116926
Consider the function \(f(x)=\cos ^2\). Then,
1 \(f\) is of period \(2 \pi\)
2 \(f\) is of period \(\sqrt{2 \pi}\)
3 \(\mathrm{f}\) is not periodic
4 \(\mathrm{f}\) is of period \(\pi\)
Explanation:
C We have, \(f(x)=\cos x^2\) Let \(T\) be the period of \(f(x)\). Then \(\mathrm{f}(\mathrm{x}+\mathrm{T})=\mathrm{f}(\mathrm{x})\) \(\Rightarrow \quad \cos (\mathrm{x}+\mathrm{T})^2=\cos \mathrm{x}^2\) But there is no value of \(\mathrm{T}\) for which \(\cos (\mathrm{x}+\mathrm{T})^2=\cos \mathrm{x}^2\)\(\therefore \mathrm{f}(\mathrm{x})\) is not periodic
WB JEE-2019
Sets, Relation and Function
116928
If \(\log _2 6+\frac{1}{2 x}=\log _2\left(2^{\frac{1}{x}}+8\right)\) then the values of \(x\) are
116929
Let \(f(x)=\sin x+\cos\) ax be periodic function. Then,
1 a is any real number
2 a is any irrational number
3 a is rational number
4 \(a=0\)
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\cos a \mathrm{x}\) \(\because\) Period of \(\sin x=\frac{2 \pi}{1}\) Period of \(\cos \mathrm{ax}=\frac{2 \pi}{\mathrm{a}}\) Hence period of \(f(x)=\) L.C.M of \(\left\{\frac{2 \pi}{1}, \frac{2 \pi}{a}\right\}\) \(=\frac{\text { L.C.M. of }(2 \pi, 2 \pi)}{\text { H.C.F. of }\{1, a\}}=\frac{2 \pi}{k}\) Where \(\mathrm{k}=\mathrm{H}\).C.F of 1 and \(\mathrm{a}\) \(\therefore \frac{1}{\mathrm{k}}=\text { integer }=\mathrm{q}(\text { say }) \neq 0 \text { and } \frac{\mathrm{a}}{\mathrm{k}}\) \(\text { Integer } \mathrm{p} \text { cosy }\) \(\therefore \quad \frac{\frac{\mathrm{a}}{\mathrm{k}}}{\frac{1}{\mathrm{k}}}=\frac{\mathrm{p}}{\mathrm{q}}\) \(\mathrm{a}=\frac{\mathrm{p}}{\mathrm{q}}\) \(\mathrm{a}=\) rational number
116926
Consider the function \(f(x)=\cos ^2\). Then,
1 \(f\) is of period \(2 \pi\)
2 \(f\) is of period \(\sqrt{2 \pi}\)
3 \(\mathrm{f}\) is not periodic
4 \(\mathrm{f}\) is of period \(\pi\)
Explanation:
C We have, \(f(x)=\cos x^2\) Let \(T\) be the period of \(f(x)\). Then \(\mathrm{f}(\mathrm{x}+\mathrm{T})=\mathrm{f}(\mathrm{x})\) \(\Rightarrow \quad \cos (\mathrm{x}+\mathrm{T})^2=\cos \mathrm{x}^2\) But there is no value of \(\mathrm{T}\) for which \(\cos (\mathrm{x}+\mathrm{T})^2=\cos \mathrm{x}^2\)\(\therefore \mathrm{f}(\mathrm{x})\) is not periodic
WB JEE-2019
Sets, Relation and Function
116928
If \(\log _2 6+\frac{1}{2 x}=\log _2\left(2^{\frac{1}{x}}+8\right)\) then the values of \(x\) are
116929
Let \(f(x)=\sin x+\cos\) ax be periodic function. Then,
1 a is any real number
2 a is any irrational number
3 a is rational number
4 \(a=0\)
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\cos a \mathrm{x}\) \(\because\) Period of \(\sin x=\frac{2 \pi}{1}\) Period of \(\cos \mathrm{ax}=\frac{2 \pi}{\mathrm{a}}\) Hence period of \(f(x)=\) L.C.M of \(\left\{\frac{2 \pi}{1}, \frac{2 \pi}{a}\right\}\) \(=\frac{\text { L.C.M. of }(2 \pi, 2 \pi)}{\text { H.C.F. of }\{1, a\}}=\frac{2 \pi}{k}\) Where \(\mathrm{k}=\mathrm{H}\).C.F of 1 and \(\mathrm{a}\) \(\therefore \frac{1}{\mathrm{k}}=\text { integer }=\mathrm{q}(\text { say }) \neq 0 \text { and } \frac{\mathrm{a}}{\mathrm{k}}\) \(\text { Integer } \mathrm{p} \text { cosy }\) \(\therefore \quad \frac{\frac{\mathrm{a}}{\mathrm{k}}}{\frac{1}{\mathrm{k}}}=\frac{\mathrm{p}}{\mathrm{q}}\) \(\mathrm{a}=\frac{\mathrm{p}}{\mathrm{q}}\) \(\mathrm{a}=\) rational number
116926
Consider the function \(f(x)=\cos ^2\). Then,
1 \(f\) is of period \(2 \pi\)
2 \(f\) is of period \(\sqrt{2 \pi}\)
3 \(\mathrm{f}\) is not periodic
4 \(\mathrm{f}\) is of period \(\pi\)
Explanation:
C We have, \(f(x)=\cos x^2\) Let \(T\) be the period of \(f(x)\). Then \(\mathrm{f}(\mathrm{x}+\mathrm{T})=\mathrm{f}(\mathrm{x})\) \(\Rightarrow \quad \cos (\mathrm{x}+\mathrm{T})^2=\cos \mathrm{x}^2\) But there is no value of \(\mathrm{T}\) for which \(\cos (\mathrm{x}+\mathrm{T})^2=\cos \mathrm{x}^2\)\(\therefore \mathrm{f}(\mathrm{x})\) is not periodic
WB JEE-2019
Sets, Relation and Function
116928
If \(\log _2 6+\frac{1}{2 x}=\log _2\left(2^{\frac{1}{x}}+8\right)\) then the values of \(x\) are
116929
Let \(f(x)=\sin x+\cos\) ax be periodic function. Then,
1 a is any real number
2 a is any irrational number
3 a is rational number
4 \(a=0\)
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\cos a \mathrm{x}\) \(\because\) Period of \(\sin x=\frac{2 \pi}{1}\) Period of \(\cos \mathrm{ax}=\frac{2 \pi}{\mathrm{a}}\) Hence period of \(f(x)=\) L.C.M of \(\left\{\frac{2 \pi}{1}, \frac{2 \pi}{a}\right\}\) \(=\frac{\text { L.C.M. of }(2 \pi, 2 \pi)}{\text { H.C.F. of }\{1, a\}}=\frac{2 \pi}{k}\) Where \(\mathrm{k}=\mathrm{H}\).C.F of 1 and \(\mathrm{a}\) \(\therefore \frac{1}{\mathrm{k}}=\text { integer }=\mathrm{q}(\text { say }) \neq 0 \text { and } \frac{\mathrm{a}}{\mathrm{k}}\) \(\text { Integer } \mathrm{p} \text { cosy }\) \(\therefore \quad \frac{\frac{\mathrm{a}}{\mathrm{k}}}{\frac{1}{\mathrm{k}}}=\frac{\mathrm{p}}{\mathrm{q}}\) \(\mathrm{a}=\frac{\mathrm{p}}{\mathrm{q}}\) \(\mathrm{a}=\) rational number
116926
Consider the function \(f(x)=\cos ^2\). Then,
1 \(f\) is of period \(2 \pi\)
2 \(f\) is of period \(\sqrt{2 \pi}\)
3 \(\mathrm{f}\) is not periodic
4 \(\mathrm{f}\) is of period \(\pi\)
Explanation:
C We have, \(f(x)=\cos x^2\) Let \(T\) be the period of \(f(x)\). Then \(\mathrm{f}(\mathrm{x}+\mathrm{T})=\mathrm{f}(\mathrm{x})\) \(\Rightarrow \quad \cos (\mathrm{x}+\mathrm{T})^2=\cos \mathrm{x}^2\) But there is no value of \(\mathrm{T}\) for which \(\cos (\mathrm{x}+\mathrm{T})^2=\cos \mathrm{x}^2\)\(\therefore \mathrm{f}(\mathrm{x})\) is not periodic
WB JEE-2019
Sets, Relation and Function
116928
If \(\log _2 6+\frac{1}{2 x}=\log _2\left(2^{\frac{1}{x}}+8\right)\) then the values of \(x\) are
116929
Let \(f(x)=\sin x+\cos\) ax be periodic function. Then,
1 a is any real number
2 a is any irrational number
3 a is rational number
4 \(a=0\)
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\cos a \mathrm{x}\) \(\because\) Period of \(\sin x=\frac{2 \pi}{1}\) Period of \(\cos \mathrm{ax}=\frac{2 \pi}{\mathrm{a}}\) Hence period of \(f(x)=\) L.C.M of \(\left\{\frac{2 \pi}{1}, \frac{2 \pi}{a}\right\}\) \(=\frac{\text { L.C.M. of }(2 \pi, 2 \pi)}{\text { H.C.F. of }\{1, a\}}=\frac{2 \pi}{k}\) Where \(\mathrm{k}=\mathrm{H}\).C.F of 1 and \(\mathrm{a}\) \(\therefore \frac{1}{\mathrm{k}}=\text { integer }=\mathrm{q}(\text { say }) \neq 0 \text { and } \frac{\mathrm{a}}{\mathrm{k}}\) \(\text { Integer } \mathrm{p} \text { cosy }\) \(\therefore \quad \frac{\frac{\mathrm{a}}{\mathrm{k}}}{\frac{1}{\mathrm{k}}}=\frac{\mathrm{p}}{\mathrm{q}}\) \(\mathrm{a}=\frac{\mathrm{p}}{\mathrm{q}}\) \(\mathrm{a}=\) rational number