NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117002
The numbers \(a_n s\) are defined by \(a_0=1, a_{n+1}=3 n^2+n+a_n,(n \geq 0)\) Then, \(a_n\) is equal to
1 \(n^3+n^2+1\)
2 \(n^3-n^2+1\)
3 \(n^3-n^2\)
4 \(n^3+n^2\)
Explanation:
C Given, \(\mathrm{a}_0=1\) and \(a_{n+1}=3 n^2+n+a_n\) \(a_n=3(n-1)^2+(n-1)+a_{n-1}\) Now, put \(\mathrm{n}=1,2,3, \ldots, \mathrm{n}\) We get, \(\mathrm{a}_1=0+\mathrm{a}_0=0+1=1\) \(\mathrm{a}_2=4+\mathrm{a}_1\) \(\mathrm{a}_3=14+\mathrm{a}_2\) \(\mathrm{a}_4=30+\mathrm{a}_3\) \(a_{n-1}=3(n-2)^2+(n-2)+a_{n-2}\) \(a_n=3(n-1)^2+(n-1)+a_{n-1}\) On adding, we get \(\mathrm{a}_{\mathrm{n}}=\left(1+4+14+30+\ldots .+3(\mathrm{n}-1)^2+(\mathrm{n}-1)\right)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma 3\left(\mathrm{n}^2+1-2 \mathrm{n}\right)+(\mathrm{n}-1)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma\left(3 \mathrm{n}^2+3-6 \mathrm{n}+\mathrm{n}-1\right)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma\left(3 \mathrm{n}^2-5 \mathrm{n}+2\right)\) \(\mathrm{a}_{\mathrm{n}}=3 \Sigma \mathrm{n}^2-5 \Sigma \mathrm{n}+2 \Sigma 1\) \(\mathrm{a}_{\mathrm{n}}=\frac{3 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{5 \mathrm{n}(\mathrm{n}+1)}{2}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{n}+1)\{2 \mathrm{n}+1-5\}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{n}+1)(2 \mathrm{n}-4)+2 \mathrm{n}\) \(= \mathrm{n}(\mathrm{n}+1)(\mathrm{n}-2)+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\left(\mathrm{n}^2+\mathrm{n}\right)(\mathrm{n}-2)+2 \mathrm{n}\) \(= \mathrm{n}^3-2 \mathrm{n}^2+\mathrm{n}^2-2 \mathrm{n}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\mathrm{n}^3-\mathrm{n}^2\)
Manipal UGET-2010
Sets, Relation and Function
117003
If \(|x-2|+|x-3|=7\), then the value of \(x\) is
1 -1
2 6
3 -1 or 6
4 None of these
Explanation:
C Given, When \(|x-2|+|x-3|=7\) \(x\lt 2\) \(-(x-2)-(x-3)=7\) \(-x+2-x+3=7\) \(-2 x=2\) \(x=-1\) Which, is less them 2 When, \(\mathrm{x} \in[2,3]\) or \(2 \leq \mathrm{x}\lt 3\) And \(|\mathrm{x}-2|=\mathrm{x}-2\) \(|\mathrm{x}-3|=-(\mathrm{x}-3)\) \(|\mathrm{x}-2|+|\mathrm{x}-3|=(\mathrm{x}-2)-(\mathrm{x}-3)=7\) \(=\mathrm{x}-2-\mathrm{x}+3=7\) \(=1 \neq 7\) So, solution is not possible - When, \(\mathrm{x} \geq 3\) \(|x-2|+|x-3|=7\) \((x-2)+(x-3)=7\) \(2 x-5=7\) \(2 x=12\) \(x=6 \geq 3\) Hence, there are two value of \(x\) \(\mathrm{x}=-1,6\)
Rajasthan PET-2006
Sets, Relation and Function
117004
If \(4^{\log _3 \sqrt{3}}+9^{\log _2 2^2}=10^{\log _x 83}(x \in R)\), then the value of \(x\) is
117002
The numbers \(a_n s\) are defined by \(a_0=1, a_{n+1}=3 n^2+n+a_n,(n \geq 0)\) Then, \(a_n\) is equal to
1 \(n^3+n^2+1\)
2 \(n^3-n^2+1\)
3 \(n^3-n^2\)
4 \(n^3+n^2\)
Explanation:
C Given, \(\mathrm{a}_0=1\) and \(a_{n+1}=3 n^2+n+a_n\) \(a_n=3(n-1)^2+(n-1)+a_{n-1}\) Now, put \(\mathrm{n}=1,2,3, \ldots, \mathrm{n}\) We get, \(\mathrm{a}_1=0+\mathrm{a}_0=0+1=1\) \(\mathrm{a}_2=4+\mathrm{a}_1\) \(\mathrm{a}_3=14+\mathrm{a}_2\) \(\mathrm{a}_4=30+\mathrm{a}_3\) \(a_{n-1}=3(n-2)^2+(n-2)+a_{n-2}\) \(a_n=3(n-1)^2+(n-1)+a_{n-1}\) On adding, we get \(\mathrm{a}_{\mathrm{n}}=\left(1+4+14+30+\ldots .+3(\mathrm{n}-1)^2+(\mathrm{n}-1)\right)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma 3\left(\mathrm{n}^2+1-2 \mathrm{n}\right)+(\mathrm{n}-1)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma\left(3 \mathrm{n}^2+3-6 \mathrm{n}+\mathrm{n}-1\right)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma\left(3 \mathrm{n}^2-5 \mathrm{n}+2\right)\) \(\mathrm{a}_{\mathrm{n}}=3 \Sigma \mathrm{n}^2-5 \Sigma \mathrm{n}+2 \Sigma 1\) \(\mathrm{a}_{\mathrm{n}}=\frac{3 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{5 \mathrm{n}(\mathrm{n}+1)}{2}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{n}+1)\{2 \mathrm{n}+1-5\}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{n}+1)(2 \mathrm{n}-4)+2 \mathrm{n}\) \(= \mathrm{n}(\mathrm{n}+1)(\mathrm{n}-2)+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\left(\mathrm{n}^2+\mathrm{n}\right)(\mathrm{n}-2)+2 \mathrm{n}\) \(= \mathrm{n}^3-2 \mathrm{n}^2+\mathrm{n}^2-2 \mathrm{n}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\mathrm{n}^3-\mathrm{n}^2\)
Manipal UGET-2010
Sets, Relation and Function
117003
If \(|x-2|+|x-3|=7\), then the value of \(x\) is
1 -1
2 6
3 -1 or 6
4 None of these
Explanation:
C Given, When \(|x-2|+|x-3|=7\) \(x\lt 2\) \(-(x-2)-(x-3)=7\) \(-x+2-x+3=7\) \(-2 x=2\) \(x=-1\) Which, is less them 2 When, \(\mathrm{x} \in[2,3]\) or \(2 \leq \mathrm{x}\lt 3\) And \(|\mathrm{x}-2|=\mathrm{x}-2\) \(|\mathrm{x}-3|=-(\mathrm{x}-3)\) \(|\mathrm{x}-2|+|\mathrm{x}-3|=(\mathrm{x}-2)-(\mathrm{x}-3)=7\) \(=\mathrm{x}-2-\mathrm{x}+3=7\) \(=1 \neq 7\) So, solution is not possible - When, \(\mathrm{x} \geq 3\) \(|x-2|+|x-3|=7\) \((x-2)+(x-3)=7\) \(2 x-5=7\) \(2 x=12\) \(x=6 \geq 3\) Hence, there are two value of \(x\) \(\mathrm{x}=-1,6\)
Rajasthan PET-2006
Sets, Relation and Function
117004
If \(4^{\log _3 \sqrt{3}}+9^{\log _2 2^2}=10^{\log _x 83}(x \in R)\), then the value of \(x\) is
117002
The numbers \(a_n s\) are defined by \(a_0=1, a_{n+1}=3 n^2+n+a_n,(n \geq 0)\) Then, \(a_n\) is equal to
1 \(n^3+n^2+1\)
2 \(n^3-n^2+1\)
3 \(n^3-n^2\)
4 \(n^3+n^2\)
Explanation:
C Given, \(\mathrm{a}_0=1\) and \(a_{n+1}=3 n^2+n+a_n\) \(a_n=3(n-1)^2+(n-1)+a_{n-1}\) Now, put \(\mathrm{n}=1,2,3, \ldots, \mathrm{n}\) We get, \(\mathrm{a}_1=0+\mathrm{a}_0=0+1=1\) \(\mathrm{a}_2=4+\mathrm{a}_1\) \(\mathrm{a}_3=14+\mathrm{a}_2\) \(\mathrm{a}_4=30+\mathrm{a}_3\) \(a_{n-1}=3(n-2)^2+(n-2)+a_{n-2}\) \(a_n=3(n-1)^2+(n-1)+a_{n-1}\) On adding, we get \(\mathrm{a}_{\mathrm{n}}=\left(1+4+14+30+\ldots .+3(\mathrm{n}-1)^2+(\mathrm{n}-1)\right)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma 3\left(\mathrm{n}^2+1-2 \mathrm{n}\right)+(\mathrm{n}-1)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma\left(3 \mathrm{n}^2+3-6 \mathrm{n}+\mathrm{n}-1\right)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma\left(3 \mathrm{n}^2-5 \mathrm{n}+2\right)\) \(\mathrm{a}_{\mathrm{n}}=3 \Sigma \mathrm{n}^2-5 \Sigma \mathrm{n}+2 \Sigma 1\) \(\mathrm{a}_{\mathrm{n}}=\frac{3 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{5 \mathrm{n}(\mathrm{n}+1)}{2}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{n}+1)\{2 \mathrm{n}+1-5\}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{n}+1)(2 \mathrm{n}-4)+2 \mathrm{n}\) \(= \mathrm{n}(\mathrm{n}+1)(\mathrm{n}-2)+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\left(\mathrm{n}^2+\mathrm{n}\right)(\mathrm{n}-2)+2 \mathrm{n}\) \(= \mathrm{n}^3-2 \mathrm{n}^2+\mathrm{n}^2-2 \mathrm{n}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\mathrm{n}^3-\mathrm{n}^2\)
Manipal UGET-2010
Sets, Relation and Function
117003
If \(|x-2|+|x-3|=7\), then the value of \(x\) is
1 -1
2 6
3 -1 or 6
4 None of these
Explanation:
C Given, When \(|x-2|+|x-3|=7\) \(x\lt 2\) \(-(x-2)-(x-3)=7\) \(-x+2-x+3=7\) \(-2 x=2\) \(x=-1\) Which, is less them 2 When, \(\mathrm{x} \in[2,3]\) or \(2 \leq \mathrm{x}\lt 3\) And \(|\mathrm{x}-2|=\mathrm{x}-2\) \(|\mathrm{x}-3|=-(\mathrm{x}-3)\) \(|\mathrm{x}-2|+|\mathrm{x}-3|=(\mathrm{x}-2)-(\mathrm{x}-3)=7\) \(=\mathrm{x}-2-\mathrm{x}+3=7\) \(=1 \neq 7\) So, solution is not possible - When, \(\mathrm{x} \geq 3\) \(|x-2|+|x-3|=7\) \((x-2)+(x-3)=7\) \(2 x-5=7\) \(2 x=12\) \(x=6 \geq 3\) Hence, there are two value of \(x\) \(\mathrm{x}=-1,6\)
Rajasthan PET-2006
Sets, Relation and Function
117004
If \(4^{\log _3 \sqrt{3}}+9^{\log _2 2^2}=10^{\log _x 83}(x \in R)\), then the value of \(x\) is
117002
The numbers \(a_n s\) are defined by \(a_0=1, a_{n+1}=3 n^2+n+a_n,(n \geq 0)\) Then, \(a_n\) is equal to
1 \(n^3+n^2+1\)
2 \(n^3-n^2+1\)
3 \(n^3-n^2\)
4 \(n^3+n^2\)
Explanation:
C Given, \(\mathrm{a}_0=1\) and \(a_{n+1}=3 n^2+n+a_n\) \(a_n=3(n-1)^2+(n-1)+a_{n-1}\) Now, put \(\mathrm{n}=1,2,3, \ldots, \mathrm{n}\) We get, \(\mathrm{a}_1=0+\mathrm{a}_0=0+1=1\) \(\mathrm{a}_2=4+\mathrm{a}_1\) \(\mathrm{a}_3=14+\mathrm{a}_2\) \(\mathrm{a}_4=30+\mathrm{a}_3\) \(a_{n-1}=3(n-2)^2+(n-2)+a_{n-2}\) \(a_n=3(n-1)^2+(n-1)+a_{n-1}\) On adding, we get \(\mathrm{a}_{\mathrm{n}}=\left(1+4+14+30+\ldots .+3(\mathrm{n}-1)^2+(\mathrm{n}-1)\right)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma 3\left(\mathrm{n}^2+1-2 \mathrm{n}\right)+(\mathrm{n}-1)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma\left(3 \mathrm{n}^2+3-6 \mathrm{n}+\mathrm{n}-1\right)\) \(\mathrm{a}_{\mathrm{n}}=\Sigma\left(3 \mathrm{n}^2-5 \mathrm{n}+2\right)\) \(\mathrm{a}_{\mathrm{n}}=3 \Sigma \mathrm{n}^2-5 \Sigma \mathrm{n}+2 \Sigma 1\) \(\mathrm{a}_{\mathrm{n}}=\frac{3 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{5 \mathrm{n}(\mathrm{n}+1)}{2}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{n}+1)\{2 \mathrm{n}+1-5\}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{n}+1)(2 \mathrm{n}-4)+2 \mathrm{n}\) \(= \mathrm{n}(\mathrm{n}+1)(\mathrm{n}-2)+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\left(\mathrm{n}^2+\mathrm{n}\right)(\mathrm{n}-2)+2 \mathrm{n}\) \(= \mathrm{n}^3-2 \mathrm{n}^2+\mathrm{n}^2-2 \mathrm{n}+2 \mathrm{n}\) \(\mathrm{a}_{\mathrm{n}}=\mathrm{n}^3-\mathrm{n}^2\)
Manipal UGET-2010
Sets, Relation and Function
117003
If \(|x-2|+|x-3|=7\), then the value of \(x\) is
1 -1
2 6
3 -1 or 6
4 None of these
Explanation:
C Given, When \(|x-2|+|x-3|=7\) \(x\lt 2\) \(-(x-2)-(x-3)=7\) \(-x+2-x+3=7\) \(-2 x=2\) \(x=-1\) Which, is less them 2 When, \(\mathrm{x} \in[2,3]\) or \(2 \leq \mathrm{x}\lt 3\) And \(|\mathrm{x}-2|=\mathrm{x}-2\) \(|\mathrm{x}-3|=-(\mathrm{x}-3)\) \(|\mathrm{x}-2|+|\mathrm{x}-3|=(\mathrm{x}-2)-(\mathrm{x}-3)=7\) \(=\mathrm{x}-2-\mathrm{x}+3=7\) \(=1 \neq 7\) So, solution is not possible - When, \(\mathrm{x} \geq 3\) \(|x-2|+|x-3|=7\) \((x-2)+(x-3)=7\) \(2 x-5=7\) \(2 x=12\) \(x=6 \geq 3\) Hence, there are two value of \(x\) \(\mathrm{x}=-1,6\)
Rajasthan PET-2006
Sets, Relation and Function
117004
If \(4^{\log _3 \sqrt{3}}+9^{\log _2 2^2}=10^{\log _x 83}(x \in R)\), then the value of \(x\) is
