116936
If \(f\) is any function, then \(\frac{1}{2}[f(x)+f(-x)]\) is always :
1 odd
2 even
3 neither even nor odd
4 one-one
Explanation:
B Given, \(f\) is any function. And let \(g(x)=\frac{1}{2}[f(x)+f(-x)]\) We know that - For even, \(f(-x)=f(x)\) And for odd, \(f(-x)=-f(x)\) Then, \(g(-x)=\frac{1}{2}[f(-x)+f(x)]\) We see that, \(g(-x)=g(x)\) Hence, the given function is always even.
BCECE-2005
Sets, Relation and Function
116860
The graph of the function \(y=f(x)\) is symmetrical about the line \(\boldsymbol{x}=2\). Then,
1 \(f(x+2)=f(x-2)\)
2 \(f(2+x)=f(2-x)\)
3 \(f(x)=f(-x)\)
4 \(f(x)=-f(-x)\)
Explanation:
B Given, The graph of the function \(y=f(x)\) Is symmetrical about the line \(\mathrm{x}=2\). We know, A function \(\mathrm{g}(\mathrm{x})\) is symmetrical about \(\mathrm{y}\)-axis means \(\mathrm{x}=0\), we can write as - \(\mathrm{g}(\mathrm{x})=\mathrm{g}(-\mathrm{x})\) It is also written as - \(g(0+x)=g(0-x)\) So, function \(y=f(x)\) which is symmetrical about the line \(\mathrm{x}=2\). Then can be written as - \(\mathrm{f}(2+\mathrm{x})=\mathrm{f}(2-\mathrm{x})\)
UPSEE-2010
Sets, Relation and Function
116861
The number of solutions of \(\log _4(x-1)=\log _2(x-\) 3) is
1 3
2 1
3 2
4 0
Explanation:
B Given, \(\log _4(x-1)=\log _2(x-3) \) \({ }_4 \log _4(x-1)={ }_4 \log _2(x-3) \) \((x-1)=(2)^{2 \log _2(x-3)} {\left[\because{ }_a \log _a(x)=x\right]}\) \((x-1)=2^{\log _2(x-3)^2} \) \((x-1)=(x-3)^2 {\left[\because x \log a=\log a^x\right]}\) \(x-1=x^2+9-6 x \) \(x^2-7 x+10=0 \) \(x^2-2 x-5 x+10=0 \) \(x(x-2)-5(x-2)=0 \) \((x-2)(x-5)=0 \) When, \(\mathrm{x}=2\) \(\log _2(x-3)=\log _2(2-3)=\log _2(-1)\) \(\log _2(-1)\) is not possible since, log does not have - ve value. So, The number of solutions of \(\log _4(x-1)=\log _2(x-3)\) is 1.
UPSEE -2008
Sets, Relation and Function
116862
If \(a\) and \(b\) are positive integers such that \(\left(a^2-b^2\right)\) is a prime number, then
A Given, \(a\) and \(b\) are positive integer such that \(\left(a^2-b^2\right)\) is a prime number. Let \(\mathrm{a}=3, \mathrm{~b}=2\) are positive integer. Then, \(a^2-b^2=3^2-2^2=9-4=5\) \(a^2-b^2=5 \text { and } a+b=5\) Where, \(\mathrm{a}^2-\mathrm{b}^2=5\) is a prime number. So, \(\quad a^2-b^2=a+b\)
116936
If \(f\) is any function, then \(\frac{1}{2}[f(x)+f(-x)]\) is always :
1 odd
2 even
3 neither even nor odd
4 one-one
Explanation:
B Given, \(f\) is any function. And let \(g(x)=\frac{1}{2}[f(x)+f(-x)]\) We know that - For even, \(f(-x)=f(x)\) And for odd, \(f(-x)=-f(x)\) Then, \(g(-x)=\frac{1}{2}[f(-x)+f(x)]\) We see that, \(g(-x)=g(x)\) Hence, the given function is always even.
BCECE-2005
Sets, Relation and Function
116860
The graph of the function \(y=f(x)\) is symmetrical about the line \(\boldsymbol{x}=2\). Then,
1 \(f(x+2)=f(x-2)\)
2 \(f(2+x)=f(2-x)\)
3 \(f(x)=f(-x)\)
4 \(f(x)=-f(-x)\)
Explanation:
B Given, The graph of the function \(y=f(x)\) Is symmetrical about the line \(\mathrm{x}=2\). We know, A function \(\mathrm{g}(\mathrm{x})\) is symmetrical about \(\mathrm{y}\)-axis means \(\mathrm{x}=0\), we can write as - \(\mathrm{g}(\mathrm{x})=\mathrm{g}(-\mathrm{x})\) It is also written as - \(g(0+x)=g(0-x)\) So, function \(y=f(x)\) which is symmetrical about the line \(\mathrm{x}=2\). Then can be written as - \(\mathrm{f}(2+\mathrm{x})=\mathrm{f}(2-\mathrm{x})\)
UPSEE-2010
Sets, Relation and Function
116861
The number of solutions of \(\log _4(x-1)=\log _2(x-\) 3) is
1 3
2 1
3 2
4 0
Explanation:
B Given, \(\log _4(x-1)=\log _2(x-3) \) \({ }_4 \log _4(x-1)={ }_4 \log _2(x-3) \) \((x-1)=(2)^{2 \log _2(x-3)} {\left[\because{ }_a \log _a(x)=x\right]}\) \((x-1)=2^{\log _2(x-3)^2} \) \((x-1)=(x-3)^2 {\left[\because x \log a=\log a^x\right]}\) \(x-1=x^2+9-6 x \) \(x^2-7 x+10=0 \) \(x^2-2 x-5 x+10=0 \) \(x(x-2)-5(x-2)=0 \) \((x-2)(x-5)=0 \) When, \(\mathrm{x}=2\) \(\log _2(x-3)=\log _2(2-3)=\log _2(-1)\) \(\log _2(-1)\) is not possible since, log does not have - ve value. So, The number of solutions of \(\log _4(x-1)=\log _2(x-3)\) is 1.
UPSEE -2008
Sets, Relation and Function
116862
If \(a\) and \(b\) are positive integers such that \(\left(a^2-b^2\right)\) is a prime number, then
A Given, \(a\) and \(b\) are positive integer such that \(\left(a^2-b^2\right)\) is a prime number. Let \(\mathrm{a}=3, \mathrm{~b}=2\) are positive integer. Then, \(a^2-b^2=3^2-2^2=9-4=5\) \(a^2-b^2=5 \text { and } a+b=5\) Where, \(\mathrm{a}^2-\mathrm{b}^2=5\) is a prime number. So, \(\quad a^2-b^2=a+b\)
116936
If \(f\) is any function, then \(\frac{1}{2}[f(x)+f(-x)]\) is always :
1 odd
2 even
3 neither even nor odd
4 one-one
Explanation:
B Given, \(f\) is any function. And let \(g(x)=\frac{1}{2}[f(x)+f(-x)]\) We know that - For even, \(f(-x)=f(x)\) And for odd, \(f(-x)=-f(x)\) Then, \(g(-x)=\frac{1}{2}[f(-x)+f(x)]\) We see that, \(g(-x)=g(x)\) Hence, the given function is always even.
BCECE-2005
Sets, Relation and Function
116860
The graph of the function \(y=f(x)\) is symmetrical about the line \(\boldsymbol{x}=2\). Then,
1 \(f(x+2)=f(x-2)\)
2 \(f(2+x)=f(2-x)\)
3 \(f(x)=f(-x)\)
4 \(f(x)=-f(-x)\)
Explanation:
B Given, The graph of the function \(y=f(x)\) Is symmetrical about the line \(\mathrm{x}=2\). We know, A function \(\mathrm{g}(\mathrm{x})\) is symmetrical about \(\mathrm{y}\)-axis means \(\mathrm{x}=0\), we can write as - \(\mathrm{g}(\mathrm{x})=\mathrm{g}(-\mathrm{x})\) It is also written as - \(g(0+x)=g(0-x)\) So, function \(y=f(x)\) which is symmetrical about the line \(\mathrm{x}=2\). Then can be written as - \(\mathrm{f}(2+\mathrm{x})=\mathrm{f}(2-\mathrm{x})\)
UPSEE-2010
Sets, Relation and Function
116861
The number of solutions of \(\log _4(x-1)=\log _2(x-\) 3) is
1 3
2 1
3 2
4 0
Explanation:
B Given, \(\log _4(x-1)=\log _2(x-3) \) \({ }_4 \log _4(x-1)={ }_4 \log _2(x-3) \) \((x-1)=(2)^{2 \log _2(x-3)} {\left[\because{ }_a \log _a(x)=x\right]}\) \((x-1)=2^{\log _2(x-3)^2} \) \((x-1)=(x-3)^2 {\left[\because x \log a=\log a^x\right]}\) \(x-1=x^2+9-6 x \) \(x^2-7 x+10=0 \) \(x^2-2 x-5 x+10=0 \) \(x(x-2)-5(x-2)=0 \) \((x-2)(x-5)=0 \) When, \(\mathrm{x}=2\) \(\log _2(x-3)=\log _2(2-3)=\log _2(-1)\) \(\log _2(-1)\) is not possible since, log does not have - ve value. So, The number of solutions of \(\log _4(x-1)=\log _2(x-3)\) is 1.
UPSEE -2008
Sets, Relation and Function
116862
If \(a\) and \(b\) are positive integers such that \(\left(a^2-b^2\right)\) is a prime number, then
A Given, \(a\) and \(b\) are positive integer such that \(\left(a^2-b^2\right)\) is a prime number. Let \(\mathrm{a}=3, \mathrm{~b}=2\) are positive integer. Then, \(a^2-b^2=3^2-2^2=9-4=5\) \(a^2-b^2=5 \text { and } a+b=5\) Where, \(\mathrm{a}^2-\mathrm{b}^2=5\) is a prime number. So, \(\quad a^2-b^2=a+b\)
116936
If \(f\) is any function, then \(\frac{1}{2}[f(x)+f(-x)]\) is always :
1 odd
2 even
3 neither even nor odd
4 one-one
Explanation:
B Given, \(f\) is any function. And let \(g(x)=\frac{1}{2}[f(x)+f(-x)]\) We know that - For even, \(f(-x)=f(x)\) And for odd, \(f(-x)=-f(x)\) Then, \(g(-x)=\frac{1}{2}[f(-x)+f(x)]\) We see that, \(g(-x)=g(x)\) Hence, the given function is always even.
BCECE-2005
Sets, Relation and Function
116860
The graph of the function \(y=f(x)\) is symmetrical about the line \(\boldsymbol{x}=2\). Then,
1 \(f(x+2)=f(x-2)\)
2 \(f(2+x)=f(2-x)\)
3 \(f(x)=f(-x)\)
4 \(f(x)=-f(-x)\)
Explanation:
B Given, The graph of the function \(y=f(x)\) Is symmetrical about the line \(\mathrm{x}=2\). We know, A function \(\mathrm{g}(\mathrm{x})\) is symmetrical about \(\mathrm{y}\)-axis means \(\mathrm{x}=0\), we can write as - \(\mathrm{g}(\mathrm{x})=\mathrm{g}(-\mathrm{x})\) It is also written as - \(g(0+x)=g(0-x)\) So, function \(y=f(x)\) which is symmetrical about the line \(\mathrm{x}=2\). Then can be written as - \(\mathrm{f}(2+\mathrm{x})=\mathrm{f}(2-\mathrm{x})\)
UPSEE-2010
Sets, Relation and Function
116861
The number of solutions of \(\log _4(x-1)=\log _2(x-\) 3) is
1 3
2 1
3 2
4 0
Explanation:
B Given, \(\log _4(x-1)=\log _2(x-3) \) \({ }_4 \log _4(x-1)={ }_4 \log _2(x-3) \) \((x-1)=(2)^{2 \log _2(x-3)} {\left[\because{ }_a \log _a(x)=x\right]}\) \((x-1)=2^{\log _2(x-3)^2} \) \((x-1)=(x-3)^2 {\left[\because x \log a=\log a^x\right]}\) \(x-1=x^2+9-6 x \) \(x^2-7 x+10=0 \) \(x^2-2 x-5 x+10=0 \) \(x(x-2)-5(x-2)=0 \) \((x-2)(x-5)=0 \) When, \(\mathrm{x}=2\) \(\log _2(x-3)=\log _2(2-3)=\log _2(-1)\) \(\log _2(-1)\) is not possible since, log does not have - ve value. So, The number of solutions of \(\log _4(x-1)=\log _2(x-3)\) is 1.
UPSEE -2008
Sets, Relation and Function
116862
If \(a\) and \(b\) are positive integers such that \(\left(a^2-b^2\right)\) is a prime number, then
A Given, \(a\) and \(b\) are positive integer such that \(\left(a^2-b^2\right)\) is a prime number. Let \(\mathrm{a}=3, \mathrm{~b}=2\) are positive integer. Then, \(a^2-b^2=3^2-2^2=9-4=5\) \(a^2-b^2=5 \text { and } a+b=5\) Where, \(\mathrm{a}^2-\mathrm{b}^2=5\) is a prime number. So, \(\quad a^2-b^2=a+b\)
