116984
If a function \(F\) is such that \(F(0)=2, F(1)=3\), \(F(n+2)=2 F(n)-F(n+1)\) for \(n \neq 0\), then \(F(5)\) is equal to
1 -7
2 -3
3 7
4 13
Explanation:
D Given \(\mathrm{F}(0)=2, \mathrm{~F}(1)=3\) \(F(n+2)=2 F(n)-F(n+1)\) Putting, \(\mathrm{n}=0\) \(F(0+2)=2 F(0)-F(0+1)\) \(F(2)=2 \times 2-3\) \(F(2)=1\) Put \(\mathrm{n}=1\) \(\mathrm{F}(3)=2 \mathrm{~F}(1)-\mathrm{F}(2)\) \(\mathrm{F}(3)=2 \times 3-1\) \(\mathrm{~F}(3)=5\) Put \(\mathrm{n}=2\) \(F(2+2)=2 F(2)-F(3)\) \(F(4)=2 \times 1-5\) \(F(4)=-3\) Put \(n=3\) \(F(5)=2 F(3)-F(4)\) \(=2 \times 5-(-3)\) \(=10+3\) \(F(5)=13\)
J and K CET-2003
Sets, Relation and Function
116986
If \(f(x)=a x^2+b x+c\) satisfies \(f(1)+2 f(2)=0\) and \(\mathbf{2 f}(\mathbf{1})+\mathbf{f}(\mathbf{2})=\mathbf{0}\), then \(3 \mathbf{a}+\mathbf{b}=\)
1 2
2 -1
3 0
4 1
Explanation:
C We have equation, \(f(x)=a x^2+b x+c\) \(f(1)+2 f(2)=0\) \(2 f(1)+f(2)=0\) Now \(f(1)=a+b+c\) \(f(2)=4 a+2 b+c\) \(f(2)-f(1)=3 a+b\) Now, \(f(1)+2 f(2)=0\) \(a+b+c+2(4 a+2 b+c)=0\) \(2(a+b+c)+4 a+2 b+c=0\) Adding these equation, \(3(a+b+c)+3(4 a+2 b+c)=0\) \(3[a+b+c+4 a+2 b+c]=0\) \(3[5 a+3 b+2 c]=0\) It can be written as \((a+b+c)-2(a+b+c)+2(4 a+2 b+c)\) \(-(4 a+2 b+c)\) \(\text { As } \quad-(a+b+c)+4 a+2 b+c=0\) \(3 a+b=0\) As \(\quad-(\mathrm{a}+\mathrm{b}+\mathrm{c})+4 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=0\) \(3 a+b=0\)
Shift-I
Sets, Relation and Function
116987
Let \(f\) be a function defined by \(f(x y)=\frac{f(x)}{y}\) for all positive real numbers \(x\) and \(y\) If \(f(30)=20\), then \(\mathbf{f}(\mathbf{4 0})=\)
1 10
2 15
3 25
4 17
Explanation:
B Given function \(f(\mathrm{xy})=\frac{f(\mathrm{x})}{\mathrm{y}}\) Put \(\mathrm{x}=1\) \(\mathrm{f}(\mathrm{y})=\frac{f(1)}{\mathrm{y}}\) Put \(\mathrm{y}=30\) \(f(30)=\frac{f(1)}{30}\) \(f(1) =f(30) \cdot 30\) \(=20 \times 30\) \(f(1) =600\) Put \(\mathrm{y}=40\) in equation \(f(40)=\frac{f(1)}{40}\) \(f(40)=\frac{600}{40}\) \(f(40)=15\)
Shift-I
Sets, Relation and Function
116988
If \(f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}\) for \(x>\) 2 , then \(f(11)\) is equal to
116984
If a function \(F\) is such that \(F(0)=2, F(1)=3\), \(F(n+2)=2 F(n)-F(n+1)\) for \(n \neq 0\), then \(F(5)\) is equal to
1 -7
2 -3
3 7
4 13
Explanation:
D Given \(\mathrm{F}(0)=2, \mathrm{~F}(1)=3\) \(F(n+2)=2 F(n)-F(n+1)\) Putting, \(\mathrm{n}=0\) \(F(0+2)=2 F(0)-F(0+1)\) \(F(2)=2 \times 2-3\) \(F(2)=1\) Put \(\mathrm{n}=1\) \(\mathrm{F}(3)=2 \mathrm{~F}(1)-\mathrm{F}(2)\) \(\mathrm{F}(3)=2 \times 3-1\) \(\mathrm{~F}(3)=5\) Put \(\mathrm{n}=2\) \(F(2+2)=2 F(2)-F(3)\) \(F(4)=2 \times 1-5\) \(F(4)=-3\) Put \(n=3\) \(F(5)=2 F(3)-F(4)\) \(=2 \times 5-(-3)\) \(=10+3\) \(F(5)=13\)
J and K CET-2003
Sets, Relation and Function
116986
If \(f(x)=a x^2+b x+c\) satisfies \(f(1)+2 f(2)=0\) and \(\mathbf{2 f}(\mathbf{1})+\mathbf{f}(\mathbf{2})=\mathbf{0}\), then \(3 \mathbf{a}+\mathbf{b}=\)
1 2
2 -1
3 0
4 1
Explanation:
C We have equation, \(f(x)=a x^2+b x+c\) \(f(1)+2 f(2)=0\) \(2 f(1)+f(2)=0\) Now \(f(1)=a+b+c\) \(f(2)=4 a+2 b+c\) \(f(2)-f(1)=3 a+b\) Now, \(f(1)+2 f(2)=0\) \(a+b+c+2(4 a+2 b+c)=0\) \(2(a+b+c)+4 a+2 b+c=0\) Adding these equation, \(3(a+b+c)+3(4 a+2 b+c)=0\) \(3[a+b+c+4 a+2 b+c]=0\) \(3[5 a+3 b+2 c]=0\) It can be written as \((a+b+c)-2(a+b+c)+2(4 a+2 b+c)\) \(-(4 a+2 b+c)\) \(\text { As } \quad-(a+b+c)+4 a+2 b+c=0\) \(3 a+b=0\) As \(\quad-(\mathrm{a}+\mathrm{b}+\mathrm{c})+4 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=0\) \(3 a+b=0\)
Shift-I
Sets, Relation and Function
116987
Let \(f\) be a function defined by \(f(x y)=\frac{f(x)}{y}\) for all positive real numbers \(x\) and \(y\) If \(f(30)=20\), then \(\mathbf{f}(\mathbf{4 0})=\)
1 10
2 15
3 25
4 17
Explanation:
B Given function \(f(\mathrm{xy})=\frac{f(\mathrm{x})}{\mathrm{y}}\) Put \(\mathrm{x}=1\) \(\mathrm{f}(\mathrm{y})=\frac{f(1)}{\mathrm{y}}\) Put \(\mathrm{y}=30\) \(f(30)=\frac{f(1)}{30}\) \(f(1) =f(30) \cdot 30\) \(=20 \times 30\) \(f(1) =600\) Put \(\mathrm{y}=40\) in equation \(f(40)=\frac{f(1)}{40}\) \(f(40)=\frac{600}{40}\) \(f(40)=15\)
Shift-I
Sets, Relation and Function
116988
If \(f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}\) for \(x>\) 2 , then \(f(11)\) is equal to
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
116984
If a function \(F\) is such that \(F(0)=2, F(1)=3\), \(F(n+2)=2 F(n)-F(n+1)\) for \(n \neq 0\), then \(F(5)\) is equal to
1 -7
2 -3
3 7
4 13
Explanation:
D Given \(\mathrm{F}(0)=2, \mathrm{~F}(1)=3\) \(F(n+2)=2 F(n)-F(n+1)\) Putting, \(\mathrm{n}=0\) \(F(0+2)=2 F(0)-F(0+1)\) \(F(2)=2 \times 2-3\) \(F(2)=1\) Put \(\mathrm{n}=1\) \(\mathrm{F}(3)=2 \mathrm{~F}(1)-\mathrm{F}(2)\) \(\mathrm{F}(3)=2 \times 3-1\) \(\mathrm{~F}(3)=5\) Put \(\mathrm{n}=2\) \(F(2+2)=2 F(2)-F(3)\) \(F(4)=2 \times 1-5\) \(F(4)=-3\) Put \(n=3\) \(F(5)=2 F(3)-F(4)\) \(=2 \times 5-(-3)\) \(=10+3\) \(F(5)=13\)
J and K CET-2003
Sets, Relation and Function
116986
If \(f(x)=a x^2+b x+c\) satisfies \(f(1)+2 f(2)=0\) and \(\mathbf{2 f}(\mathbf{1})+\mathbf{f}(\mathbf{2})=\mathbf{0}\), then \(3 \mathbf{a}+\mathbf{b}=\)
1 2
2 -1
3 0
4 1
Explanation:
C We have equation, \(f(x)=a x^2+b x+c\) \(f(1)+2 f(2)=0\) \(2 f(1)+f(2)=0\) Now \(f(1)=a+b+c\) \(f(2)=4 a+2 b+c\) \(f(2)-f(1)=3 a+b\) Now, \(f(1)+2 f(2)=0\) \(a+b+c+2(4 a+2 b+c)=0\) \(2(a+b+c)+4 a+2 b+c=0\) Adding these equation, \(3(a+b+c)+3(4 a+2 b+c)=0\) \(3[a+b+c+4 a+2 b+c]=0\) \(3[5 a+3 b+2 c]=0\) It can be written as \((a+b+c)-2(a+b+c)+2(4 a+2 b+c)\) \(-(4 a+2 b+c)\) \(\text { As } \quad-(a+b+c)+4 a+2 b+c=0\) \(3 a+b=0\) As \(\quad-(\mathrm{a}+\mathrm{b}+\mathrm{c})+4 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=0\) \(3 a+b=0\)
Shift-I
Sets, Relation and Function
116987
Let \(f\) be a function defined by \(f(x y)=\frac{f(x)}{y}\) for all positive real numbers \(x\) and \(y\) If \(f(30)=20\), then \(\mathbf{f}(\mathbf{4 0})=\)
1 10
2 15
3 25
4 17
Explanation:
B Given function \(f(\mathrm{xy})=\frac{f(\mathrm{x})}{\mathrm{y}}\) Put \(\mathrm{x}=1\) \(\mathrm{f}(\mathrm{y})=\frac{f(1)}{\mathrm{y}}\) Put \(\mathrm{y}=30\) \(f(30)=\frac{f(1)}{30}\) \(f(1) =f(30) \cdot 30\) \(=20 \times 30\) \(f(1) =600\) Put \(\mathrm{y}=40\) in equation \(f(40)=\frac{f(1)}{40}\) \(f(40)=\frac{600}{40}\) \(f(40)=15\)
Shift-I
Sets, Relation and Function
116988
If \(f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}\) for \(x>\) 2 , then \(f(11)\) is equal to
116984
If a function \(F\) is such that \(F(0)=2, F(1)=3\), \(F(n+2)=2 F(n)-F(n+1)\) for \(n \neq 0\), then \(F(5)\) is equal to
1 -7
2 -3
3 7
4 13
Explanation:
D Given \(\mathrm{F}(0)=2, \mathrm{~F}(1)=3\) \(F(n+2)=2 F(n)-F(n+1)\) Putting, \(\mathrm{n}=0\) \(F(0+2)=2 F(0)-F(0+1)\) \(F(2)=2 \times 2-3\) \(F(2)=1\) Put \(\mathrm{n}=1\) \(\mathrm{F}(3)=2 \mathrm{~F}(1)-\mathrm{F}(2)\) \(\mathrm{F}(3)=2 \times 3-1\) \(\mathrm{~F}(3)=5\) Put \(\mathrm{n}=2\) \(F(2+2)=2 F(2)-F(3)\) \(F(4)=2 \times 1-5\) \(F(4)=-3\) Put \(n=3\) \(F(5)=2 F(3)-F(4)\) \(=2 \times 5-(-3)\) \(=10+3\) \(F(5)=13\)
J and K CET-2003
Sets, Relation and Function
116986
If \(f(x)=a x^2+b x+c\) satisfies \(f(1)+2 f(2)=0\) and \(\mathbf{2 f}(\mathbf{1})+\mathbf{f}(\mathbf{2})=\mathbf{0}\), then \(3 \mathbf{a}+\mathbf{b}=\)
1 2
2 -1
3 0
4 1
Explanation:
C We have equation, \(f(x)=a x^2+b x+c\) \(f(1)+2 f(2)=0\) \(2 f(1)+f(2)=0\) Now \(f(1)=a+b+c\) \(f(2)=4 a+2 b+c\) \(f(2)-f(1)=3 a+b\) Now, \(f(1)+2 f(2)=0\) \(a+b+c+2(4 a+2 b+c)=0\) \(2(a+b+c)+4 a+2 b+c=0\) Adding these equation, \(3(a+b+c)+3(4 a+2 b+c)=0\) \(3[a+b+c+4 a+2 b+c]=0\) \(3[5 a+3 b+2 c]=0\) It can be written as \((a+b+c)-2(a+b+c)+2(4 a+2 b+c)\) \(-(4 a+2 b+c)\) \(\text { As } \quad-(a+b+c)+4 a+2 b+c=0\) \(3 a+b=0\) As \(\quad-(\mathrm{a}+\mathrm{b}+\mathrm{c})+4 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=0\) \(3 a+b=0\)
Shift-I
Sets, Relation and Function
116987
Let \(f\) be a function defined by \(f(x y)=\frac{f(x)}{y}\) for all positive real numbers \(x\) and \(y\) If \(f(30)=20\), then \(\mathbf{f}(\mathbf{4 0})=\)
1 10
2 15
3 25
4 17
Explanation:
B Given function \(f(\mathrm{xy})=\frac{f(\mathrm{x})}{\mathrm{y}}\) Put \(\mathrm{x}=1\) \(\mathrm{f}(\mathrm{y})=\frac{f(1)}{\mathrm{y}}\) Put \(\mathrm{y}=30\) \(f(30)=\frac{f(1)}{30}\) \(f(1) =f(30) \cdot 30\) \(=20 \times 30\) \(f(1) =600\) Put \(\mathrm{y}=40\) in equation \(f(40)=\frac{f(1)}{40}\) \(f(40)=\frac{600}{40}\) \(f(40)=15\)
Shift-I
Sets, Relation and Function
116988
If \(f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}\) for \(x>\) 2 , then \(f(11)\) is equal to