116970
If \(f(x)=e^x g(x), g(0)=2, g^{\prime}(0)=1\), then \(f(0)\) is equal to
1 1
2 3
3 2
4 0
Explanation:
B We have, \(f(x)=e^x g(x)\) Now differentiating, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{e}^{\mathrm{x}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}\left(\mathrm{g}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})\right)\) \(\mathrm{f}^{\prime}(0)=\mathrm{e}^0\left(\mathrm{~g}(0)+\mathrm{g}^{\prime}(0)\right)\) \(\mathrm{g}(0)=2 \text { and } \mathrm{g}^{\prime}(0)=1\) \(\mathrm{f}^{\prime}(0)=\mathrm{e}^0(2+1)\) \(\mathrm{f}^{\prime}(0)=1(3)\) \(\mathrm{f}^{\prime}(0)=3\)
Jamia Millia Islamia-2010
Sets, Relation and Function
116971
If \(f(x)=\left|\log _e\right| x||\), then \(f_0(x)\) equals
1 \(\frac{1}{|\mathbf{x}|} \mathrm{x} \neq 0\)
2 \(\frac{1}{x}\) for \(|x|>1\) and \(\frac{-1}{x}\) for \(|x|\lt 1\)
3 \(\frac{-1}{\mathrm{x}}\) for \(|\mathrm{x}|>1\) and \(\frac{1}{\mathrm{x}}\) for \(|\mathrm{x}|\lt 1\)
4 \(\frac{1}{\mathrm{x}}\) for \(\mathrm{x}>0\) and \(-\frac{1}{\mathrm{x}}\) for \(\mathrm{x}\lt 0\)
Explanation:
B \(\mathrm{f}(\mathrm{x})=\left|\log _{\mathrm{e}}\right| \mathrm{x}||\) Let, \(\quad y=f(x)=\left|\log _e\right| x||\) Thus, for \(x>1\) \(f(x)=\log _e x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) For, \(\quad \mathrm{x}\lt -1\) \(f(x)=\log _e(-x)\) \(f(x)=\log _e(-x)\) \(f^{\prime}(x)=-\frac{1}{x}(-1)=\frac{1}{x}\) For, \(\quad x \in(0,1)\) or \(0\lt x\lt 1\) \(\mathrm{f}(\mathrm{x})=-\log _{\mathrm{e}} \mathrm{x}\) \(\left\{\because-\log _{\mathrm{e}} \mathrm{x} x \in(0,1)\right\}\) \(f^{\prime}(x)=-\frac{1}{x}\) For, \(\quad x \in(-1,0)\) \(\mathrm{f}(\mathrm{x})=-\log (-\mathrm{x})\) \(\mathrm{f}^{\prime}(\mathrm{x})=-\left(-\frac{1}{\mathrm{x}}(-1)\right)\) \(\mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{\mathrm{x}}\) \(\left\{\because-\log _{\mathrm{e}} \mathrm{x} x \in(0,1)\right\}\) For, Then, \(f(x)=\left|\log _e\right| x||\) \(\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})= \begin{cases}\frac{1}{\mathrm{x}} & \mathrm{x}>1 \\ \frac{1}{\mathrm{x}} & \mathrm{x}\lt -1 \\ -\frac{1}{\mathrm{x}} & 0\lt \mathrm{x}\lt 1 \\ -\frac{1}{\mathrm{x}} & -1\lt \mathrm{x}\lt 0\end{cases} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{aligned} \frac{1}{\mathrm{x}} & |\mathrm{x}|>1 \\ -\frac{1}{\mathrm{x}} & |\mathrm{x}|\lt 1\end{aligned}\right.\end{aligned}\)
Jamia Millia Islamia-2009
Sets, Relation and Function
116972
If \(f(x+2 y, x-2 y)=x y\), then \(f(x, y)\) equals
1 \(\frac{x^2-y^2}{8}\)
2 \(\frac{x^2-y^2}{4}\)
3 \(\frac{x^2+y^2}{4}\)
4 \(\frac{x^2-y^2}{2}\)
Explanation:
A We have, \(f(x+2 y, x-2 y)=x y\) Let, \(\quad x+2 y=U\) And, \(x-2 y=V\) Now adding, we get- \(2 \mathrm{x}=\mathrm{U}+\mathrm{V}\) \(\mathrm{x}=\frac{\mathrm{U}+\mathrm{V}}{2}\) And subtracting, we get- \(4 \mathrm{y}=\mathrm{U}-\mathrm{V}\) \(\mathrm{y}=\frac{\mathrm{U}+\mathrm{V}}{4}\) Then \(f(x, y)=\left(\frac{U+V}{2}\right)\left(\frac{U-V}{4}\right)=\frac{U^2-V^2}{8}\)Hence, \(f(x, y)=\frac{x^2-y^2}{8}\)
Jamia Millia Islamia-2009
Sets, Relation and Function
116973
If \(f(x)\) is an odd periodic function with period 2 , then \(f(4)\) equals
1 0
2 2
3 4
4 -4
Explanation:
A Given, \(f(x)=\) odd periodic function Period \(=2\) \(\therefore \quad \mathrm{f}(-\mathrm{x}) =-\mathrm{f}(\mathrm{x})\) \(\mathrm{f}(\mathrm{x}+2) =\mathrm{f}(\mathrm{x})\) \(\mathrm{f}(2) =\mathrm{f}(0)\) \(\mathrm{f}(-2) =\mathrm{f}(-2+2)\) \(\mathrm{f}(-2) =\mathrm{f}(0)\) Now, \(\mathrm{f}(0) =\mathrm{f}(-2)=-\mathrm{f}(2)=-\mathrm{f}(0)\) \(2 \mathrm{f}(0) =0\) \(\mathrm{f}(0) =0\) \(\mathrm{f}(4)=\mathrm{f}(2) =\mathrm{f}(0)=0\)
116970
If \(f(x)=e^x g(x), g(0)=2, g^{\prime}(0)=1\), then \(f(0)\) is equal to
1 1
2 3
3 2
4 0
Explanation:
B We have, \(f(x)=e^x g(x)\) Now differentiating, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{e}^{\mathrm{x}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}\left(\mathrm{g}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})\right)\) \(\mathrm{f}^{\prime}(0)=\mathrm{e}^0\left(\mathrm{~g}(0)+\mathrm{g}^{\prime}(0)\right)\) \(\mathrm{g}(0)=2 \text { and } \mathrm{g}^{\prime}(0)=1\) \(\mathrm{f}^{\prime}(0)=\mathrm{e}^0(2+1)\) \(\mathrm{f}^{\prime}(0)=1(3)\) \(\mathrm{f}^{\prime}(0)=3\)
Jamia Millia Islamia-2010
Sets, Relation and Function
116971
If \(f(x)=\left|\log _e\right| x||\), then \(f_0(x)\) equals
1 \(\frac{1}{|\mathbf{x}|} \mathrm{x} \neq 0\)
2 \(\frac{1}{x}\) for \(|x|>1\) and \(\frac{-1}{x}\) for \(|x|\lt 1\)
3 \(\frac{-1}{\mathrm{x}}\) for \(|\mathrm{x}|>1\) and \(\frac{1}{\mathrm{x}}\) for \(|\mathrm{x}|\lt 1\)
4 \(\frac{1}{\mathrm{x}}\) for \(\mathrm{x}>0\) and \(-\frac{1}{\mathrm{x}}\) for \(\mathrm{x}\lt 0\)
Explanation:
B \(\mathrm{f}(\mathrm{x})=\left|\log _{\mathrm{e}}\right| \mathrm{x}||\) Let, \(\quad y=f(x)=\left|\log _e\right| x||\) Thus, for \(x>1\) \(f(x)=\log _e x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) For, \(\quad \mathrm{x}\lt -1\) \(f(x)=\log _e(-x)\) \(f(x)=\log _e(-x)\) \(f^{\prime}(x)=-\frac{1}{x}(-1)=\frac{1}{x}\) For, \(\quad x \in(0,1)\) or \(0\lt x\lt 1\) \(\mathrm{f}(\mathrm{x})=-\log _{\mathrm{e}} \mathrm{x}\) \(\left\{\because-\log _{\mathrm{e}} \mathrm{x} x \in(0,1)\right\}\) \(f^{\prime}(x)=-\frac{1}{x}\) For, \(\quad x \in(-1,0)\) \(\mathrm{f}(\mathrm{x})=-\log (-\mathrm{x})\) \(\mathrm{f}^{\prime}(\mathrm{x})=-\left(-\frac{1}{\mathrm{x}}(-1)\right)\) \(\mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{\mathrm{x}}\) \(\left\{\because-\log _{\mathrm{e}} \mathrm{x} x \in(0,1)\right\}\) For, Then, \(f(x)=\left|\log _e\right| x||\) \(\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})= \begin{cases}\frac{1}{\mathrm{x}} & \mathrm{x}>1 \\ \frac{1}{\mathrm{x}} & \mathrm{x}\lt -1 \\ -\frac{1}{\mathrm{x}} & 0\lt \mathrm{x}\lt 1 \\ -\frac{1}{\mathrm{x}} & -1\lt \mathrm{x}\lt 0\end{cases} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{aligned} \frac{1}{\mathrm{x}} & |\mathrm{x}|>1 \\ -\frac{1}{\mathrm{x}} & |\mathrm{x}|\lt 1\end{aligned}\right.\end{aligned}\)
Jamia Millia Islamia-2009
Sets, Relation and Function
116972
If \(f(x+2 y, x-2 y)=x y\), then \(f(x, y)\) equals
1 \(\frac{x^2-y^2}{8}\)
2 \(\frac{x^2-y^2}{4}\)
3 \(\frac{x^2+y^2}{4}\)
4 \(\frac{x^2-y^2}{2}\)
Explanation:
A We have, \(f(x+2 y, x-2 y)=x y\) Let, \(\quad x+2 y=U\) And, \(x-2 y=V\) Now adding, we get- \(2 \mathrm{x}=\mathrm{U}+\mathrm{V}\) \(\mathrm{x}=\frac{\mathrm{U}+\mathrm{V}}{2}\) And subtracting, we get- \(4 \mathrm{y}=\mathrm{U}-\mathrm{V}\) \(\mathrm{y}=\frac{\mathrm{U}+\mathrm{V}}{4}\) Then \(f(x, y)=\left(\frac{U+V}{2}\right)\left(\frac{U-V}{4}\right)=\frac{U^2-V^2}{8}\)Hence, \(f(x, y)=\frac{x^2-y^2}{8}\)
Jamia Millia Islamia-2009
Sets, Relation and Function
116973
If \(f(x)\) is an odd periodic function with period 2 , then \(f(4)\) equals
1 0
2 2
3 4
4 -4
Explanation:
A Given, \(f(x)=\) odd periodic function Period \(=2\) \(\therefore \quad \mathrm{f}(-\mathrm{x}) =-\mathrm{f}(\mathrm{x})\) \(\mathrm{f}(\mathrm{x}+2) =\mathrm{f}(\mathrm{x})\) \(\mathrm{f}(2) =\mathrm{f}(0)\) \(\mathrm{f}(-2) =\mathrm{f}(-2+2)\) \(\mathrm{f}(-2) =\mathrm{f}(0)\) Now, \(\mathrm{f}(0) =\mathrm{f}(-2)=-\mathrm{f}(2)=-\mathrm{f}(0)\) \(2 \mathrm{f}(0) =0\) \(\mathrm{f}(0) =0\) \(\mathrm{f}(4)=\mathrm{f}(2) =\mathrm{f}(0)=0\)
116970
If \(f(x)=e^x g(x), g(0)=2, g^{\prime}(0)=1\), then \(f(0)\) is equal to
1 1
2 3
3 2
4 0
Explanation:
B We have, \(f(x)=e^x g(x)\) Now differentiating, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{e}^{\mathrm{x}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}\left(\mathrm{g}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})\right)\) \(\mathrm{f}^{\prime}(0)=\mathrm{e}^0\left(\mathrm{~g}(0)+\mathrm{g}^{\prime}(0)\right)\) \(\mathrm{g}(0)=2 \text { and } \mathrm{g}^{\prime}(0)=1\) \(\mathrm{f}^{\prime}(0)=\mathrm{e}^0(2+1)\) \(\mathrm{f}^{\prime}(0)=1(3)\) \(\mathrm{f}^{\prime}(0)=3\)
Jamia Millia Islamia-2010
Sets, Relation and Function
116971
If \(f(x)=\left|\log _e\right| x||\), then \(f_0(x)\) equals
1 \(\frac{1}{|\mathbf{x}|} \mathrm{x} \neq 0\)
2 \(\frac{1}{x}\) for \(|x|>1\) and \(\frac{-1}{x}\) for \(|x|\lt 1\)
3 \(\frac{-1}{\mathrm{x}}\) for \(|\mathrm{x}|>1\) and \(\frac{1}{\mathrm{x}}\) for \(|\mathrm{x}|\lt 1\)
4 \(\frac{1}{\mathrm{x}}\) for \(\mathrm{x}>0\) and \(-\frac{1}{\mathrm{x}}\) for \(\mathrm{x}\lt 0\)
Explanation:
B \(\mathrm{f}(\mathrm{x})=\left|\log _{\mathrm{e}}\right| \mathrm{x}||\) Let, \(\quad y=f(x)=\left|\log _e\right| x||\) Thus, for \(x>1\) \(f(x)=\log _e x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) For, \(\quad \mathrm{x}\lt -1\) \(f(x)=\log _e(-x)\) \(f(x)=\log _e(-x)\) \(f^{\prime}(x)=-\frac{1}{x}(-1)=\frac{1}{x}\) For, \(\quad x \in(0,1)\) or \(0\lt x\lt 1\) \(\mathrm{f}(\mathrm{x})=-\log _{\mathrm{e}} \mathrm{x}\) \(\left\{\because-\log _{\mathrm{e}} \mathrm{x} x \in(0,1)\right\}\) \(f^{\prime}(x)=-\frac{1}{x}\) For, \(\quad x \in(-1,0)\) \(\mathrm{f}(\mathrm{x})=-\log (-\mathrm{x})\) \(\mathrm{f}^{\prime}(\mathrm{x})=-\left(-\frac{1}{\mathrm{x}}(-1)\right)\) \(\mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{\mathrm{x}}\) \(\left\{\because-\log _{\mathrm{e}} \mathrm{x} x \in(0,1)\right\}\) For, Then, \(f(x)=\left|\log _e\right| x||\) \(\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})= \begin{cases}\frac{1}{\mathrm{x}} & \mathrm{x}>1 \\ \frac{1}{\mathrm{x}} & \mathrm{x}\lt -1 \\ -\frac{1}{\mathrm{x}} & 0\lt \mathrm{x}\lt 1 \\ -\frac{1}{\mathrm{x}} & -1\lt \mathrm{x}\lt 0\end{cases} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{aligned} \frac{1}{\mathrm{x}} & |\mathrm{x}|>1 \\ -\frac{1}{\mathrm{x}} & |\mathrm{x}|\lt 1\end{aligned}\right.\end{aligned}\)
Jamia Millia Islamia-2009
Sets, Relation and Function
116972
If \(f(x+2 y, x-2 y)=x y\), then \(f(x, y)\) equals
1 \(\frac{x^2-y^2}{8}\)
2 \(\frac{x^2-y^2}{4}\)
3 \(\frac{x^2+y^2}{4}\)
4 \(\frac{x^2-y^2}{2}\)
Explanation:
A We have, \(f(x+2 y, x-2 y)=x y\) Let, \(\quad x+2 y=U\) And, \(x-2 y=V\) Now adding, we get- \(2 \mathrm{x}=\mathrm{U}+\mathrm{V}\) \(\mathrm{x}=\frac{\mathrm{U}+\mathrm{V}}{2}\) And subtracting, we get- \(4 \mathrm{y}=\mathrm{U}-\mathrm{V}\) \(\mathrm{y}=\frac{\mathrm{U}+\mathrm{V}}{4}\) Then \(f(x, y)=\left(\frac{U+V}{2}\right)\left(\frac{U-V}{4}\right)=\frac{U^2-V^2}{8}\)Hence, \(f(x, y)=\frac{x^2-y^2}{8}\)
Jamia Millia Islamia-2009
Sets, Relation and Function
116973
If \(f(x)\) is an odd periodic function with period 2 , then \(f(4)\) equals
1 0
2 2
3 4
4 -4
Explanation:
A Given, \(f(x)=\) odd periodic function Period \(=2\) \(\therefore \quad \mathrm{f}(-\mathrm{x}) =-\mathrm{f}(\mathrm{x})\) \(\mathrm{f}(\mathrm{x}+2) =\mathrm{f}(\mathrm{x})\) \(\mathrm{f}(2) =\mathrm{f}(0)\) \(\mathrm{f}(-2) =\mathrm{f}(-2+2)\) \(\mathrm{f}(-2) =\mathrm{f}(0)\) Now, \(\mathrm{f}(0) =\mathrm{f}(-2)=-\mathrm{f}(2)=-\mathrm{f}(0)\) \(2 \mathrm{f}(0) =0\) \(\mathrm{f}(0) =0\) \(\mathrm{f}(4)=\mathrm{f}(2) =\mathrm{f}(0)=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
116970
If \(f(x)=e^x g(x), g(0)=2, g^{\prime}(0)=1\), then \(f(0)\) is equal to
1 1
2 3
3 2
4 0
Explanation:
B We have, \(f(x)=e^x g(x)\) Now differentiating, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{e}^{\mathrm{x}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}\left(\mathrm{g}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})\right)\) \(\mathrm{f}^{\prime}(0)=\mathrm{e}^0\left(\mathrm{~g}(0)+\mathrm{g}^{\prime}(0)\right)\) \(\mathrm{g}(0)=2 \text { and } \mathrm{g}^{\prime}(0)=1\) \(\mathrm{f}^{\prime}(0)=\mathrm{e}^0(2+1)\) \(\mathrm{f}^{\prime}(0)=1(3)\) \(\mathrm{f}^{\prime}(0)=3\)
Jamia Millia Islamia-2010
Sets, Relation and Function
116971
If \(f(x)=\left|\log _e\right| x||\), then \(f_0(x)\) equals
1 \(\frac{1}{|\mathbf{x}|} \mathrm{x} \neq 0\)
2 \(\frac{1}{x}\) for \(|x|>1\) and \(\frac{-1}{x}\) for \(|x|\lt 1\)
3 \(\frac{-1}{\mathrm{x}}\) for \(|\mathrm{x}|>1\) and \(\frac{1}{\mathrm{x}}\) for \(|\mathrm{x}|\lt 1\)
4 \(\frac{1}{\mathrm{x}}\) for \(\mathrm{x}>0\) and \(-\frac{1}{\mathrm{x}}\) for \(\mathrm{x}\lt 0\)
Explanation:
B \(\mathrm{f}(\mathrm{x})=\left|\log _{\mathrm{e}}\right| \mathrm{x}||\) Let, \(\quad y=f(x)=\left|\log _e\right| x||\) Thus, for \(x>1\) \(f(x)=\log _e x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) For, \(\quad \mathrm{x}\lt -1\) \(f(x)=\log _e(-x)\) \(f(x)=\log _e(-x)\) \(f^{\prime}(x)=-\frac{1}{x}(-1)=\frac{1}{x}\) For, \(\quad x \in(0,1)\) or \(0\lt x\lt 1\) \(\mathrm{f}(\mathrm{x})=-\log _{\mathrm{e}} \mathrm{x}\) \(\left\{\because-\log _{\mathrm{e}} \mathrm{x} x \in(0,1)\right\}\) \(f^{\prime}(x)=-\frac{1}{x}\) For, \(\quad x \in(-1,0)\) \(\mathrm{f}(\mathrm{x})=-\log (-\mathrm{x})\) \(\mathrm{f}^{\prime}(\mathrm{x})=-\left(-\frac{1}{\mathrm{x}}(-1)\right)\) \(\mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{\mathrm{x}}\) \(\left\{\because-\log _{\mathrm{e}} \mathrm{x} x \in(0,1)\right\}\) For, Then, \(f(x)=\left|\log _e\right| x||\) \(\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})= \begin{cases}\frac{1}{\mathrm{x}} & \mathrm{x}>1 \\ \frac{1}{\mathrm{x}} & \mathrm{x}\lt -1 \\ -\frac{1}{\mathrm{x}} & 0\lt \mathrm{x}\lt 1 \\ -\frac{1}{\mathrm{x}} & -1\lt \mathrm{x}\lt 0\end{cases} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{aligned} \frac{1}{\mathrm{x}} & |\mathrm{x}|>1 \\ -\frac{1}{\mathrm{x}} & |\mathrm{x}|\lt 1\end{aligned}\right.\end{aligned}\)
Jamia Millia Islamia-2009
Sets, Relation and Function
116972
If \(f(x+2 y, x-2 y)=x y\), then \(f(x, y)\) equals
1 \(\frac{x^2-y^2}{8}\)
2 \(\frac{x^2-y^2}{4}\)
3 \(\frac{x^2+y^2}{4}\)
4 \(\frac{x^2-y^2}{2}\)
Explanation:
A We have, \(f(x+2 y, x-2 y)=x y\) Let, \(\quad x+2 y=U\) And, \(x-2 y=V\) Now adding, we get- \(2 \mathrm{x}=\mathrm{U}+\mathrm{V}\) \(\mathrm{x}=\frac{\mathrm{U}+\mathrm{V}}{2}\) And subtracting, we get- \(4 \mathrm{y}=\mathrm{U}-\mathrm{V}\) \(\mathrm{y}=\frac{\mathrm{U}+\mathrm{V}}{4}\) Then \(f(x, y)=\left(\frac{U+V}{2}\right)\left(\frac{U-V}{4}\right)=\frac{U^2-V^2}{8}\)Hence, \(f(x, y)=\frac{x^2-y^2}{8}\)
Jamia Millia Islamia-2009
Sets, Relation and Function
116973
If \(f(x)\) is an odd periodic function with period 2 , then \(f(4)\) equals
1 0
2 2
3 4
4 -4
Explanation:
A Given, \(f(x)=\) odd periodic function Period \(=2\) \(\therefore \quad \mathrm{f}(-\mathrm{x}) =-\mathrm{f}(\mathrm{x})\) \(\mathrm{f}(\mathrm{x}+2) =\mathrm{f}(\mathrm{x})\) \(\mathrm{f}(2) =\mathrm{f}(0)\) \(\mathrm{f}(-2) =\mathrm{f}(-2+2)\) \(\mathrm{f}(-2) =\mathrm{f}(0)\) Now, \(\mathrm{f}(0) =\mathrm{f}(-2)=-\mathrm{f}(2)=-\mathrm{f}(0)\) \(2 \mathrm{f}(0) =0\) \(\mathrm{f}(0) =0\) \(\mathrm{f}(4)=\mathrm{f}(2) =\mathrm{f}(0)=0\)
