NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
116931
The period of the function \(f(x)=|\sin x|-|\cos x|\)
1 \(\pi / 2\)
2 \(\pi\)
3 \(2 \pi\)
4 None of these
Explanation:
B Given, the period of the function is- \(\mathrm{f}(\mathrm{x})=|\sin \mathrm{x}|-|\cos \mathrm{x}|\) \(\therefore \mathrm{f}(\mathrm{x}+\pi)=|\sin (\mathrm{x}+\pi)|-|\cos (\mathrm{x}+\pi)|\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=|\sin \mathrm{x}|-|\cos \mathrm{x}|\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=|\sin \mathrm{x}|-|\cos \mathrm{x}|=\mathrm{f}(\mathrm{x}), \text { for all } \mathrm{x} \in \mathrm{R}\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=\mathrm{f}(\mathrm{x}) \text { for all } \mathrm{x} \in \mathrm{R}\)So, \(\mathrm{f}(\mathrm{x})\) is periodic with period \(\pi\).
BITSAT-2016
Sets, Relation and Function
116932
If \(f(x)=\cot ^{-1}\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)\), then \(f^{\prime}(1)\) is equal to
1 -1
2 1
3 \(\log 2\)
4 \(-\log 2\)
Explanation:
A Given, \(f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)\) Firstly, differentiate the \(\mathrm{x}^{\mathrm{x}}\) and \(\mathrm{x}^{-\mathrm{x}}\) Let, \(\mathrm{x}^{\mathrm{x}}=\mathrm{u}\) and \(\mathrm{x}^{-\mathrm{x}}=\mathrm{v}\) Taking log on both side, we get \(\log \mathrm{u}=\mathrm{x} \log \mathrm{x}\) and \(\log \mathrm{v}=-\mathrm{x} \log \mathrm{x}\) Then, differentiate - \(\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{x}+\log x\) \(\text { And, } \frac{1}{v} \frac{d v}{d x}=-x \cdot \frac{1}{x}-\log x\) \(\frac{d u}{d x}=u(1+\log x) \text { and } \frac{d v}{d x}=v(-1-\log x)\) \(\frac{d u}{d x}=x^x(1+\log x) \text { and } \frac{d v}{d x}=-x^{-x}(1+\log x)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)^2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)\) \(f^{\prime}(x)=\frac{-4}{x^{2 x}+x^{-2 x}+4-2}\left[\frac{1}{2}\left[\left(x^x+x^{-x}\right)(1+\log x)\right]\right]\) \(f^{\prime}(x)=\frac{-4}{x^{2 x}+x^{-2 x}+2}\left[\frac{1}{2}\left[\left(x^x+x^{-x}\right)(1+\log x)\right]\right]\) Hence, \(f^{\prime}(1)=\frac{-2}{1+1+2}[2\{1+\log 1\}]\) \(f^{\prime}(1)=\frac{-4}{4}\) \(f^{\prime}(1)=-1\)
UPSEE - 2008
Sets, Relation and Function
116933
If \(2 \log (x+1)-\log \left(x^2-1\right)=\log 2\), then \(x=\)
116931
The period of the function \(f(x)=|\sin x|-|\cos x|\)
1 \(\pi / 2\)
2 \(\pi\)
3 \(2 \pi\)
4 None of these
Explanation:
B Given, the period of the function is- \(\mathrm{f}(\mathrm{x})=|\sin \mathrm{x}|-|\cos \mathrm{x}|\) \(\therefore \mathrm{f}(\mathrm{x}+\pi)=|\sin (\mathrm{x}+\pi)|-|\cos (\mathrm{x}+\pi)|\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=|\sin \mathrm{x}|-|\cos \mathrm{x}|\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=|\sin \mathrm{x}|-|\cos \mathrm{x}|=\mathrm{f}(\mathrm{x}), \text { for all } \mathrm{x} \in \mathrm{R}\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=\mathrm{f}(\mathrm{x}) \text { for all } \mathrm{x} \in \mathrm{R}\)So, \(\mathrm{f}(\mathrm{x})\) is periodic with period \(\pi\).
BITSAT-2016
Sets, Relation and Function
116932
If \(f(x)=\cot ^{-1}\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)\), then \(f^{\prime}(1)\) is equal to
1 -1
2 1
3 \(\log 2\)
4 \(-\log 2\)
Explanation:
A Given, \(f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)\) Firstly, differentiate the \(\mathrm{x}^{\mathrm{x}}\) and \(\mathrm{x}^{-\mathrm{x}}\) Let, \(\mathrm{x}^{\mathrm{x}}=\mathrm{u}\) and \(\mathrm{x}^{-\mathrm{x}}=\mathrm{v}\) Taking log on both side, we get \(\log \mathrm{u}=\mathrm{x} \log \mathrm{x}\) and \(\log \mathrm{v}=-\mathrm{x} \log \mathrm{x}\) Then, differentiate - \(\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{x}+\log x\) \(\text { And, } \frac{1}{v} \frac{d v}{d x}=-x \cdot \frac{1}{x}-\log x\) \(\frac{d u}{d x}=u(1+\log x) \text { and } \frac{d v}{d x}=v(-1-\log x)\) \(\frac{d u}{d x}=x^x(1+\log x) \text { and } \frac{d v}{d x}=-x^{-x}(1+\log x)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)^2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)\) \(f^{\prime}(x)=\frac{-4}{x^{2 x}+x^{-2 x}+4-2}\left[\frac{1}{2}\left[\left(x^x+x^{-x}\right)(1+\log x)\right]\right]\) \(f^{\prime}(x)=\frac{-4}{x^{2 x}+x^{-2 x}+2}\left[\frac{1}{2}\left[\left(x^x+x^{-x}\right)(1+\log x)\right]\right]\) Hence, \(f^{\prime}(1)=\frac{-2}{1+1+2}[2\{1+\log 1\}]\) \(f^{\prime}(1)=\frac{-4}{4}\) \(f^{\prime}(1)=-1\)
UPSEE - 2008
Sets, Relation and Function
116933
If \(2 \log (x+1)-\log \left(x^2-1\right)=\log 2\), then \(x=\)
116931
The period of the function \(f(x)=|\sin x|-|\cos x|\)
1 \(\pi / 2\)
2 \(\pi\)
3 \(2 \pi\)
4 None of these
Explanation:
B Given, the period of the function is- \(\mathrm{f}(\mathrm{x})=|\sin \mathrm{x}|-|\cos \mathrm{x}|\) \(\therefore \mathrm{f}(\mathrm{x}+\pi)=|\sin (\mathrm{x}+\pi)|-|\cos (\mathrm{x}+\pi)|\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=|\sin \mathrm{x}|-|\cos \mathrm{x}|\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=|\sin \mathrm{x}|-|\cos \mathrm{x}|=\mathrm{f}(\mathrm{x}), \text { for all } \mathrm{x} \in \mathrm{R}\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=\mathrm{f}(\mathrm{x}) \text { for all } \mathrm{x} \in \mathrm{R}\)So, \(\mathrm{f}(\mathrm{x})\) is periodic with period \(\pi\).
BITSAT-2016
Sets, Relation and Function
116932
If \(f(x)=\cot ^{-1}\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)\), then \(f^{\prime}(1)\) is equal to
1 -1
2 1
3 \(\log 2\)
4 \(-\log 2\)
Explanation:
A Given, \(f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)\) Firstly, differentiate the \(\mathrm{x}^{\mathrm{x}}\) and \(\mathrm{x}^{-\mathrm{x}}\) Let, \(\mathrm{x}^{\mathrm{x}}=\mathrm{u}\) and \(\mathrm{x}^{-\mathrm{x}}=\mathrm{v}\) Taking log on both side, we get \(\log \mathrm{u}=\mathrm{x} \log \mathrm{x}\) and \(\log \mathrm{v}=-\mathrm{x} \log \mathrm{x}\) Then, differentiate - \(\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{x}+\log x\) \(\text { And, } \frac{1}{v} \frac{d v}{d x}=-x \cdot \frac{1}{x}-\log x\) \(\frac{d u}{d x}=u(1+\log x) \text { and } \frac{d v}{d x}=v(-1-\log x)\) \(\frac{d u}{d x}=x^x(1+\log x) \text { and } \frac{d v}{d x}=-x^{-x}(1+\log x)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)^2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)\) \(f^{\prime}(x)=\frac{-4}{x^{2 x}+x^{-2 x}+4-2}\left[\frac{1}{2}\left[\left(x^x+x^{-x}\right)(1+\log x)\right]\right]\) \(f^{\prime}(x)=\frac{-4}{x^{2 x}+x^{-2 x}+2}\left[\frac{1}{2}\left[\left(x^x+x^{-x}\right)(1+\log x)\right]\right]\) Hence, \(f^{\prime}(1)=\frac{-2}{1+1+2}[2\{1+\log 1\}]\) \(f^{\prime}(1)=\frac{-4}{4}\) \(f^{\prime}(1)=-1\)
UPSEE - 2008
Sets, Relation and Function
116933
If \(2 \log (x+1)-\log \left(x^2-1\right)=\log 2\), then \(x=\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
116931
The period of the function \(f(x)=|\sin x|-|\cos x|\)
1 \(\pi / 2\)
2 \(\pi\)
3 \(2 \pi\)
4 None of these
Explanation:
B Given, the period of the function is- \(\mathrm{f}(\mathrm{x})=|\sin \mathrm{x}|-|\cos \mathrm{x}|\) \(\therefore \mathrm{f}(\mathrm{x}+\pi)=|\sin (\mathrm{x}+\pi)|-|\cos (\mathrm{x}+\pi)|\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=|\sin \mathrm{x}|-|\cos \mathrm{x}|\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=|\sin \mathrm{x}|-|\cos \mathrm{x}|=\mathrm{f}(\mathrm{x}), \text { for all } \mathrm{x} \in \mathrm{R}\) \(\Rightarrow \mathrm{f}(\mathrm{x}+\pi)=\mathrm{f}(\mathrm{x}) \text { for all } \mathrm{x} \in \mathrm{R}\)So, \(\mathrm{f}(\mathrm{x})\) is periodic with period \(\pi\).
BITSAT-2016
Sets, Relation and Function
116932
If \(f(x)=\cot ^{-1}\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)\), then \(f^{\prime}(1)\) is equal to
1 -1
2 1
3 \(\log 2\)
4 \(-\log 2\)
Explanation:
A Given, \(f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)\) Firstly, differentiate the \(\mathrm{x}^{\mathrm{x}}\) and \(\mathrm{x}^{-\mathrm{x}}\) Let, \(\mathrm{x}^{\mathrm{x}}=\mathrm{u}\) and \(\mathrm{x}^{-\mathrm{x}}=\mathrm{v}\) Taking log on both side, we get \(\log \mathrm{u}=\mathrm{x} \log \mathrm{x}\) and \(\log \mathrm{v}=-\mathrm{x} \log \mathrm{x}\) Then, differentiate - \(\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{x}+\log x\) \(\text { And, } \frac{1}{v} \frac{d v}{d x}=-x \cdot \frac{1}{x}-\log x\) \(\frac{d u}{d x}=u(1+\log x) \text { and } \frac{d v}{d x}=v(-1-\log x)\) \(\frac{d u}{d x}=x^x(1+\log x) \text { and } \frac{d v}{d x}=-x^{-x}(1+\log x)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)^2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{\mathrm{x}}-\mathrm{x}^{-\mathrm{x}}}{2}\right)\) \(f^{\prime}(x)=\frac{-4}{x^{2 x}+x^{-2 x}+4-2}\left[\frac{1}{2}\left[\left(x^x+x^{-x}\right)(1+\log x)\right]\right]\) \(f^{\prime}(x)=\frac{-4}{x^{2 x}+x^{-2 x}+2}\left[\frac{1}{2}\left[\left(x^x+x^{-x}\right)(1+\log x)\right]\right]\) Hence, \(f^{\prime}(1)=\frac{-2}{1+1+2}[2\{1+\log 1\}]\) \(f^{\prime}(1)=\frac{-4}{4}\) \(f^{\prime}(1)=-1\)
UPSEE - 2008
Sets, Relation and Function
116933
If \(2 \log (x+1)-\log \left(x^2-1\right)=\log 2\), then \(x=\)