117014
If \(f:[\mathrm{R} \rightarrow[\mathrm{R}\) is such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in[R . f(1)=7\) and \(\sum_{\mathrm{r}=1}^{\mathrm{n}} f(\mathrm{r})=14114\), then \(\mathrm{n}=\)
1 9
2 13
3 63
4 62
Explanation:
C Given relation \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) Such that \(f(x+y)=f(x)+f(y)\) And \(f(1)=7\) Now, \(\sum_{r=1}^n f(r)=f(1)+f(2)+f(3) \ldots \ldots . f(n)\) \(=f(1)+2 f(1)+3 f(1)+4 f(1) \ldots \ldots . n f(1)\) \(=7+2.7+3.7+4.7 \ldots \ldots . n .7\) \(=7[1+2+3+4+5+\ldots \ldots . n]\) \(=\frac{7(n(n+1))}{2}\) Given, \(\sum_{r=1}^n f(r)=14114\) \(7 \frac{n(n+1)}{2}=14114\) \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}=2016\) \(\mathrm{n}(\mathrm{n}+1)=4032=63 \times 64\) \(\mathrm{n}=63\)
Shift-I
Sets, Relation and Function
117016
Let \(f(n)=A(-2)^n+B(-3)^n \forall A, B \in R\) and \(n \in N\) \(-\{1,2\}\). If \(\mathbf{f}(\mathbf{n})+\mathbf{a f}(\mathbf{n}-1)+\mathbf{b f}(\mathbf{n}-2)=0\), then \((\mathbf{a}+\mathbf{b})(\mathbf{b}-\mathbf{a})=\)
1 0
2 5
3 7
4 11
Explanation:
D Given, \(f(n)=A(-2)^n+B(-3)^n f(n)+a f(n-1)+b f(n-2)=0\) \(\therefore A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}+B(-3)^{n-1}\right)\) \(+b\left(A(-2)^{n-2}+B(-3)^{n-2}=0\right.\) \(\therefore\) It is possible only \(4-2 a+b=0 \text { and } 9-3 a+b=0\) Solving, we get \(\mathrm{a}=5, \mathrm{~b}=6\) \(\therefore \quad(a+b)(b-a)=(5+6)(6-5)=11\)
Shift-II
Sets, Relation and Function
117017
The set of all real values of \(x\) for which \(f(x)\) \(=\log _2\left(2^x-2\right)+\sqrt{1-x}\) is also real is
1 \(\mathrm{R}\)
2 \((1, \infty)\)
3 \((-\infty, 1]\)
4 \(\phi\)
Explanation:
D Given, \(\mathrm{f}(\mathrm{x}) =\log _2\left(2^{\mathrm{x}}-2\right)+\sqrt{1-\mathrm{x}}\) \(=\mathrm{f}_1(\mathrm{x})+\mathrm{f}_2(\mathrm{x})\) Now, the value of \(f_1(x)\) will be real if. \(\left(2^x-2\right)>0\) \(\text { Or } 2^x-2>0\) \(2^x>2\) \(\text { Or } x>1\) Similarly \(\sqrt{1-\mathrm{x}}=\mathrm{f}_2(\mathrm{x})\) will be real- \(\sqrt{1-x} \geq 0=x \leq 1\) Hence, real points of \(f(x)=f_1(x)+f_2(x)\) \(=\mathrm{f}_1(\mathrm{x}) \mathrm{U} \mathrm{f}_2(\mathrm{x})\) \(=\phi\)
TS EAMCET-19.07.2022
Sets, Relation and Function
117018
If \(f(1)=3\), and \(f(n+1)-f(n)=3\left(4^n-1\right)\), then \(\forall \mathbf{n} \in \mathbf{N}, \mathbf{f}(\mathbf{n})=\)
1 \(4^{\mathrm{n}}-1\)
2 \(4^{\mathrm{n}}-5 \mathrm{n}+4\)
3 \(4^n-3 n+2\)
4 \(4^n+4 n-5\)
Explanation:
C We have, \(\mathrm{f}(1)=3\) \(f(n+1)-f(n) =3\left(4^n-1\right)\) \(f(2)-f(1) =3(4-1)\) \(f(3)-f(2) =3\left(4^2-1\right)\) \(f(n)-f(n-2) =3\left(4^{n-1}-1\right)\) Adding, we get \(f(n)-f(1) =3\left(4+4^2-4^{n-1}\right)-3(n-1)\) \(f(n)-3 =3\left(\frac{4\left(4^{n-1}-1\right)}{4-1}\right)-3 n+3\) \(f(n)-3 =4^n-4-3 n+3\) \(f(n) =4^n-3 n+2\)
117014
If \(f:[\mathrm{R} \rightarrow[\mathrm{R}\) is such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in[R . f(1)=7\) and \(\sum_{\mathrm{r}=1}^{\mathrm{n}} f(\mathrm{r})=14114\), then \(\mathrm{n}=\)
1 9
2 13
3 63
4 62
Explanation:
C Given relation \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) Such that \(f(x+y)=f(x)+f(y)\) And \(f(1)=7\) Now, \(\sum_{r=1}^n f(r)=f(1)+f(2)+f(3) \ldots \ldots . f(n)\) \(=f(1)+2 f(1)+3 f(1)+4 f(1) \ldots \ldots . n f(1)\) \(=7+2.7+3.7+4.7 \ldots \ldots . n .7\) \(=7[1+2+3+4+5+\ldots \ldots . n]\) \(=\frac{7(n(n+1))}{2}\) Given, \(\sum_{r=1}^n f(r)=14114\) \(7 \frac{n(n+1)}{2}=14114\) \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}=2016\) \(\mathrm{n}(\mathrm{n}+1)=4032=63 \times 64\) \(\mathrm{n}=63\)
Shift-I
Sets, Relation and Function
117016
Let \(f(n)=A(-2)^n+B(-3)^n \forall A, B \in R\) and \(n \in N\) \(-\{1,2\}\). If \(\mathbf{f}(\mathbf{n})+\mathbf{a f}(\mathbf{n}-1)+\mathbf{b f}(\mathbf{n}-2)=0\), then \((\mathbf{a}+\mathbf{b})(\mathbf{b}-\mathbf{a})=\)
1 0
2 5
3 7
4 11
Explanation:
D Given, \(f(n)=A(-2)^n+B(-3)^n f(n)+a f(n-1)+b f(n-2)=0\) \(\therefore A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}+B(-3)^{n-1}\right)\) \(+b\left(A(-2)^{n-2}+B(-3)^{n-2}=0\right.\) \(\therefore\) It is possible only \(4-2 a+b=0 \text { and } 9-3 a+b=0\) Solving, we get \(\mathrm{a}=5, \mathrm{~b}=6\) \(\therefore \quad(a+b)(b-a)=(5+6)(6-5)=11\)
Shift-II
Sets, Relation and Function
117017
The set of all real values of \(x\) for which \(f(x)\) \(=\log _2\left(2^x-2\right)+\sqrt{1-x}\) is also real is
1 \(\mathrm{R}\)
2 \((1, \infty)\)
3 \((-\infty, 1]\)
4 \(\phi\)
Explanation:
D Given, \(\mathrm{f}(\mathrm{x}) =\log _2\left(2^{\mathrm{x}}-2\right)+\sqrt{1-\mathrm{x}}\) \(=\mathrm{f}_1(\mathrm{x})+\mathrm{f}_2(\mathrm{x})\) Now, the value of \(f_1(x)\) will be real if. \(\left(2^x-2\right)>0\) \(\text { Or } 2^x-2>0\) \(2^x>2\) \(\text { Or } x>1\) Similarly \(\sqrt{1-\mathrm{x}}=\mathrm{f}_2(\mathrm{x})\) will be real- \(\sqrt{1-x} \geq 0=x \leq 1\) Hence, real points of \(f(x)=f_1(x)+f_2(x)\) \(=\mathrm{f}_1(\mathrm{x}) \mathrm{U} \mathrm{f}_2(\mathrm{x})\) \(=\phi\)
TS EAMCET-19.07.2022
Sets, Relation and Function
117018
If \(f(1)=3\), and \(f(n+1)-f(n)=3\left(4^n-1\right)\), then \(\forall \mathbf{n} \in \mathbf{N}, \mathbf{f}(\mathbf{n})=\)
1 \(4^{\mathrm{n}}-1\)
2 \(4^{\mathrm{n}}-5 \mathrm{n}+4\)
3 \(4^n-3 n+2\)
4 \(4^n+4 n-5\)
Explanation:
C We have, \(\mathrm{f}(1)=3\) \(f(n+1)-f(n) =3\left(4^n-1\right)\) \(f(2)-f(1) =3(4-1)\) \(f(3)-f(2) =3\left(4^2-1\right)\) \(f(n)-f(n-2) =3\left(4^{n-1}-1\right)\) Adding, we get \(f(n)-f(1) =3\left(4+4^2-4^{n-1}\right)-3(n-1)\) \(f(n)-3 =3\left(\frac{4\left(4^{n-1}-1\right)}{4-1}\right)-3 n+3\) \(f(n)-3 =4^n-4-3 n+3\) \(f(n) =4^n-3 n+2\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117014
If \(f:[\mathrm{R} \rightarrow[\mathrm{R}\) is such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in[R . f(1)=7\) and \(\sum_{\mathrm{r}=1}^{\mathrm{n}} f(\mathrm{r})=14114\), then \(\mathrm{n}=\)
1 9
2 13
3 63
4 62
Explanation:
C Given relation \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) Such that \(f(x+y)=f(x)+f(y)\) And \(f(1)=7\) Now, \(\sum_{r=1}^n f(r)=f(1)+f(2)+f(3) \ldots \ldots . f(n)\) \(=f(1)+2 f(1)+3 f(1)+4 f(1) \ldots \ldots . n f(1)\) \(=7+2.7+3.7+4.7 \ldots \ldots . n .7\) \(=7[1+2+3+4+5+\ldots \ldots . n]\) \(=\frac{7(n(n+1))}{2}\) Given, \(\sum_{r=1}^n f(r)=14114\) \(7 \frac{n(n+1)}{2}=14114\) \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}=2016\) \(\mathrm{n}(\mathrm{n}+1)=4032=63 \times 64\) \(\mathrm{n}=63\)
Shift-I
Sets, Relation and Function
117016
Let \(f(n)=A(-2)^n+B(-3)^n \forall A, B \in R\) and \(n \in N\) \(-\{1,2\}\). If \(\mathbf{f}(\mathbf{n})+\mathbf{a f}(\mathbf{n}-1)+\mathbf{b f}(\mathbf{n}-2)=0\), then \((\mathbf{a}+\mathbf{b})(\mathbf{b}-\mathbf{a})=\)
1 0
2 5
3 7
4 11
Explanation:
D Given, \(f(n)=A(-2)^n+B(-3)^n f(n)+a f(n-1)+b f(n-2)=0\) \(\therefore A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}+B(-3)^{n-1}\right)\) \(+b\left(A(-2)^{n-2}+B(-3)^{n-2}=0\right.\) \(\therefore\) It is possible only \(4-2 a+b=0 \text { and } 9-3 a+b=0\) Solving, we get \(\mathrm{a}=5, \mathrm{~b}=6\) \(\therefore \quad(a+b)(b-a)=(5+6)(6-5)=11\)
Shift-II
Sets, Relation and Function
117017
The set of all real values of \(x\) for which \(f(x)\) \(=\log _2\left(2^x-2\right)+\sqrt{1-x}\) is also real is
1 \(\mathrm{R}\)
2 \((1, \infty)\)
3 \((-\infty, 1]\)
4 \(\phi\)
Explanation:
D Given, \(\mathrm{f}(\mathrm{x}) =\log _2\left(2^{\mathrm{x}}-2\right)+\sqrt{1-\mathrm{x}}\) \(=\mathrm{f}_1(\mathrm{x})+\mathrm{f}_2(\mathrm{x})\) Now, the value of \(f_1(x)\) will be real if. \(\left(2^x-2\right)>0\) \(\text { Or } 2^x-2>0\) \(2^x>2\) \(\text { Or } x>1\) Similarly \(\sqrt{1-\mathrm{x}}=\mathrm{f}_2(\mathrm{x})\) will be real- \(\sqrt{1-x} \geq 0=x \leq 1\) Hence, real points of \(f(x)=f_1(x)+f_2(x)\) \(=\mathrm{f}_1(\mathrm{x}) \mathrm{U} \mathrm{f}_2(\mathrm{x})\) \(=\phi\)
TS EAMCET-19.07.2022
Sets, Relation and Function
117018
If \(f(1)=3\), and \(f(n+1)-f(n)=3\left(4^n-1\right)\), then \(\forall \mathbf{n} \in \mathbf{N}, \mathbf{f}(\mathbf{n})=\)
1 \(4^{\mathrm{n}}-1\)
2 \(4^{\mathrm{n}}-5 \mathrm{n}+4\)
3 \(4^n-3 n+2\)
4 \(4^n+4 n-5\)
Explanation:
C We have, \(\mathrm{f}(1)=3\) \(f(n+1)-f(n) =3\left(4^n-1\right)\) \(f(2)-f(1) =3(4-1)\) \(f(3)-f(2) =3\left(4^2-1\right)\) \(f(n)-f(n-2) =3\left(4^{n-1}-1\right)\) Adding, we get \(f(n)-f(1) =3\left(4+4^2-4^{n-1}\right)-3(n-1)\) \(f(n)-3 =3\left(\frac{4\left(4^{n-1}-1\right)}{4-1}\right)-3 n+3\) \(f(n)-3 =4^n-4-3 n+3\) \(f(n) =4^n-3 n+2\)
117014
If \(f:[\mathrm{R} \rightarrow[\mathrm{R}\) is such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in[R . f(1)=7\) and \(\sum_{\mathrm{r}=1}^{\mathrm{n}} f(\mathrm{r})=14114\), then \(\mathrm{n}=\)
1 9
2 13
3 63
4 62
Explanation:
C Given relation \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) Such that \(f(x+y)=f(x)+f(y)\) And \(f(1)=7\) Now, \(\sum_{r=1}^n f(r)=f(1)+f(2)+f(3) \ldots \ldots . f(n)\) \(=f(1)+2 f(1)+3 f(1)+4 f(1) \ldots \ldots . n f(1)\) \(=7+2.7+3.7+4.7 \ldots \ldots . n .7\) \(=7[1+2+3+4+5+\ldots \ldots . n]\) \(=\frac{7(n(n+1))}{2}\) Given, \(\sum_{r=1}^n f(r)=14114\) \(7 \frac{n(n+1)}{2}=14114\) \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}=2016\) \(\mathrm{n}(\mathrm{n}+1)=4032=63 \times 64\) \(\mathrm{n}=63\)
Shift-I
Sets, Relation and Function
117016
Let \(f(n)=A(-2)^n+B(-3)^n \forall A, B \in R\) and \(n \in N\) \(-\{1,2\}\). If \(\mathbf{f}(\mathbf{n})+\mathbf{a f}(\mathbf{n}-1)+\mathbf{b f}(\mathbf{n}-2)=0\), then \((\mathbf{a}+\mathbf{b})(\mathbf{b}-\mathbf{a})=\)
1 0
2 5
3 7
4 11
Explanation:
D Given, \(f(n)=A(-2)^n+B(-3)^n f(n)+a f(n-1)+b f(n-2)=0\) \(\therefore A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}+B(-3)^{n-1}\right)\) \(+b\left(A(-2)^{n-2}+B(-3)^{n-2}=0\right.\) \(\therefore\) It is possible only \(4-2 a+b=0 \text { and } 9-3 a+b=0\) Solving, we get \(\mathrm{a}=5, \mathrm{~b}=6\) \(\therefore \quad(a+b)(b-a)=(5+6)(6-5)=11\)
Shift-II
Sets, Relation and Function
117017
The set of all real values of \(x\) for which \(f(x)\) \(=\log _2\left(2^x-2\right)+\sqrt{1-x}\) is also real is
1 \(\mathrm{R}\)
2 \((1, \infty)\)
3 \((-\infty, 1]\)
4 \(\phi\)
Explanation:
D Given, \(\mathrm{f}(\mathrm{x}) =\log _2\left(2^{\mathrm{x}}-2\right)+\sqrt{1-\mathrm{x}}\) \(=\mathrm{f}_1(\mathrm{x})+\mathrm{f}_2(\mathrm{x})\) Now, the value of \(f_1(x)\) will be real if. \(\left(2^x-2\right)>0\) \(\text { Or } 2^x-2>0\) \(2^x>2\) \(\text { Or } x>1\) Similarly \(\sqrt{1-\mathrm{x}}=\mathrm{f}_2(\mathrm{x})\) will be real- \(\sqrt{1-x} \geq 0=x \leq 1\) Hence, real points of \(f(x)=f_1(x)+f_2(x)\) \(=\mathrm{f}_1(\mathrm{x}) \mathrm{U} \mathrm{f}_2(\mathrm{x})\) \(=\phi\)
TS EAMCET-19.07.2022
Sets, Relation and Function
117018
If \(f(1)=3\), and \(f(n+1)-f(n)=3\left(4^n-1\right)\), then \(\forall \mathbf{n} \in \mathbf{N}, \mathbf{f}(\mathbf{n})=\)
1 \(4^{\mathrm{n}}-1\)
2 \(4^{\mathrm{n}}-5 \mathrm{n}+4\)
3 \(4^n-3 n+2\)
4 \(4^n+4 n-5\)
Explanation:
C We have, \(\mathrm{f}(1)=3\) \(f(n+1)-f(n) =3\left(4^n-1\right)\) \(f(2)-f(1) =3(4-1)\) \(f(3)-f(2) =3\left(4^2-1\right)\) \(f(n)-f(n-2) =3\left(4^{n-1}-1\right)\) Adding, we get \(f(n)-f(1) =3\left(4+4^2-4^{n-1}\right)-3(n-1)\) \(f(n)-3 =3\left(\frac{4\left(4^{n-1}-1\right)}{4-1}\right)-3 n+3\) \(f(n)-3 =4^n-4-3 n+3\) \(f(n) =4^n-3 n+2\)