117321
If \(f(x)=\log _e\left(6-\left|x^2+x-6\right|\right)\), then domain of \(f(x)\) has how many integral values of \(x\) ?
1 5
2 4
3 Infinite
4 None of these
Explanation:
D Given, \(f(\mathrm{x})=\log _{\mathrm{e}}\left(6-\left|\mathrm{x}^2+\mathrm{x}-6\right|\right)\) The function \(f(\mathrm{x})\) is defined, if - \(\left(6-\left|\mathrm{x}^2+\mathrm{x}-6\right|\right)>0\) \(\Rightarrow\) \(\left|\mathrm{x}^2+\mathrm{x}-6\right|\lt 6\) \(\Rightarrow\) \(-6\lt x^2+x-6\lt 6\) \(\text { if }\) \(\mathrm{x}^2+\mathrm{x}-6\lt 6\) \(\Rightarrow \quad \mathrm{x}^2+\mathrm{x}-12\lt 0\) \(\Rightarrow\) \((\mathrm{x}+4)(\mathrm{x}-3)\lt 0\) \(\therefore \quad \mathrm{x} \in(-4,3)\) Now, if \(-6\lt x^2+x-6\) \(\therefore \quad \mathrm{x} \in(0, \infty)\) From equation (i) and (ii), we get - \(\mathrm{x} \in(0,3)\) \(\because \mathrm{f}(\mathrm{x})\) has only two integral values. \(\therefore \quad \mathrm{x}=1,2\)
CG PET- 2015
Sets, Relation and Function
117322
The domain of the function \(\cos ^{-1}\left(\frac{2 \sin ^{-1}\left(\frac{1}{4 \mathrm{x}^2-1}\right)}{\pi}\right)\) is:
D Given, the function - \(\cos ^{-1}\left[\frac{2 \sin ^{-1}\left(\frac{1}{4 \mathrm{x}^2-1}\right)}{\pi}\right] \text { we have }\) To find interval in which \(x\) lies. Here, the max value of \(\sin \theta\) is 1 and minimum value of \(\sin \theta=-1\) \(\therefore \frac{1}{4 \mathrm{x}^2-1} \leq 1\) \(\Rightarrow \left.\left(4 \mathrm{x}^2-1\right) \geq 1 \quad\left(\because\left(\frac{1}{4 \mathrm{x}^2-1}\right)^{-1}\right) \geq(1)^{-1}\right)\) \(\Rightarrow 4 \mathrm{x}^2-1-1 \geq 0\) \(\Rightarrow 4 \mathrm{x}^2-2 \geq 0\) \(\Rightarrow 2 \mathrm{x}^2-1 \geq 0\) \(\Rightarrow \mathrm{x}^2 \geq \frac{1}{2}\) \(\Rightarrow \mathrm{x} \leq-\frac{1}{\sqrt{2}} \text { and } \mathrm{x} \geq \frac{1}{\sqrt{2}}\) \(\therefore\) The required interval is \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right)\) We also know \(\mathrm{x}=0\) \(\Rightarrow \sin ^{-1}\left(\frac{1}{-1}\right)=\sin ^{-1}(-1)\) which is also possible. \(\therefore\) The required interval is - \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}\)
Shift-I
Sets, Relation and Function
117323
The domain of the function \(f(x)=\sin ^{-1}\left[2 x^2-3\right]\) \(+\log _2\left(\log _{\frac{1}{2}}\left(x^2-5 x+5\right)\right)\), where \([t]\) is the greatest integer function is :
117324
The range of the function \(f(x)=\sqrt{3-x}+\sqrt{2+x}\) is :
1 \([2 \sqrt{2}, \sqrt{11}]\)
2 \([\sqrt{5}, \sqrt{10}]\)
3 \([\sqrt{5}, \sqrt{13}]\)
4 \([\sqrt{2}, \sqrt{7}]\)
Explanation:
B We have to find the range of \(f(x)=\sqrt{3-x}+\sqrt{2+x}\), The function is possible for \(3-x \geq 0\) and \(2+x \geq 0\) or \(\quad x \in[-2,3]\) Let \(\quad y=\sqrt{3-x}+\sqrt{2+x}\) \(\Rightarrow \quad \mathrm{y}^2=3-\mathrm{x}+2+\mathrm{x}+2 \sqrt{(3-\mathrm{x})(2+\mathrm{x})}\) \(\Rightarrow \quad y^2=5+2 \cdot \sqrt{(3-x)(2+x)}\) \(\therefore \quad \frac{\mathrm{y}^2-5}{2}=\sqrt{(3-\mathrm{x})(2+\mathrm{x})}\) \(\Rightarrow \quad=6-2 \mathrm{x}+3 \mathrm{x}-\mathrm{x}^2\) \(=6+x-x^2\) \(\therefore \mathrm{x}^2-\mathrm{x}+\left[\frac{\mathrm{y}^2-5}{2}\right]^2-6=0\) \(\mathrm{x} \text { is real } \Rightarrow \text { Discriminate, } \mathrm{D}, \geq 0\) \(\therefore 1-4\left[\left[\frac{\mathrm{y}^2-5}{2}\right]^2-6\right] \geq 0\) \(\Rightarrow \quad 25 \geq 4\left(\frac{\mathrm{y}^2-5}{2}\right)^2 .\) \(\Rightarrow \quad\left(\frac{\mathrm{y}^2-5}{2}\right)^2 \leq \frac{25}{4}\) \(\therefore \frac{y^2-5}{2} \leq \frac{5}{2}\) \(\Rightarrow \mathrm{y}^2 \leq 10\) \(\text { or } y^2 \leq \sqrt{10}\) Also, we know \(x \in[-2,3]\) \(\therefore \quad f(-2)=\sqrt{5}\) \(f(3)=\sqrt{5}\)\(\therefore\) The range is \((\sqrt{5}, \sqrt{10})\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117321
If \(f(x)=\log _e\left(6-\left|x^2+x-6\right|\right)\), then domain of \(f(x)\) has how many integral values of \(x\) ?
1 5
2 4
3 Infinite
4 None of these
Explanation:
D Given, \(f(\mathrm{x})=\log _{\mathrm{e}}\left(6-\left|\mathrm{x}^2+\mathrm{x}-6\right|\right)\) The function \(f(\mathrm{x})\) is defined, if - \(\left(6-\left|\mathrm{x}^2+\mathrm{x}-6\right|\right)>0\) \(\Rightarrow\) \(\left|\mathrm{x}^2+\mathrm{x}-6\right|\lt 6\) \(\Rightarrow\) \(-6\lt x^2+x-6\lt 6\) \(\text { if }\) \(\mathrm{x}^2+\mathrm{x}-6\lt 6\) \(\Rightarrow \quad \mathrm{x}^2+\mathrm{x}-12\lt 0\) \(\Rightarrow\) \((\mathrm{x}+4)(\mathrm{x}-3)\lt 0\) \(\therefore \quad \mathrm{x} \in(-4,3)\) Now, if \(-6\lt x^2+x-6\) \(\therefore \quad \mathrm{x} \in(0, \infty)\) From equation (i) and (ii), we get - \(\mathrm{x} \in(0,3)\) \(\because \mathrm{f}(\mathrm{x})\) has only two integral values. \(\therefore \quad \mathrm{x}=1,2\)
CG PET- 2015
Sets, Relation and Function
117322
The domain of the function \(\cos ^{-1}\left(\frac{2 \sin ^{-1}\left(\frac{1}{4 \mathrm{x}^2-1}\right)}{\pi}\right)\) is:
D Given, the function - \(\cos ^{-1}\left[\frac{2 \sin ^{-1}\left(\frac{1}{4 \mathrm{x}^2-1}\right)}{\pi}\right] \text { we have }\) To find interval in which \(x\) lies. Here, the max value of \(\sin \theta\) is 1 and minimum value of \(\sin \theta=-1\) \(\therefore \frac{1}{4 \mathrm{x}^2-1} \leq 1\) \(\Rightarrow \left.\left(4 \mathrm{x}^2-1\right) \geq 1 \quad\left(\because\left(\frac{1}{4 \mathrm{x}^2-1}\right)^{-1}\right) \geq(1)^{-1}\right)\) \(\Rightarrow 4 \mathrm{x}^2-1-1 \geq 0\) \(\Rightarrow 4 \mathrm{x}^2-2 \geq 0\) \(\Rightarrow 2 \mathrm{x}^2-1 \geq 0\) \(\Rightarrow \mathrm{x}^2 \geq \frac{1}{2}\) \(\Rightarrow \mathrm{x} \leq-\frac{1}{\sqrt{2}} \text { and } \mathrm{x} \geq \frac{1}{\sqrt{2}}\) \(\therefore\) The required interval is \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right)\) We also know \(\mathrm{x}=0\) \(\Rightarrow \sin ^{-1}\left(\frac{1}{-1}\right)=\sin ^{-1}(-1)\) which is also possible. \(\therefore\) The required interval is - \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}\)
Shift-I
Sets, Relation and Function
117323
The domain of the function \(f(x)=\sin ^{-1}\left[2 x^2-3\right]\) \(+\log _2\left(\log _{\frac{1}{2}}\left(x^2-5 x+5\right)\right)\), where \([t]\) is the greatest integer function is :
117324
The range of the function \(f(x)=\sqrt{3-x}+\sqrt{2+x}\) is :
1 \([2 \sqrt{2}, \sqrt{11}]\)
2 \([\sqrt{5}, \sqrt{10}]\)
3 \([\sqrt{5}, \sqrt{13}]\)
4 \([\sqrt{2}, \sqrt{7}]\)
Explanation:
B We have to find the range of \(f(x)=\sqrt{3-x}+\sqrt{2+x}\), The function is possible for \(3-x \geq 0\) and \(2+x \geq 0\) or \(\quad x \in[-2,3]\) Let \(\quad y=\sqrt{3-x}+\sqrt{2+x}\) \(\Rightarrow \quad \mathrm{y}^2=3-\mathrm{x}+2+\mathrm{x}+2 \sqrt{(3-\mathrm{x})(2+\mathrm{x})}\) \(\Rightarrow \quad y^2=5+2 \cdot \sqrt{(3-x)(2+x)}\) \(\therefore \quad \frac{\mathrm{y}^2-5}{2}=\sqrt{(3-\mathrm{x})(2+\mathrm{x})}\) \(\Rightarrow \quad=6-2 \mathrm{x}+3 \mathrm{x}-\mathrm{x}^2\) \(=6+x-x^2\) \(\therefore \mathrm{x}^2-\mathrm{x}+\left[\frac{\mathrm{y}^2-5}{2}\right]^2-6=0\) \(\mathrm{x} \text { is real } \Rightarrow \text { Discriminate, } \mathrm{D}, \geq 0\) \(\therefore 1-4\left[\left[\frac{\mathrm{y}^2-5}{2}\right]^2-6\right] \geq 0\) \(\Rightarrow \quad 25 \geq 4\left(\frac{\mathrm{y}^2-5}{2}\right)^2 .\) \(\Rightarrow \quad\left(\frac{\mathrm{y}^2-5}{2}\right)^2 \leq \frac{25}{4}\) \(\therefore \frac{y^2-5}{2} \leq \frac{5}{2}\) \(\Rightarrow \mathrm{y}^2 \leq 10\) \(\text { or } y^2 \leq \sqrt{10}\) Also, we know \(x \in[-2,3]\) \(\therefore \quad f(-2)=\sqrt{5}\) \(f(3)=\sqrt{5}\)\(\therefore\) The range is \((\sqrt{5}, \sqrt{10})\)
117321
If \(f(x)=\log _e\left(6-\left|x^2+x-6\right|\right)\), then domain of \(f(x)\) has how many integral values of \(x\) ?
1 5
2 4
3 Infinite
4 None of these
Explanation:
D Given, \(f(\mathrm{x})=\log _{\mathrm{e}}\left(6-\left|\mathrm{x}^2+\mathrm{x}-6\right|\right)\) The function \(f(\mathrm{x})\) is defined, if - \(\left(6-\left|\mathrm{x}^2+\mathrm{x}-6\right|\right)>0\) \(\Rightarrow\) \(\left|\mathrm{x}^2+\mathrm{x}-6\right|\lt 6\) \(\Rightarrow\) \(-6\lt x^2+x-6\lt 6\) \(\text { if }\) \(\mathrm{x}^2+\mathrm{x}-6\lt 6\) \(\Rightarrow \quad \mathrm{x}^2+\mathrm{x}-12\lt 0\) \(\Rightarrow\) \((\mathrm{x}+4)(\mathrm{x}-3)\lt 0\) \(\therefore \quad \mathrm{x} \in(-4,3)\) Now, if \(-6\lt x^2+x-6\) \(\therefore \quad \mathrm{x} \in(0, \infty)\) From equation (i) and (ii), we get - \(\mathrm{x} \in(0,3)\) \(\because \mathrm{f}(\mathrm{x})\) has only two integral values. \(\therefore \quad \mathrm{x}=1,2\)
CG PET- 2015
Sets, Relation and Function
117322
The domain of the function \(\cos ^{-1}\left(\frac{2 \sin ^{-1}\left(\frac{1}{4 \mathrm{x}^2-1}\right)}{\pi}\right)\) is:
D Given, the function - \(\cos ^{-1}\left[\frac{2 \sin ^{-1}\left(\frac{1}{4 \mathrm{x}^2-1}\right)}{\pi}\right] \text { we have }\) To find interval in which \(x\) lies. Here, the max value of \(\sin \theta\) is 1 and minimum value of \(\sin \theta=-1\) \(\therefore \frac{1}{4 \mathrm{x}^2-1} \leq 1\) \(\Rightarrow \left.\left(4 \mathrm{x}^2-1\right) \geq 1 \quad\left(\because\left(\frac{1}{4 \mathrm{x}^2-1}\right)^{-1}\right) \geq(1)^{-1}\right)\) \(\Rightarrow 4 \mathrm{x}^2-1-1 \geq 0\) \(\Rightarrow 4 \mathrm{x}^2-2 \geq 0\) \(\Rightarrow 2 \mathrm{x}^2-1 \geq 0\) \(\Rightarrow \mathrm{x}^2 \geq \frac{1}{2}\) \(\Rightarrow \mathrm{x} \leq-\frac{1}{\sqrt{2}} \text { and } \mathrm{x} \geq \frac{1}{\sqrt{2}}\) \(\therefore\) The required interval is \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right)\) We also know \(\mathrm{x}=0\) \(\Rightarrow \sin ^{-1}\left(\frac{1}{-1}\right)=\sin ^{-1}(-1)\) which is also possible. \(\therefore\) The required interval is - \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}\)
Shift-I
Sets, Relation and Function
117323
The domain of the function \(f(x)=\sin ^{-1}\left[2 x^2-3\right]\) \(+\log _2\left(\log _{\frac{1}{2}}\left(x^2-5 x+5\right)\right)\), where \([t]\) is the greatest integer function is :
117324
The range of the function \(f(x)=\sqrt{3-x}+\sqrt{2+x}\) is :
1 \([2 \sqrt{2}, \sqrt{11}]\)
2 \([\sqrt{5}, \sqrt{10}]\)
3 \([\sqrt{5}, \sqrt{13}]\)
4 \([\sqrt{2}, \sqrt{7}]\)
Explanation:
B We have to find the range of \(f(x)=\sqrt{3-x}+\sqrt{2+x}\), The function is possible for \(3-x \geq 0\) and \(2+x \geq 0\) or \(\quad x \in[-2,3]\) Let \(\quad y=\sqrt{3-x}+\sqrt{2+x}\) \(\Rightarrow \quad \mathrm{y}^2=3-\mathrm{x}+2+\mathrm{x}+2 \sqrt{(3-\mathrm{x})(2+\mathrm{x})}\) \(\Rightarrow \quad y^2=5+2 \cdot \sqrt{(3-x)(2+x)}\) \(\therefore \quad \frac{\mathrm{y}^2-5}{2}=\sqrt{(3-\mathrm{x})(2+\mathrm{x})}\) \(\Rightarrow \quad=6-2 \mathrm{x}+3 \mathrm{x}-\mathrm{x}^2\) \(=6+x-x^2\) \(\therefore \mathrm{x}^2-\mathrm{x}+\left[\frac{\mathrm{y}^2-5}{2}\right]^2-6=0\) \(\mathrm{x} \text { is real } \Rightarrow \text { Discriminate, } \mathrm{D}, \geq 0\) \(\therefore 1-4\left[\left[\frac{\mathrm{y}^2-5}{2}\right]^2-6\right] \geq 0\) \(\Rightarrow \quad 25 \geq 4\left(\frac{\mathrm{y}^2-5}{2}\right)^2 .\) \(\Rightarrow \quad\left(\frac{\mathrm{y}^2-5}{2}\right)^2 \leq \frac{25}{4}\) \(\therefore \frac{y^2-5}{2} \leq \frac{5}{2}\) \(\Rightarrow \mathrm{y}^2 \leq 10\) \(\text { or } y^2 \leq \sqrt{10}\) Also, we know \(x \in[-2,3]\) \(\therefore \quad f(-2)=\sqrt{5}\) \(f(3)=\sqrt{5}\)\(\therefore\) The range is \((\sqrt{5}, \sqrt{10})\)
117321
If \(f(x)=\log _e\left(6-\left|x^2+x-6\right|\right)\), then domain of \(f(x)\) has how many integral values of \(x\) ?
1 5
2 4
3 Infinite
4 None of these
Explanation:
D Given, \(f(\mathrm{x})=\log _{\mathrm{e}}\left(6-\left|\mathrm{x}^2+\mathrm{x}-6\right|\right)\) The function \(f(\mathrm{x})\) is defined, if - \(\left(6-\left|\mathrm{x}^2+\mathrm{x}-6\right|\right)>0\) \(\Rightarrow\) \(\left|\mathrm{x}^2+\mathrm{x}-6\right|\lt 6\) \(\Rightarrow\) \(-6\lt x^2+x-6\lt 6\) \(\text { if }\) \(\mathrm{x}^2+\mathrm{x}-6\lt 6\) \(\Rightarrow \quad \mathrm{x}^2+\mathrm{x}-12\lt 0\) \(\Rightarrow\) \((\mathrm{x}+4)(\mathrm{x}-3)\lt 0\) \(\therefore \quad \mathrm{x} \in(-4,3)\) Now, if \(-6\lt x^2+x-6\) \(\therefore \quad \mathrm{x} \in(0, \infty)\) From equation (i) and (ii), we get - \(\mathrm{x} \in(0,3)\) \(\because \mathrm{f}(\mathrm{x})\) has only two integral values. \(\therefore \quad \mathrm{x}=1,2\)
CG PET- 2015
Sets, Relation and Function
117322
The domain of the function \(\cos ^{-1}\left(\frac{2 \sin ^{-1}\left(\frac{1}{4 \mathrm{x}^2-1}\right)}{\pi}\right)\) is:
D Given, the function - \(\cos ^{-1}\left[\frac{2 \sin ^{-1}\left(\frac{1}{4 \mathrm{x}^2-1}\right)}{\pi}\right] \text { we have }\) To find interval in which \(x\) lies. Here, the max value of \(\sin \theta\) is 1 and minimum value of \(\sin \theta=-1\) \(\therefore \frac{1}{4 \mathrm{x}^2-1} \leq 1\) \(\Rightarrow \left.\left(4 \mathrm{x}^2-1\right) \geq 1 \quad\left(\because\left(\frac{1}{4 \mathrm{x}^2-1}\right)^{-1}\right) \geq(1)^{-1}\right)\) \(\Rightarrow 4 \mathrm{x}^2-1-1 \geq 0\) \(\Rightarrow 4 \mathrm{x}^2-2 \geq 0\) \(\Rightarrow 2 \mathrm{x}^2-1 \geq 0\) \(\Rightarrow \mathrm{x}^2 \geq \frac{1}{2}\) \(\Rightarrow \mathrm{x} \leq-\frac{1}{\sqrt{2}} \text { and } \mathrm{x} \geq \frac{1}{\sqrt{2}}\) \(\therefore\) The required interval is \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right)\) We also know \(\mathrm{x}=0\) \(\Rightarrow \sin ^{-1}\left(\frac{1}{-1}\right)=\sin ^{-1}(-1)\) which is also possible. \(\therefore\) The required interval is - \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}\)
Shift-I
Sets, Relation and Function
117323
The domain of the function \(f(x)=\sin ^{-1}\left[2 x^2-3\right]\) \(+\log _2\left(\log _{\frac{1}{2}}\left(x^2-5 x+5\right)\right)\), where \([t]\) is the greatest integer function is :
117324
The range of the function \(f(x)=\sqrt{3-x}+\sqrt{2+x}\) is :
1 \([2 \sqrt{2}, \sqrt{11}]\)
2 \([\sqrt{5}, \sqrt{10}]\)
3 \([\sqrt{5}, \sqrt{13}]\)
4 \([\sqrt{2}, \sqrt{7}]\)
Explanation:
B We have to find the range of \(f(x)=\sqrt{3-x}+\sqrt{2+x}\), The function is possible for \(3-x \geq 0\) and \(2+x \geq 0\) or \(\quad x \in[-2,3]\) Let \(\quad y=\sqrt{3-x}+\sqrt{2+x}\) \(\Rightarrow \quad \mathrm{y}^2=3-\mathrm{x}+2+\mathrm{x}+2 \sqrt{(3-\mathrm{x})(2+\mathrm{x})}\) \(\Rightarrow \quad y^2=5+2 \cdot \sqrt{(3-x)(2+x)}\) \(\therefore \quad \frac{\mathrm{y}^2-5}{2}=\sqrt{(3-\mathrm{x})(2+\mathrm{x})}\) \(\Rightarrow \quad=6-2 \mathrm{x}+3 \mathrm{x}-\mathrm{x}^2\) \(=6+x-x^2\) \(\therefore \mathrm{x}^2-\mathrm{x}+\left[\frac{\mathrm{y}^2-5}{2}\right]^2-6=0\) \(\mathrm{x} \text { is real } \Rightarrow \text { Discriminate, } \mathrm{D}, \geq 0\) \(\therefore 1-4\left[\left[\frac{\mathrm{y}^2-5}{2}\right]^2-6\right] \geq 0\) \(\Rightarrow \quad 25 \geq 4\left(\frac{\mathrm{y}^2-5}{2}\right)^2 .\) \(\Rightarrow \quad\left(\frac{\mathrm{y}^2-5}{2}\right)^2 \leq \frac{25}{4}\) \(\therefore \frac{y^2-5}{2} \leq \frac{5}{2}\) \(\Rightarrow \mathrm{y}^2 \leq 10\) \(\text { or } y^2 \leq \sqrt{10}\) Also, we know \(x \in[-2,3]\) \(\therefore \quad f(-2)=\sqrt{5}\) \(f(3)=\sqrt{5}\)\(\therefore\) The range is \((\sqrt{5}, \sqrt{10})\)
