 2
4 infinitely many
Explanation:
B We have given equation. \(5^{2 \mathrm{x}-1}+5^{\mathrm{x}+1}=250\) \(\frac{5^{2 \mathrm{x}}}{5}+5^{\mathrm{x}} \cdot 5=250\) \(5^{2 \mathrm{x}}+25 \cdot 5^{\mathrm{x}}=1250\) Let, \(5^{\mathrm{x}}=\mathrm{t}\) \(\mathrm{t}^2+25 \mathrm{t}-1250=0\) \(\mathrm{t}=\frac{-25 \pm \sqrt{(25)^2-4 \times 1 \times(-1250)}}{2 \times 1}\) \(\mathrm{t}=\frac{-25 \pm \sqrt{625+5000}}{2}\) \(\mathrm{t}=\frac{-25 \pm 75}{2}\) \(\mathrm{t}=\frac{-25+75}{2} \text { or } \mathrm{t}=\frac{-25-75}{2}\) \(\mathrm{t}=25 \text { and } \mathrm{t}=-50\) \(5^{\mathrm{x}}=5^2 \quad 5^{\mathrm{x}}=-50\) \(\mathrm{x}=2\) \(5^{\mathrm{x}}=-50 \text { can not express in } 5^{\mathrm{x}}\) Hence, the number of solution is one.
116980
If \(f(x)= \begin{cases}\frac{\sin [x]}{[x]}, & {[x] \neq 0} \\ 0, & {[x]=0}\end{cases}\) Where \([x]\) denotes the greatest integer less than or equal to \(x\), then \(\lim _{x \rightarrow 0} f(x)\) equals
116981
For which of the following values of ' \(x\) ', does the function \(f(x)=\log\) \(\left[\left\{\sqrt{\left(25-\mathbf{x}^2\right)}\right\} /(2-x)\right]\) have the real values?
1 \(-5\lt x\lt 5\)
2 \(-5\lt x\lt 2\)
3 \(x>-2\)
4 \(x\lt 2\)
Explanation:
B Given function, \(f(x)=\log \left[\frac{\left\{\sqrt{25-x^2}\right\}}{2-x}\right]\) \(f(x)=\log \left(\sqrt{25-x^2}\right)-\log (2-x)\) \(\sqrt{25-x^2}>0 \text { and } 2-x>0\) \(=25-x^2>0,-x>-2\) \(=x^2-25\lt 0, x\lt 2\) \(=(x-5)(x+5)\lt 0 \quad x \in(-\infty, 2)\) \(=x \in(-5,5), x \in(-5,2)\) \(=-5\lt x\lt 2\)
J and K CET-2019
Sets, Relation and Function
116982
Let function \(f(x)=(x-1)^2(x+1)^3\). Then which of the following is false?
1 There exists a point where \(f(x)\) has a maximum value
2 There exists a point where \(f(x)\) has a minimum value
3 There exists a point where \(f(x)\) has neither maximum nor minimum value
4 All of the above
Explanation:
D We have, \(f(x)=(x-1)^2(x+1)^3\) Now, differentiating we get - \(f^{\prime}(x)=(x-1)^2 3(x+1)^2+(x+1)^3 2(x-1)\) For maxima or minima- \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3(\mathrm{x}-1)^2(\mathrm{x}+1)^2+2(\mathrm{x}+1)^3(\mathrm{x}-1)=0\) \((\mathrm{x}-1)(\mathrm{x}+1)^2(5 \mathrm{x}-1)=0\) \(\mathrm{x}=-1,1, \frac{1}{5}\) Again differentiating \(w\). r. \(t x\) we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(5 \mathrm{x}^2-6 \mathrm{x}+1\right) 2(\mathrm{x}+1)+(\mathrm{x}+1)^2(10 \mathrm{x}-6)\) \(=2\left[5 \mathrm{x}^3-\mathrm{x}^2-5 \mathrm{x}+1\right]+2\left[5 \mathrm{x}^3+7 \mathrm{x}^2-\mathrm{x}-3\right]\) \(f^{\prime \prime}(\mathrm{x})=2\left[10 \mathrm{x}^3+6 \mathrm{x}^2-6 \mathrm{x}-2\right]\) \(\text { at } \quad \mathrm{x}=-1\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})=0\) \(\text { at } \mathrm{x}=\frac{1}{5}\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})\lt 0\) \(\text { at } \quad \mathrm{x}=1\) \(f^{\prime \prime}(x)>0\) At, \(\mathrm{x}=-1 \mathrm{f}(\mathrm{x})\) has neither maximum nor minimum at \(x=-1 \mathrm{f}(\mathrm{x})\) has minimum value. \(x=\frac{1}{5} f(x)\) has maximum value.
J and K CET-2015
Sets, Relation and Function
116983
The number of the solutions of the equation \(\mathbf{5}^{2 \mathrm{x}-1}+\mathbf{5}^{\mathrm{x}+1}=\mathbf{2 5 0}\) is/are
1 0
2 1
3 2
4 infinitely many
Explanation:
B We have given equation. \(5^{2 \mathrm{x}-1}+5^{\mathrm{x}+1}=250\) \(\frac{5^{2 \mathrm{x}}}{5}+5^{\mathrm{x}} \cdot 5=250\) \(5^{2 \mathrm{x}}+25 \cdot 5^{\mathrm{x}}=1250\) Let, \(5^{\mathrm{x}}=\mathrm{t}\) \(\mathrm{t}^2+25 \mathrm{t}-1250=0\) \(\mathrm{t}=\frac{-25 \pm \sqrt{(25)^2-4 \times 1 \times(-1250)}}{2 \times 1}\) \(\mathrm{t}=\frac{-25 \pm \sqrt{625+5000}}{2}\) \(\mathrm{t}=\frac{-25 \pm 75}{2}\) \(\mathrm{t}=\frac{-25+75}{2} \text { or } \mathrm{t}=\frac{-25-75}{2}\) \(\mathrm{t}=25 \text { and } \mathrm{t}=-50\) \(5^{\mathrm{x}}=5^2 \quad 5^{\mathrm{x}}=-50\) \(\mathrm{x}=2\) \(5^{\mathrm{x}}=-50 \text { can not express in } 5^{\mathrm{x}}\) Hence, the number of solution is one.
