117051
If \(F: R \rightarrow R\) is defined by \(f(x)=2 x+|x|\), then \(\mathbf{f}(\mathbf{2 x})+\mathbf{f}(-\mathbf{x})-\mathbf{f}(\mathbf{x})\) is equal to
1 \(2 x\)
2 \(2|\mathrm{x}|\)
3 \(-2 x\)
4 \(-2|x|\)
Explanation:
B We have, \(f(x)=2 x+|x|\) \(\text { Then, } f(2 x)+f(-x)-f(x)\) \(\Rightarrow 2(2 x)+|2 x|+2(-x)+|-x|-2 x-|x|\) \(\Rightarrow 4 x+|2 x|-2 x+|-x|-2 x-|x|\) \(\Rightarrow |2 x|+|-x|-|x|\) \(\text { If } x>0, 2 x+x-x\) \(=2 x\) If \(x\lt 0\), \(=-2 \mathrm{x}+(-\mathrm{x})+\mathrm{x}\) \(=-2 \mathrm{x}\)So, the value of \(f(2 x)+f(-x)-f(x)=2|x|\).
EAMCET-2000
Sets, Relation and Function
117053
If \(f:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty]\) be a function defined by \(y=\sin \left(\frac{x}{2}\right)\) then \(f\) is
1 Injective
2 surjective
3 bijective
4 None of these
Explanation:
A Given, \(y=\sin \left(\frac{x}{2}\right)\) And, \(\quad \mathrm{f}:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)\) Then, \(\quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) \(0 \leq \frac{x}{2} \leq \frac{\pi}{4}\) \(\sin 0 \leq \sin \frac{x}{2} \leq \sin \frac{\pi}{4}\) \(0 \leq \sin \frac{x}{2} \leq \frac{1}{\sqrt{2}}\) \(\left(0, \frac{1}{\sqrt{2}}\right) \subset[0, \infty)\) So, the function is an injective but is not surjective as for \(0 \leq \mathrm{x} \leq \frac{\pi}{2}\) where, \(\sin \frac{\mathrm{x}}{2}\) gives unique image .
JCECE-2013
Sets, Relation and Function
117054
Let \(f: R \rightarrow\) satisfy \(f(x) f(y)=f(x y)\) for all real number \(x\) and \(y\). If \(f(2)=4\), then \(f\left(\frac{1}{2}\right)=\)
1 0
2 \(\frac{1}{4}\)
3 \(\frac{1}{2}\)
4 1
5 2
Explanation:
B Given, \(\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})=\mathrm{f}(\mathrm{xy})\) Taking value \(\mathrm{x}=1, \mathrm{y}=1\) \(\mathrm{f}(1) \mathrm{f}(1)=\mathrm{f}(1 \times 1) \Rightarrow \mathrm{f}(1)^2=\mathrm{f}(1) \Rightarrow \mathrm{f}(1)=1\) Now taking \(\mathrm{x}=2 \mathrm{y}=\frac{1}{2}\) \(\mathrm{f}(2) \times \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}\left(2 \times \frac{1}{2}\right)=\mathrm{f}(1)\) \(4 \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}(1) (\because \mathrm{f}(2)=4 \text { given })\) \(\mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{4} \mathrm{f}(1)\) On putting the value of \(f(1)\), we get - \(\mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{4} \times 1=\frac{1}{4}\)
Kerala CEE-2018
Sets, Relation and Function
117055
If \(f(x)=a \log |\mathbf{x}|+b x^2+x\) has its extremun values at \(x=-1\) and \(x=2\), then
1 \(\mathrm{a}=2, \mathrm{~b}=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-2 \mathrm{~b}=\frac{1}{2}\)
4 None of these
Explanation:
B Given, \(f(x)=a \log |x|+b x^2+x\) \(f^{\prime}(x)=\frac{a}{x}+2 b x+1\) \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Now, \(\left.\quad \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=-1}=\frac{\mathrm{a}}{-1}+2 \mathrm{~b} \times-1+1\) \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=-1}=-\mathrm{a}-2 \mathrm{~b}+1\) And \(\left.\quad \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=2}=\frac{\mathrm{a}}{2}+2 \mathrm{~b} \times 2+1\) \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=2}=\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1\) Since, for extremum \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Then, \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) Then, \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) and \(-\mathrm{a}-2 \mathrm{~b}+1=0\) \(\frac{a}{2}+4 b=-1\) \(-a-2 b=-1\) \(a+8 b=-2\) \(a+2 b=1\) From subtracting equation (i) by equation, (ii) we get - \(6 b=-3\) \(b=\frac{-3}{6}=\frac{-1}{2}\) Then, \(a+8 \times-\frac{1}{2}=-2\) \(a-4=-2\) \(a=-2+4\) \(a=2\)So, \(\quad \mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
117051
If \(F: R \rightarrow R\) is defined by \(f(x)=2 x+|x|\), then \(\mathbf{f}(\mathbf{2 x})+\mathbf{f}(-\mathbf{x})-\mathbf{f}(\mathbf{x})\) is equal to
1 \(2 x\)
2 \(2|\mathrm{x}|\)
3 \(-2 x\)
4 \(-2|x|\)
Explanation:
B We have, \(f(x)=2 x+|x|\) \(\text { Then, } f(2 x)+f(-x)-f(x)\) \(\Rightarrow 2(2 x)+|2 x|+2(-x)+|-x|-2 x-|x|\) \(\Rightarrow 4 x+|2 x|-2 x+|-x|-2 x-|x|\) \(\Rightarrow |2 x|+|-x|-|x|\) \(\text { If } x>0, 2 x+x-x\) \(=2 x\) If \(x\lt 0\), \(=-2 \mathrm{x}+(-\mathrm{x})+\mathrm{x}\) \(=-2 \mathrm{x}\)So, the value of \(f(2 x)+f(-x)-f(x)=2|x|\).
EAMCET-2000
Sets, Relation and Function
117053
If \(f:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty]\) be a function defined by \(y=\sin \left(\frac{x}{2}\right)\) then \(f\) is
1 Injective
2 surjective
3 bijective
4 None of these
Explanation:
A Given, \(y=\sin \left(\frac{x}{2}\right)\) And, \(\quad \mathrm{f}:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)\) Then, \(\quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) \(0 \leq \frac{x}{2} \leq \frac{\pi}{4}\) \(\sin 0 \leq \sin \frac{x}{2} \leq \sin \frac{\pi}{4}\) \(0 \leq \sin \frac{x}{2} \leq \frac{1}{\sqrt{2}}\) \(\left(0, \frac{1}{\sqrt{2}}\right) \subset[0, \infty)\) So, the function is an injective but is not surjective as for \(0 \leq \mathrm{x} \leq \frac{\pi}{2}\) where, \(\sin \frac{\mathrm{x}}{2}\) gives unique image .
JCECE-2013
Sets, Relation and Function
117054
Let \(f: R \rightarrow\) satisfy \(f(x) f(y)=f(x y)\) for all real number \(x\) and \(y\). If \(f(2)=4\), then \(f\left(\frac{1}{2}\right)=\)
1 0
2 \(\frac{1}{4}\)
3 \(\frac{1}{2}\)
4 1
5 2
Explanation:
B Given, \(\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})=\mathrm{f}(\mathrm{xy})\) Taking value \(\mathrm{x}=1, \mathrm{y}=1\) \(\mathrm{f}(1) \mathrm{f}(1)=\mathrm{f}(1 \times 1) \Rightarrow \mathrm{f}(1)^2=\mathrm{f}(1) \Rightarrow \mathrm{f}(1)=1\) Now taking \(\mathrm{x}=2 \mathrm{y}=\frac{1}{2}\) \(\mathrm{f}(2) \times \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}\left(2 \times \frac{1}{2}\right)=\mathrm{f}(1)\) \(4 \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}(1) (\because \mathrm{f}(2)=4 \text { given })\) \(\mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{4} \mathrm{f}(1)\) On putting the value of \(f(1)\), we get - \(\mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{4} \times 1=\frac{1}{4}\)
Kerala CEE-2018
Sets, Relation and Function
117055
If \(f(x)=a \log |\mathbf{x}|+b x^2+x\) has its extremun values at \(x=-1\) and \(x=2\), then
1 \(\mathrm{a}=2, \mathrm{~b}=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-2 \mathrm{~b}=\frac{1}{2}\)
4 None of these
Explanation:
B Given, \(f(x)=a \log |x|+b x^2+x\) \(f^{\prime}(x)=\frac{a}{x}+2 b x+1\) \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Now, \(\left.\quad \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=-1}=\frac{\mathrm{a}}{-1}+2 \mathrm{~b} \times-1+1\) \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=-1}=-\mathrm{a}-2 \mathrm{~b}+1\) And \(\left.\quad \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=2}=\frac{\mathrm{a}}{2}+2 \mathrm{~b} \times 2+1\) \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=2}=\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1\) Since, for extremum \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Then, \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) Then, \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) and \(-\mathrm{a}-2 \mathrm{~b}+1=0\) \(\frac{a}{2}+4 b=-1\) \(-a-2 b=-1\) \(a+8 b=-2\) \(a+2 b=1\) From subtracting equation (i) by equation, (ii) we get - \(6 b=-3\) \(b=\frac{-3}{6}=\frac{-1}{2}\) Then, \(a+8 \times-\frac{1}{2}=-2\) \(a-4=-2\) \(a=-2+4\) \(a=2\)So, \(\quad \mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
117051
If \(F: R \rightarrow R\) is defined by \(f(x)=2 x+|x|\), then \(\mathbf{f}(\mathbf{2 x})+\mathbf{f}(-\mathbf{x})-\mathbf{f}(\mathbf{x})\) is equal to
1 \(2 x\)
2 \(2|\mathrm{x}|\)
3 \(-2 x\)
4 \(-2|x|\)
Explanation:
B We have, \(f(x)=2 x+|x|\) \(\text { Then, } f(2 x)+f(-x)-f(x)\) \(\Rightarrow 2(2 x)+|2 x|+2(-x)+|-x|-2 x-|x|\) \(\Rightarrow 4 x+|2 x|-2 x+|-x|-2 x-|x|\) \(\Rightarrow |2 x|+|-x|-|x|\) \(\text { If } x>0, 2 x+x-x\) \(=2 x\) If \(x\lt 0\), \(=-2 \mathrm{x}+(-\mathrm{x})+\mathrm{x}\) \(=-2 \mathrm{x}\)So, the value of \(f(2 x)+f(-x)-f(x)=2|x|\).
EAMCET-2000
Sets, Relation and Function
117053
If \(f:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty]\) be a function defined by \(y=\sin \left(\frac{x}{2}\right)\) then \(f\) is
1 Injective
2 surjective
3 bijective
4 None of these
Explanation:
A Given, \(y=\sin \left(\frac{x}{2}\right)\) And, \(\quad \mathrm{f}:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)\) Then, \(\quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) \(0 \leq \frac{x}{2} \leq \frac{\pi}{4}\) \(\sin 0 \leq \sin \frac{x}{2} \leq \sin \frac{\pi}{4}\) \(0 \leq \sin \frac{x}{2} \leq \frac{1}{\sqrt{2}}\) \(\left(0, \frac{1}{\sqrt{2}}\right) \subset[0, \infty)\) So, the function is an injective but is not surjective as for \(0 \leq \mathrm{x} \leq \frac{\pi}{2}\) where, \(\sin \frac{\mathrm{x}}{2}\) gives unique image .
JCECE-2013
Sets, Relation and Function
117054
Let \(f: R \rightarrow\) satisfy \(f(x) f(y)=f(x y)\) for all real number \(x\) and \(y\). If \(f(2)=4\), then \(f\left(\frac{1}{2}\right)=\)
1 0
2 \(\frac{1}{4}\)
3 \(\frac{1}{2}\)
4 1
5 2
Explanation:
B Given, \(\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})=\mathrm{f}(\mathrm{xy})\) Taking value \(\mathrm{x}=1, \mathrm{y}=1\) \(\mathrm{f}(1) \mathrm{f}(1)=\mathrm{f}(1 \times 1) \Rightarrow \mathrm{f}(1)^2=\mathrm{f}(1) \Rightarrow \mathrm{f}(1)=1\) Now taking \(\mathrm{x}=2 \mathrm{y}=\frac{1}{2}\) \(\mathrm{f}(2) \times \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}\left(2 \times \frac{1}{2}\right)=\mathrm{f}(1)\) \(4 \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}(1) (\because \mathrm{f}(2)=4 \text { given })\) \(\mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{4} \mathrm{f}(1)\) On putting the value of \(f(1)\), we get - \(\mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{4} \times 1=\frac{1}{4}\)
Kerala CEE-2018
Sets, Relation and Function
117055
If \(f(x)=a \log |\mathbf{x}|+b x^2+x\) has its extremun values at \(x=-1\) and \(x=2\), then
1 \(\mathrm{a}=2, \mathrm{~b}=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-2 \mathrm{~b}=\frac{1}{2}\)
4 None of these
Explanation:
B Given, \(f(x)=a \log |x|+b x^2+x\) \(f^{\prime}(x)=\frac{a}{x}+2 b x+1\) \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Now, \(\left.\quad \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=-1}=\frac{\mathrm{a}}{-1}+2 \mathrm{~b} \times-1+1\) \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=-1}=-\mathrm{a}-2 \mathrm{~b}+1\) And \(\left.\quad \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=2}=\frac{\mathrm{a}}{2}+2 \mathrm{~b} \times 2+1\) \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=2}=\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1\) Since, for extremum \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Then, \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) Then, \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) and \(-\mathrm{a}-2 \mathrm{~b}+1=0\) \(\frac{a}{2}+4 b=-1\) \(-a-2 b=-1\) \(a+8 b=-2\) \(a+2 b=1\) From subtracting equation (i) by equation, (ii) we get - \(6 b=-3\) \(b=\frac{-3}{6}=\frac{-1}{2}\) Then, \(a+8 \times-\frac{1}{2}=-2\) \(a-4=-2\) \(a=-2+4\) \(a=2\)So, \(\quad \mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117051
If \(F: R \rightarrow R\) is defined by \(f(x)=2 x+|x|\), then \(\mathbf{f}(\mathbf{2 x})+\mathbf{f}(-\mathbf{x})-\mathbf{f}(\mathbf{x})\) is equal to
1 \(2 x\)
2 \(2|\mathrm{x}|\)
3 \(-2 x\)
4 \(-2|x|\)
Explanation:
B We have, \(f(x)=2 x+|x|\) \(\text { Then, } f(2 x)+f(-x)-f(x)\) \(\Rightarrow 2(2 x)+|2 x|+2(-x)+|-x|-2 x-|x|\) \(\Rightarrow 4 x+|2 x|-2 x+|-x|-2 x-|x|\) \(\Rightarrow |2 x|+|-x|-|x|\) \(\text { If } x>0, 2 x+x-x\) \(=2 x\) If \(x\lt 0\), \(=-2 \mathrm{x}+(-\mathrm{x})+\mathrm{x}\) \(=-2 \mathrm{x}\)So, the value of \(f(2 x)+f(-x)-f(x)=2|x|\).
EAMCET-2000
Sets, Relation and Function
117053
If \(f:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty]\) be a function defined by \(y=\sin \left(\frac{x}{2}\right)\) then \(f\) is
1 Injective
2 surjective
3 bijective
4 None of these
Explanation:
A Given, \(y=\sin \left(\frac{x}{2}\right)\) And, \(\quad \mathrm{f}:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)\) Then, \(\quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) \(0 \leq \frac{x}{2} \leq \frac{\pi}{4}\) \(\sin 0 \leq \sin \frac{x}{2} \leq \sin \frac{\pi}{4}\) \(0 \leq \sin \frac{x}{2} \leq \frac{1}{\sqrt{2}}\) \(\left(0, \frac{1}{\sqrt{2}}\right) \subset[0, \infty)\) So, the function is an injective but is not surjective as for \(0 \leq \mathrm{x} \leq \frac{\pi}{2}\) where, \(\sin \frac{\mathrm{x}}{2}\) gives unique image .
JCECE-2013
Sets, Relation and Function
117054
Let \(f: R \rightarrow\) satisfy \(f(x) f(y)=f(x y)\) for all real number \(x\) and \(y\). If \(f(2)=4\), then \(f\left(\frac{1}{2}\right)=\)
1 0
2 \(\frac{1}{4}\)
3 \(\frac{1}{2}\)
4 1
5 2
Explanation:
B Given, \(\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})=\mathrm{f}(\mathrm{xy})\) Taking value \(\mathrm{x}=1, \mathrm{y}=1\) \(\mathrm{f}(1) \mathrm{f}(1)=\mathrm{f}(1 \times 1) \Rightarrow \mathrm{f}(1)^2=\mathrm{f}(1) \Rightarrow \mathrm{f}(1)=1\) Now taking \(\mathrm{x}=2 \mathrm{y}=\frac{1}{2}\) \(\mathrm{f}(2) \times \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}\left(2 \times \frac{1}{2}\right)=\mathrm{f}(1)\) \(4 \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{f}(1) (\because \mathrm{f}(2)=4 \text { given })\) \(\mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{4} \mathrm{f}(1)\) On putting the value of \(f(1)\), we get - \(\mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{4} \times 1=\frac{1}{4}\)
Kerala CEE-2018
Sets, Relation and Function
117055
If \(f(x)=a \log |\mathbf{x}|+b x^2+x\) has its extremun values at \(x=-1\) and \(x=2\), then
1 \(\mathrm{a}=2, \mathrm{~b}=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-2 \mathrm{~b}=\frac{1}{2}\)
4 None of these
Explanation:
B Given, \(f(x)=a \log |x|+b x^2+x\) \(f^{\prime}(x)=\frac{a}{x}+2 b x+1\) \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Now, \(\left.\quad \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=-1}=\frac{\mathrm{a}}{-1}+2 \mathrm{~b} \times-1+1\) \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=-1}=-\mathrm{a}-2 \mathrm{~b}+1\) And \(\left.\quad \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=2}=\frac{\mathrm{a}}{2}+2 \mathrm{~b} \times 2+1\) \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=2}=\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1\) Since, for extremum \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Then, \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) Then, \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) and \(-\mathrm{a}-2 \mathrm{~b}+1=0\) \(\frac{a}{2}+4 b=-1\) \(-a-2 b=-1\) \(a+8 b=-2\) \(a+2 b=1\) From subtracting equation (i) by equation, (ii) we get - \(6 b=-3\) \(b=\frac{-3}{6}=\frac{-1}{2}\) Then, \(a+8 \times-\frac{1}{2}=-2\) \(a-4=-2\) \(a=-2+4\) \(a=2\)So, \(\quad \mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
