116885
Let \(f: N \rightarrow R\) be a function such that \(f(x+y)=\) \(2 f(x) f(y)\) for natural numbers \(x\) and \(y\). If \(f(1)\) \(=2\), then the value of \(\alpha\) for which \(\sum_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)\) holds, is
1 2
2 3
3 4
4 6
Explanation:
C Given, \(f: N \rightarrow R, f(x+y)=2 f(x) f(y)\) \(f(1)=2\) \(\sum_{k=1}^{10} f(\alpha+k)=2 f(\alpha) \sum_{k=1}^{10} f(k)\) \(=2 f(\alpha)\{f(1)+f(2)+\ldots .+f(10)\}\) Form equation (i), \(f(10)=2^9 f^{10}(1)=2^{19}\) \(\mathrm{f}(\alpha)=2^{2 \alpha-1} ; \alpha \in \mathrm{N}\) Form equation (ii) \(\sum_{\mathrm{k}=1}^{10} \mathrm{f}(\alpha+\mathrm{k})=2\left(2^{2 \alpha-1}\right)\left(2+2^3+2^5+\ldots+2^{19}\right)\) \(\frac{512}{3}\left(2^{20}-1\right)=2^{2 \alpha}\left(2 \cdot \frac{\left(2^{20}-1\right)}{3}\right)\) \(\frac{512}{3}\left(2^{20}-1\right)=\frac{2^{2 \alpha+1}}{3}\left(2^{20}-1\right)\) Comparing both side, we get- \(2^{2 \alpha+1}=512\) \(2^{2 \alpha+1}=2^9\) \(2 \alpha+1=9\) \(2 \alpha=8\)Hence, \(\alpha=4\)
Shift-I
Sets, Relation and Function
116886
The remainder when \(3^{2022}\) is divided by 5 is
1 1
2 2
3 3
4 4
Explanation:
D Given, \(3^{2022}\) \(\quad=\left(3^2\right)^{1011}\) \(\quad=(9)^{1011}\) \(\quad=(10-1)^{1011}\) \(\quad={ }^{1011} \mathrm{C}_0 \cdot 10^{1011-1011} \mathrm{C}_1 \cdot 10^{1010}+\ldots . .+{ }^{1011} \mathrm{C}_{1010}\) \(10^1-{ }^{1011} \mathrm{C}_{1011}\) \(=10 \mathrm{k}-1, \text { where } \mathrm{k}=\text { integer }\) \(=10 \mathrm{k}-1-4+4\) \(=10 \mathrm{k}-5+4\) \(=5(2 \mathrm{k}-1)+4\)So, when it is divided by 5 , remainder will be ' 4 '
Shift-I
Sets, Relation and Function
116887
Let \(f(x)=a x^2+b x+c\) be such that \(f(1)=3, f(-\) 2) \(=\lambda\) and \(f(3)=4\). If \(f(0)+f(1)+f(-2)+f(3)=\) 14 then \(\lambda\) is equal to
1 -4
2 \(\frac{13}{2}\)
3 \(\frac{23}{2}\)
4 4
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}\) Then, \(\begin{gathered} f(1)=a+b+c=3 \\ f(-2)=4 a-2 b+c=\lambda \\ f(3)=9 a+3 b+c=4 \end{gathered}\) \(\begin{array}{ll} \because & \mathrm{f}(0)+\mathrm{f}(1)+\mathrm{f}(-2)+\mathrm{f}(3)=14 \\ \therefore & \mathrm{c}+3+\lambda+4=14 \\ & \mathrm{c}+\lambda=7 \\ & \lambda=7-\mathrm{c} \end{array}\) Solving (i) and (ii):- \(\begin{gathered} 2 a+2 b+2 c=6 \\ 4 a-2 b+c=\lambda \\ \hline 6 a+3 c=6+\lambda \end{gathered}\) From (ii) and (iii):- \(\begin{gathered} 12 a-6 b+3 c=3 \lambda \\ 18 a+6 b+2 c=8 \\ \hline 30 a+5 c=3 \lambda+8 \end{gathered}\) Now, we have- \(6 a+3 c=6+\lambda\) \(30 a+5 c=3 \lambda+8 .\) Solving (iv) and (v), we get - \(30 a+15 c=30+5 \lambda\) \(30 a+5 c=8+3 \lambda\) \(\therefore \quad c \frac{22}{10}+\frac{\lambda}{5}\) Then, \(\lambda=7-\frac{22}{10}-\frac{\lambda}{5}\) Or \(\quad \frac{6}{5} \lambda=\frac{70-22}{10}=\frac{48}{10}\) So, \(\quad \lambda=\frac{48}{10} \times \frac{5}{6}=\frac{8}{2}=4\)
Shift-II
Sets, Relation and Function
116888
If \(x^2+y^2+z^2 \neq 0, x=c y+b z, y=a z+c x\) and \(z=\) \(b x+a y\), then \(a^2+b^2+c^2+2 a b c\) is equal to
1 1
2 2
3 \(a+b+c\)
4 \(a b+b c+c a\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2 \neq 0\) And the system of equation- \(x-c y-b z=0\) \(c x-y+a z=0\) \(b x+a y-z=0\) This can be written as:- \(\left[\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=0\) We should have, \(\left|\begin{array}{ccc}1 & -\mathrm{c} & -\mathrm{b} \\ \mathrm{c} & -1 & \mathrm{a} \\ \mathrm{b} & \mathrm{a} & -1\end{array}\right|=0\) \(\Rightarrow 1\left(1-\mathrm{a}^2\right)+\mathrm{c}(-\mathrm{c}-\mathrm{ab})-\mathrm{b}(\mathrm{ac}+\mathrm{b})=0\) Or \(1-a^2-c^2-a b c-a b c-b^2=0\) So, \(\quad a^2+b^2+c^2+2 a b c=1\)
116885
Let \(f: N \rightarrow R\) be a function such that \(f(x+y)=\) \(2 f(x) f(y)\) for natural numbers \(x\) and \(y\). If \(f(1)\) \(=2\), then the value of \(\alpha\) for which \(\sum_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)\) holds, is
1 2
2 3
3 4
4 6
Explanation:
C Given, \(f: N \rightarrow R, f(x+y)=2 f(x) f(y)\) \(f(1)=2\) \(\sum_{k=1}^{10} f(\alpha+k)=2 f(\alpha) \sum_{k=1}^{10} f(k)\) \(=2 f(\alpha)\{f(1)+f(2)+\ldots .+f(10)\}\) Form equation (i), \(f(10)=2^9 f^{10}(1)=2^{19}\) \(\mathrm{f}(\alpha)=2^{2 \alpha-1} ; \alpha \in \mathrm{N}\) Form equation (ii) \(\sum_{\mathrm{k}=1}^{10} \mathrm{f}(\alpha+\mathrm{k})=2\left(2^{2 \alpha-1}\right)\left(2+2^3+2^5+\ldots+2^{19}\right)\) \(\frac{512}{3}\left(2^{20}-1\right)=2^{2 \alpha}\left(2 \cdot \frac{\left(2^{20}-1\right)}{3}\right)\) \(\frac{512}{3}\left(2^{20}-1\right)=\frac{2^{2 \alpha+1}}{3}\left(2^{20}-1\right)\) Comparing both side, we get- \(2^{2 \alpha+1}=512\) \(2^{2 \alpha+1}=2^9\) \(2 \alpha+1=9\) \(2 \alpha=8\)Hence, \(\alpha=4\)
Shift-I
Sets, Relation and Function
116886
The remainder when \(3^{2022}\) is divided by 5 is
1 1
2 2
3 3
4 4
Explanation:
D Given, \(3^{2022}\) \(\quad=\left(3^2\right)^{1011}\) \(\quad=(9)^{1011}\) \(\quad=(10-1)^{1011}\) \(\quad={ }^{1011} \mathrm{C}_0 \cdot 10^{1011-1011} \mathrm{C}_1 \cdot 10^{1010}+\ldots . .+{ }^{1011} \mathrm{C}_{1010}\) \(10^1-{ }^{1011} \mathrm{C}_{1011}\) \(=10 \mathrm{k}-1, \text { where } \mathrm{k}=\text { integer }\) \(=10 \mathrm{k}-1-4+4\) \(=10 \mathrm{k}-5+4\) \(=5(2 \mathrm{k}-1)+4\)So, when it is divided by 5 , remainder will be ' 4 '
Shift-I
Sets, Relation and Function
116887
Let \(f(x)=a x^2+b x+c\) be such that \(f(1)=3, f(-\) 2) \(=\lambda\) and \(f(3)=4\). If \(f(0)+f(1)+f(-2)+f(3)=\) 14 then \(\lambda\) is equal to
1 -4
2 \(\frac{13}{2}\)
3 \(\frac{23}{2}\)
4 4
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}\) Then, \(\begin{gathered} f(1)=a+b+c=3 \\ f(-2)=4 a-2 b+c=\lambda \\ f(3)=9 a+3 b+c=4 \end{gathered}\) \(\begin{array}{ll} \because & \mathrm{f}(0)+\mathrm{f}(1)+\mathrm{f}(-2)+\mathrm{f}(3)=14 \\ \therefore & \mathrm{c}+3+\lambda+4=14 \\ & \mathrm{c}+\lambda=7 \\ & \lambda=7-\mathrm{c} \end{array}\) Solving (i) and (ii):- \(\begin{gathered} 2 a+2 b+2 c=6 \\ 4 a-2 b+c=\lambda \\ \hline 6 a+3 c=6+\lambda \end{gathered}\) From (ii) and (iii):- \(\begin{gathered} 12 a-6 b+3 c=3 \lambda \\ 18 a+6 b+2 c=8 \\ \hline 30 a+5 c=3 \lambda+8 \end{gathered}\) Now, we have- \(6 a+3 c=6+\lambda\) \(30 a+5 c=3 \lambda+8 .\) Solving (iv) and (v), we get - \(30 a+15 c=30+5 \lambda\) \(30 a+5 c=8+3 \lambda\) \(\therefore \quad c \frac{22}{10}+\frac{\lambda}{5}\) Then, \(\lambda=7-\frac{22}{10}-\frac{\lambda}{5}\) Or \(\quad \frac{6}{5} \lambda=\frac{70-22}{10}=\frac{48}{10}\) So, \(\quad \lambda=\frac{48}{10} \times \frac{5}{6}=\frac{8}{2}=4\)
Shift-II
Sets, Relation and Function
116888
If \(x^2+y^2+z^2 \neq 0, x=c y+b z, y=a z+c x\) and \(z=\) \(b x+a y\), then \(a^2+b^2+c^2+2 a b c\) is equal to
1 1
2 2
3 \(a+b+c\)
4 \(a b+b c+c a\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2 \neq 0\) And the system of equation- \(x-c y-b z=0\) \(c x-y+a z=0\) \(b x+a y-z=0\) This can be written as:- \(\left[\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=0\) We should have, \(\left|\begin{array}{ccc}1 & -\mathrm{c} & -\mathrm{b} \\ \mathrm{c} & -1 & \mathrm{a} \\ \mathrm{b} & \mathrm{a} & -1\end{array}\right|=0\) \(\Rightarrow 1\left(1-\mathrm{a}^2\right)+\mathrm{c}(-\mathrm{c}-\mathrm{ab})-\mathrm{b}(\mathrm{ac}+\mathrm{b})=0\) Or \(1-a^2-c^2-a b c-a b c-b^2=0\) So, \(\quad a^2+b^2+c^2+2 a b c=1\)
116885
Let \(f: N \rightarrow R\) be a function such that \(f(x+y)=\) \(2 f(x) f(y)\) for natural numbers \(x\) and \(y\). If \(f(1)\) \(=2\), then the value of \(\alpha\) for which \(\sum_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)\) holds, is
1 2
2 3
3 4
4 6
Explanation:
C Given, \(f: N \rightarrow R, f(x+y)=2 f(x) f(y)\) \(f(1)=2\) \(\sum_{k=1}^{10} f(\alpha+k)=2 f(\alpha) \sum_{k=1}^{10} f(k)\) \(=2 f(\alpha)\{f(1)+f(2)+\ldots .+f(10)\}\) Form equation (i), \(f(10)=2^9 f^{10}(1)=2^{19}\) \(\mathrm{f}(\alpha)=2^{2 \alpha-1} ; \alpha \in \mathrm{N}\) Form equation (ii) \(\sum_{\mathrm{k}=1}^{10} \mathrm{f}(\alpha+\mathrm{k})=2\left(2^{2 \alpha-1}\right)\left(2+2^3+2^5+\ldots+2^{19}\right)\) \(\frac{512}{3}\left(2^{20}-1\right)=2^{2 \alpha}\left(2 \cdot \frac{\left(2^{20}-1\right)}{3}\right)\) \(\frac{512}{3}\left(2^{20}-1\right)=\frac{2^{2 \alpha+1}}{3}\left(2^{20}-1\right)\) Comparing both side, we get- \(2^{2 \alpha+1}=512\) \(2^{2 \alpha+1}=2^9\) \(2 \alpha+1=9\) \(2 \alpha=8\)Hence, \(\alpha=4\)
Shift-I
Sets, Relation and Function
116886
The remainder when \(3^{2022}\) is divided by 5 is
1 1
2 2
3 3
4 4
Explanation:
D Given, \(3^{2022}\) \(\quad=\left(3^2\right)^{1011}\) \(\quad=(9)^{1011}\) \(\quad=(10-1)^{1011}\) \(\quad={ }^{1011} \mathrm{C}_0 \cdot 10^{1011-1011} \mathrm{C}_1 \cdot 10^{1010}+\ldots . .+{ }^{1011} \mathrm{C}_{1010}\) \(10^1-{ }^{1011} \mathrm{C}_{1011}\) \(=10 \mathrm{k}-1, \text { where } \mathrm{k}=\text { integer }\) \(=10 \mathrm{k}-1-4+4\) \(=10 \mathrm{k}-5+4\) \(=5(2 \mathrm{k}-1)+4\)So, when it is divided by 5 , remainder will be ' 4 '
Shift-I
Sets, Relation and Function
116887
Let \(f(x)=a x^2+b x+c\) be such that \(f(1)=3, f(-\) 2) \(=\lambda\) and \(f(3)=4\). If \(f(0)+f(1)+f(-2)+f(3)=\) 14 then \(\lambda\) is equal to
1 -4
2 \(\frac{13}{2}\)
3 \(\frac{23}{2}\)
4 4
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}\) Then, \(\begin{gathered} f(1)=a+b+c=3 \\ f(-2)=4 a-2 b+c=\lambda \\ f(3)=9 a+3 b+c=4 \end{gathered}\) \(\begin{array}{ll} \because & \mathrm{f}(0)+\mathrm{f}(1)+\mathrm{f}(-2)+\mathrm{f}(3)=14 \\ \therefore & \mathrm{c}+3+\lambda+4=14 \\ & \mathrm{c}+\lambda=7 \\ & \lambda=7-\mathrm{c} \end{array}\) Solving (i) and (ii):- \(\begin{gathered} 2 a+2 b+2 c=6 \\ 4 a-2 b+c=\lambda \\ \hline 6 a+3 c=6+\lambda \end{gathered}\) From (ii) and (iii):- \(\begin{gathered} 12 a-6 b+3 c=3 \lambda \\ 18 a+6 b+2 c=8 \\ \hline 30 a+5 c=3 \lambda+8 \end{gathered}\) Now, we have- \(6 a+3 c=6+\lambda\) \(30 a+5 c=3 \lambda+8 .\) Solving (iv) and (v), we get - \(30 a+15 c=30+5 \lambda\) \(30 a+5 c=8+3 \lambda\) \(\therefore \quad c \frac{22}{10}+\frac{\lambda}{5}\) Then, \(\lambda=7-\frac{22}{10}-\frac{\lambda}{5}\) Or \(\quad \frac{6}{5} \lambda=\frac{70-22}{10}=\frac{48}{10}\) So, \(\quad \lambda=\frac{48}{10} \times \frac{5}{6}=\frac{8}{2}=4\)
Shift-II
Sets, Relation and Function
116888
If \(x^2+y^2+z^2 \neq 0, x=c y+b z, y=a z+c x\) and \(z=\) \(b x+a y\), then \(a^2+b^2+c^2+2 a b c\) is equal to
1 1
2 2
3 \(a+b+c\)
4 \(a b+b c+c a\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2 \neq 0\) And the system of equation- \(x-c y-b z=0\) \(c x-y+a z=0\) \(b x+a y-z=0\) This can be written as:- \(\left[\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=0\) We should have, \(\left|\begin{array}{ccc}1 & -\mathrm{c} & -\mathrm{b} \\ \mathrm{c} & -1 & \mathrm{a} \\ \mathrm{b} & \mathrm{a} & -1\end{array}\right|=0\) \(\Rightarrow 1\left(1-\mathrm{a}^2\right)+\mathrm{c}(-\mathrm{c}-\mathrm{ab})-\mathrm{b}(\mathrm{ac}+\mathrm{b})=0\) Or \(1-a^2-c^2-a b c-a b c-b^2=0\) So, \(\quad a^2+b^2+c^2+2 a b c=1\)
116885
Let \(f: N \rightarrow R\) be a function such that \(f(x+y)=\) \(2 f(x) f(y)\) for natural numbers \(x\) and \(y\). If \(f(1)\) \(=2\), then the value of \(\alpha\) for which \(\sum_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)\) holds, is
1 2
2 3
3 4
4 6
Explanation:
C Given, \(f: N \rightarrow R, f(x+y)=2 f(x) f(y)\) \(f(1)=2\) \(\sum_{k=1}^{10} f(\alpha+k)=2 f(\alpha) \sum_{k=1}^{10} f(k)\) \(=2 f(\alpha)\{f(1)+f(2)+\ldots .+f(10)\}\) Form equation (i), \(f(10)=2^9 f^{10}(1)=2^{19}\) \(\mathrm{f}(\alpha)=2^{2 \alpha-1} ; \alpha \in \mathrm{N}\) Form equation (ii) \(\sum_{\mathrm{k}=1}^{10} \mathrm{f}(\alpha+\mathrm{k})=2\left(2^{2 \alpha-1}\right)\left(2+2^3+2^5+\ldots+2^{19}\right)\) \(\frac{512}{3}\left(2^{20}-1\right)=2^{2 \alpha}\left(2 \cdot \frac{\left(2^{20}-1\right)}{3}\right)\) \(\frac{512}{3}\left(2^{20}-1\right)=\frac{2^{2 \alpha+1}}{3}\left(2^{20}-1\right)\) Comparing both side, we get- \(2^{2 \alpha+1}=512\) \(2^{2 \alpha+1}=2^9\) \(2 \alpha+1=9\) \(2 \alpha=8\)Hence, \(\alpha=4\)
Shift-I
Sets, Relation and Function
116886
The remainder when \(3^{2022}\) is divided by 5 is
1 1
2 2
3 3
4 4
Explanation:
D Given, \(3^{2022}\) \(\quad=\left(3^2\right)^{1011}\) \(\quad=(9)^{1011}\) \(\quad=(10-1)^{1011}\) \(\quad={ }^{1011} \mathrm{C}_0 \cdot 10^{1011-1011} \mathrm{C}_1 \cdot 10^{1010}+\ldots . .+{ }^{1011} \mathrm{C}_{1010}\) \(10^1-{ }^{1011} \mathrm{C}_{1011}\) \(=10 \mathrm{k}-1, \text { where } \mathrm{k}=\text { integer }\) \(=10 \mathrm{k}-1-4+4\) \(=10 \mathrm{k}-5+4\) \(=5(2 \mathrm{k}-1)+4\)So, when it is divided by 5 , remainder will be ' 4 '
Shift-I
Sets, Relation and Function
116887
Let \(f(x)=a x^2+b x+c\) be such that \(f(1)=3, f(-\) 2) \(=\lambda\) and \(f(3)=4\). If \(f(0)+f(1)+f(-2)+f(3)=\) 14 then \(\lambda\) is equal to
1 -4
2 \(\frac{13}{2}\)
3 \(\frac{23}{2}\)
4 4
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}\) Then, \(\begin{gathered} f(1)=a+b+c=3 \\ f(-2)=4 a-2 b+c=\lambda \\ f(3)=9 a+3 b+c=4 \end{gathered}\) \(\begin{array}{ll} \because & \mathrm{f}(0)+\mathrm{f}(1)+\mathrm{f}(-2)+\mathrm{f}(3)=14 \\ \therefore & \mathrm{c}+3+\lambda+4=14 \\ & \mathrm{c}+\lambda=7 \\ & \lambda=7-\mathrm{c} \end{array}\) Solving (i) and (ii):- \(\begin{gathered} 2 a+2 b+2 c=6 \\ 4 a-2 b+c=\lambda \\ \hline 6 a+3 c=6+\lambda \end{gathered}\) From (ii) and (iii):- \(\begin{gathered} 12 a-6 b+3 c=3 \lambda \\ 18 a+6 b+2 c=8 \\ \hline 30 a+5 c=3 \lambda+8 \end{gathered}\) Now, we have- \(6 a+3 c=6+\lambda\) \(30 a+5 c=3 \lambda+8 .\) Solving (iv) and (v), we get - \(30 a+15 c=30+5 \lambda\) \(30 a+5 c=8+3 \lambda\) \(\therefore \quad c \frac{22}{10}+\frac{\lambda}{5}\) Then, \(\lambda=7-\frac{22}{10}-\frac{\lambda}{5}\) Or \(\quad \frac{6}{5} \lambda=\frac{70-22}{10}=\frac{48}{10}\) So, \(\quad \lambda=\frac{48}{10} \times \frac{5}{6}=\frac{8}{2}=4\)
Shift-II
Sets, Relation and Function
116888
If \(x^2+y^2+z^2 \neq 0, x=c y+b z, y=a z+c x\) and \(z=\) \(b x+a y\), then \(a^2+b^2+c^2+2 a b c\) is equal to
1 1
2 2
3 \(a+b+c\)
4 \(a b+b c+c a\)
Explanation:
A Given, \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2 \neq 0\) And the system of equation- \(x-c y-b z=0\) \(c x-y+a z=0\) \(b x+a y-z=0\) This can be written as:- \(\left[\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=0\) We should have, \(\left|\begin{array}{ccc}1 & -\mathrm{c} & -\mathrm{b} \\ \mathrm{c} & -1 & \mathrm{a} \\ \mathrm{b} & \mathrm{a} & -1\end{array}\right|=0\) \(\Rightarrow 1\left(1-\mathrm{a}^2\right)+\mathrm{c}(-\mathrm{c}-\mathrm{ab})-\mathrm{b}(\mathrm{ac}+\mathrm{b})=0\) Or \(1-a^2-c^2-a b c-a b c-b^2=0\) So, \(\quad a^2+b^2+c^2+2 a b c=1\)
