117031
The distinct linear functions which map \([-1,1\) ] onto [0,2 ] are
1 \(f(x)=x+1, g(x)=-x+1\)
2 \(f(x)=x-1, g(x)=x+1\)
3 \(f(x)=-x-1, g(x)=x-1\)
4 none of these
Explanation:
A Let, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) be the required linear function, then \(f(x)\) is either strictly increasing or strictly decreasing. \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) or \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[-1,1]\) So, \(\mathrm{a}>0\) or \(\mathrm{a}\lt 0\) Case-I When \(\mathrm{a}>0\) \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) is strictly increasing and maps \([-1,1]\) onto \([0,2]\). \(\therefore \mathrm{f}(-1)=0 \text { and } \mathrm{f}(1)=2\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(0=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(2=\mathrm{a}+\mathrm{b}\) On solving equation (i) and (ii), we get - \(\mathrm{a}=1, \mathrm{~b}=1\) \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}+1\) Case-II When \(\mathrm{a}\lt 0\) In this case \(f(x)\) is strictly decreasing and maps \([-1,1]\) onto [0,2]. Therefore, \(\mathrm{f}(-1)=2\) and \(\mathrm{f}(1)=0\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(2=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(0=\mathrm{a}+\mathrm{b}\) On solving equation (iii) and (iv), we get - \(\mathrm{a}=-1, \mathrm{~b}=1\) \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+1\)Hence, the distinct function are \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) and \(\mathrm{f}(\mathrm{x})=-\) \(\mathrm{x}+1\).
SRMJEEE-2007
Sets, Relation and Function
117032
Let \(f(x)=2 x^n+\lambda, \lambda \in R, n \in N\), and \(f(4)=133\), \(f(5)=255\). Then the sum of all the positive integer divisors of \(\{(\mathbf{f}(3)-\mathbf{f}(2)\}\) is
1 59
2 60
3 61
4 58
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{\mathrm{n}}+\lambda, \lambda \in \mathrm{R} \text { and } \mathrm{n} \in \mathrm{N}\) \(\mathrm{f}(4)=133\) \(\mathrm{f}(5)=255\) \(\mathrm{f}(4)=133=2 \times(4)^{\mathrm{n}}+\lambda\) \(\mathrm{f}(5)=255=2 \times(5)^{\mathrm{n}}+\lambda\) Now, subtracting the equation- \(2\left\{(5)^{\mathrm{n}}-(4)^{\mathrm{n}}\right\}=255-133\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=\frac{122}{2}\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=61\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=(5)^3-(4)^3\) \(\mathrm{n}=3\) From equation (i) - \(2 \times(4)^3+\lambda=133\) \(\lambda=133-128\) \(\lambda=5\) Now, \(\mathrm{f}(3)-\mathrm{f}(2)\) \(=\left\{2(3)^3+\lambda\right)-\left(2(2)^3+\lambda\right\}\) \(=2\left(3^3-2^3\right)=2(27-8)=38\) The number of divisor is \(1,2,19,38\) Sum of divisor \(1+2+19+38=60\)
Shift-II
Sets, Relation and Function
117033
Let \(f(n)=2^{n+1}, g(n)=1+(n+1) 2^n\) for all \(n \in N\). Then
1 \(f(n)>g(n)\)
2 f(n) \(\lt \) g(n)
3 \(f(n)\) and \(g(n)\) are not comparable
4 \(f(n)>g(n)\) if \(n\) be even and \(f(n)\lt g(n)\) if \(n\) be odd.
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=2^{\mathrm{n}+1}\) and \(\mathrm{g}(\mathrm{n})=1+(\mathrm{n}+1) 2^{\mathrm{n}} \forall \mathrm{n} \in \mathrm{N}\) Now, \(\mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n}) =(\mathrm{n}+1) 2^{\mathrm{n}}-2^{\mathrm{n}+1}+1\) \(=1+2^{\mathrm{n}}[\mathrm{n}+1-2]\) \(=1+2^{\mathrm{n}}(\mathrm{n}-1) \text { always positive. }\)\(\therefore \quad \mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n})>0\) So, \(\quad \mathrm{g}(\mathrm{n})>\mathrm{f}(\mathrm{n})\)
117036
Let \(T\) \& \(U\) be the set of all orthogonal matrices of order 3 over \(R\) \& the set of all non-singular matrices of order 3 over \(R\) respectively. Let \(A=\{-1,0,1\}\), then
1 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\).
2 There does not exist bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\)
3 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}\) but no between \(\mathrm{A} \& \mathrm{U}\).
4 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{U}\) but not between \(\mathrm{A} \& \mathrm{~T}\).
Explanation:
: \(\mathrm{T}=\text { Orthogonal Matrices }\) \(\mathrm{T}=\{+1,-1\}\) \(\mathrm{U}=\{\mathrm{R}-0\}\) \(\mathrm{A}=\{-1,0,1\}\) Ans: b: \(\mathrm{T}=\) Orthogonal Matrices \(\mathrm{T}=\{+1,-1\}\) Bijective \(\rightarrow\) On to and one-one \(\mathrm{A} \rightarrow \mathrm{T}\) Bijective mapping is not possible \(\mathrm{A} \rightarrow \mathrm{U}\) Bijective mapping is not possible
117031
The distinct linear functions which map \([-1,1\) ] onto [0,2 ] are
1 \(f(x)=x+1, g(x)=-x+1\)
2 \(f(x)=x-1, g(x)=x+1\)
3 \(f(x)=-x-1, g(x)=x-1\)
4 none of these
Explanation:
A Let, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) be the required linear function, then \(f(x)\) is either strictly increasing or strictly decreasing. \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) or \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[-1,1]\) So, \(\mathrm{a}>0\) or \(\mathrm{a}\lt 0\) Case-I When \(\mathrm{a}>0\) \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) is strictly increasing and maps \([-1,1]\) onto \([0,2]\). \(\therefore \mathrm{f}(-1)=0 \text { and } \mathrm{f}(1)=2\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(0=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(2=\mathrm{a}+\mathrm{b}\) On solving equation (i) and (ii), we get - \(\mathrm{a}=1, \mathrm{~b}=1\) \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}+1\) Case-II When \(\mathrm{a}\lt 0\) In this case \(f(x)\) is strictly decreasing and maps \([-1,1]\) onto [0,2]. Therefore, \(\mathrm{f}(-1)=2\) and \(\mathrm{f}(1)=0\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(2=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(0=\mathrm{a}+\mathrm{b}\) On solving equation (iii) and (iv), we get - \(\mathrm{a}=-1, \mathrm{~b}=1\) \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+1\)Hence, the distinct function are \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) and \(\mathrm{f}(\mathrm{x})=-\) \(\mathrm{x}+1\).
SRMJEEE-2007
Sets, Relation and Function
117032
Let \(f(x)=2 x^n+\lambda, \lambda \in R, n \in N\), and \(f(4)=133\), \(f(5)=255\). Then the sum of all the positive integer divisors of \(\{(\mathbf{f}(3)-\mathbf{f}(2)\}\) is
1 59
2 60
3 61
4 58
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{\mathrm{n}}+\lambda, \lambda \in \mathrm{R} \text { and } \mathrm{n} \in \mathrm{N}\) \(\mathrm{f}(4)=133\) \(\mathrm{f}(5)=255\) \(\mathrm{f}(4)=133=2 \times(4)^{\mathrm{n}}+\lambda\) \(\mathrm{f}(5)=255=2 \times(5)^{\mathrm{n}}+\lambda\) Now, subtracting the equation- \(2\left\{(5)^{\mathrm{n}}-(4)^{\mathrm{n}}\right\}=255-133\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=\frac{122}{2}\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=61\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=(5)^3-(4)^3\) \(\mathrm{n}=3\) From equation (i) - \(2 \times(4)^3+\lambda=133\) \(\lambda=133-128\) \(\lambda=5\) Now, \(\mathrm{f}(3)-\mathrm{f}(2)\) \(=\left\{2(3)^3+\lambda\right)-\left(2(2)^3+\lambda\right\}\) \(=2\left(3^3-2^3\right)=2(27-8)=38\) The number of divisor is \(1,2,19,38\) Sum of divisor \(1+2+19+38=60\)
Shift-II
Sets, Relation and Function
117033
Let \(f(n)=2^{n+1}, g(n)=1+(n+1) 2^n\) for all \(n \in N\). Then
1 \(f(n)>g(n)\)
2 f(n) \(\lt \) g(n)
3 \(f(n)\) and \(g(n)\) are not comparable
4 \(f(n)>g(n)\) if \(n\) be even and \(f(n)\lt g(n)\) if \(n\) be odd.
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=2^{\mathrm{n}+1}\) and \(\mathrm{g}(\mathrm{n})=1+(\mathrm{n}+1) 2^{\mathrm{n}} \forall \mathrm{n} \in \mathrm{N}\) Now, \(\mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n}) =(\mathrm{n}+1) 2^{\mathrm{n}}-2^{\mathrm{n}+1}+1\) \(=1+2^{\mathrm{n}}[\mathrm{n}+1-2]\) \(=1+2^{\mathrm{n}}(\mathrm{n}-1) \text { always positive. }\)\(\therefore \quad \mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n})>0\) So, \(\quad \mathrm{g}(\mathrm{n})>\mathrm{f}(\mathrm{n})\)
117036
Let \(T\) \& \(U\) be the set of all orthogonal matrices of order 3 over \(R\) \& the set of all non-singular matrices of order 3 over \(R\) respectively. Let \(A=\{-1,0,1\}\), then
1 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\).
2 There does not exist bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\)
3 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}\) but no between \(\mathrm{A} \& \mathrm{U}\).
4 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{U}\) but not between \(\mathrm{A} \& \mathrm{~T}\).
Explanation:
: \(\mathrm{T}=\text { Orthogonal Matrices }\) \(\mathrm{T}=\{+1,-1\}\) \(\mathrm{U}=\{\mathrm{R}-0\}\) \(\mathrm{A}=\{-1,0,1\}\) Ans: b: \(\mathrm{T}=\) Orthogonal Matrices \(\mathrm{T}=\{+1,-1\}\) Bijective \(\rightarrow\) On to and one-one \(\mathrm{A} \rightarrow \mathrm{T}\) Bijective mapping is not possible \(\mathrm{A} \rightarrow \mathrm{U}\) Bijective mapping is not possible
117031
The distinct linear functions which map \([-1,1\) ] onto [0,2 ] are
1 \(f(x)=x+1, g(x)=-x+1\)
2 \(f(x)=x-1, g(x)=x+1\)
3 \(f(x)=-x-1, g(x)=x-1\)
4 none of these
Explanation:
A Let, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) be the required linear function, then \(f(x)\) is either strictly increasing or strictly decreasing. \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) or \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[-1,1]\) So, \(\mathrm{a}>0\) or \(\mathrm{a}\lt 0\) Case-I When \(\mathrm{a}>0\) \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) is strictly increasing and maps \([-1,1]\) onto \([0,2]\). \(\therefore \mathrm{f}(-1)=0 \text { and } \mathrm{f}(1)=2\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(0=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(2=\mathrm{a}+\mathrm{b}\) On solving equation (i) and (ii), we get - \(\mathrm{a}=1, \mathrm{~b}=1\) \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}+1\) Case-II When \(\mathrm{a}\lt 0\) In this case \(f(x)\) is strictly decreasing and maps \([-1,1]\) onto [0,2]. Therefore, \(\mathrm{f}(-1)=2\) and \(\mathrm{f}(1)=0\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(2=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(0=\mathrm{a}+\mathrm{b}\) On solving equation (iii) and (iv), we get - \(\mathrm{a}=-1, \mathrm{~b}=1\) \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+1\)Hence, the distinct function are \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) and \(\mathrm{f}(\mathrm{x})=-\) \(\mathrm{x}+1\).
SRMJEEE-2007
Sets, Relation and Function
117032
Let \(f(x)=2 x^n+\lambda, \lambda \in R, n \in N\), and \(f(4)=133\), \(f(5)=255\). Then the sum of all the positive integer divisors of \(\{(\mathbf{f}(3)-\mathbf{f}(2)\}\) is
1 59
2 60
3 61
4 58
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{\mathrm{n}}+\lambda, \lambda \in \mathrm{R} \text { and } \mathrm{n} \in \mathrm{N}\) \(\mathrm{f}(4)=133\) \(\mathrm{f}(5)=255\) \(\mathrm{f}(4)=133=2 \times(4)^{\mathrm{n}}+\lambda\) \(\mathrm{f}(5)=255=2 \times(5)^{\mathrm{n}}+\lambda\) Now, subtracting the equation- \(2\left\{(5)^{\mathrm{n}}-(4)^{\mathrm{n}}\right\}=255-133\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=\frac{122}{2}\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=61\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=(5)^3-(4)^3\) \(\mathrm{n}=3\) From equation (i) - \(2 \times(4)^3+\lambda=133\) \(\lambda=133-128\) \(\lambda=5\) Now, \(\mathrm{f}(3)-\mathrm{f}(2)\) \(=\left\{2(3)^3+\lambda\right)-\left(2(2)^3+\lambda\right\}\) \(=2\left(3^3-2^3\right)=2(27-8)=38\) The number of divisor is \(1,2,19,38\) Sum of divisor \(1+2+19+38=60\)
Shift-II
Sets, Relation and Function
117033
Let \(f(n)=2^{n+1}, g(n)=1+(n+1) 2^n\) for all \(n \in N\). Then
1 \(f(n)>g(n)\)
2 f(n) \(\lt \) g(n)
3 \(f(n)\) and \(g(n)\) are not comparable
4 \(f(n)>g(n)\) if \(n\) be even and \(f(n)\lt g(n)\) if \(n\) be odd.
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=2^{\mathrm{n}+1}\) and \(\mathrm{g}(\mathrm{n})=1+(\mathrm{n}+1) 2^{\mathrm{n}} \forall \mathrm{n} \in \mathrm{N}\) Now, \(\mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n}) =(\mathrm{n}+1) 2^{\mathrm{n}}-2^{\mathrm{n}+1}+1\) \(=1+2^{\mathrm{n}}[\mathrm{n}+1-2]\) \(=1+2^{\mathrm{n}}(\mathrm{n}-1) \text { always positive. }\)\(\therefore \quad \mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n})>0\) So, \(\quad \mathrm{g}(\mathrm{n})>\mathrm{f}(\mathrm{n})\)
117036
Let \(T\) \& \(U\) be the set of all orthogonal matrices of order 3 over \(R\) \& the set of all non-singular matrices of order 3 over \(R\) respectively. Let \(A=\{-1,0,1\}\), then
1 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\).
2 There does not exist bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\)
3 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}\) but no between \(\mathrm{A} \& \mathrm{U}\).
4 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{U}\) but not between \(\mathrm{A} \& \mathrm{~T}\).
Explanation:
: \(\mathrm{T}=\text { Orthogonal Matrices }\) \(\mathrm{T}=\{+1,-1\}\) \(\mathrm{U}=\{\mathrm{R}-0\}\) \(\mathrm{A}=\{-1,0,1\}\) Ans: b: \(\mathrm{T}=\) Orthogonal Matrices \(\mathrm{T}=\{+1,-1\}\) Bijective \(\rightarrow\) On to and one-one \(\mathrm{A} \rightarrow \mathrm{T}\) Bijective mapping is not possible \(\mathrm{A} \rightarrow \mathrm{U}\) Bijective mapping is not possible
117031
The distinct linear functions which map \([-1,1\) ] onto [0,2 ] are
1 \(f(x)=x+1, g(x)=-x+1\)
2 \(f(x)=x-1, g(x)=x+1\)
3 \(f(x)=-x-1, g(x)=x-1\)
4 none of these
Explanation:
A Let, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) be the required linear function, then \(f(x)\) is either strictly increasing or strictly decreasing. \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) or \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[-1,1]\) So, \(\mathrm{a}>0\) or \(\mathrm{a}\lt 0\) Case-I When \(\mathrm{a}>0\) \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) is strictly increasing and maps \([-1,1]\) onto \([0,2]\). \(\therefore \mathrm{f}(-1)=0 \text { and } \mathrm{f}(1)=2\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(0=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(2=\mathrm{a}+\mathrm{b}\) On solving equation (i) and (ii), we get - \(\mathrm{a}=1, \mathrm{~b}=1\) \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}+1\) Case-II When \(\mathrm{a}\lt 0\) In this case \(f(x)\) is strictly decreasing and maps \([-1,1]\) onto [0,2]. Therefore, \(\mathrm{f}(-1)=2\) and \(\mathrm{f}(1)=0\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(2=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(0=\mathrm{a}+\mathrm{b}\) On solving equation (iii) and (iv), we get - \(\mathrm{a}=-1, \mathrm{~b}=1\) \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+1\)Hence, the distinct function are \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) and \(\mathrm{f}(\mathrm{x})=-\) \(\mathrm{x}+1\).
SRMJEEE-2007
Sets, Relation and Function
117032
Let \(f(x)=2 x^n+\lambda, \lambda \in R, n \in N\), and \(f(4)=133\), \(f(5)=255\). Then the sum of all the positive integer divisors of \(\{(\mathbf{f}(3)-\mathbf{f}(2)\}\) is
1 59
2 60
3 61
4 58
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{\mathrm{n}}+\lambda, \lambda \in \mathrm{R} \text { and } \mathrm{n} \in \mathrm{N}\) \(\mathrm{f}(4)=133\) \(\mathrm{f}(5)=255\) \(\mathrm{f}(4)=133=2 \times(4)^{\mathrm{n}}+\lambda\) \(\mathrm{f}(5)=255=2 \times(5)^{\mathrm{n}}+\lambda\) Now, subtracting the equation- \(2\left\{(5)^{\mathrm{n}}-(4)^{\mathrm{n}}\right\}=255-133\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=\frac{122}{2}\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=61\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=(5)^3-(4)^3\) \(\mathrm{n}=3\) From equation (i) - \(2 \times(4)^3+\lambda=133\) \(\lambda=133-128\) \(\lambda=5\) Now, \(\mathrm{f}(3)-\mathrm{f}(2)\) \(=\left\{2(3)^3+\lambda\right)-\left(2(2)^3+\lambda\right\}\) \(=2\left(3^3-2^3\right)=2(27-8)=38\) The number of divisor is \(1,2,19,38\) Sum of divisor \(1+2+19+38=60\)
Shift-II
Sets, Relation and Function
117033
Let \(f(n)=2^{n+1}, g(n)=1+(n+1) 2^n\) for all \(n \in N\). Then
1 \(f(n)>g(n)\)
2 f(n) \(\lt \) g(n)
3 \(f(n)\) and \(g(n)\) are not comparable
4 \(f(n)>g(n)\) if \(n\) be even and \(f(n)\lt g(n)\) if \(n\) be odd.
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=2^{\mathrm{n}+1}\) and \(\mathrm{g}(\mathrm{n})=1+(\mathrm{n}+1) 2^{\mathrm{n}} \forall \mathrm{n} \in \mathrm{N}\) Now, \(\mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n}) =(\mathrm{n}+1) 2^{\mathrm{n}}-2^{\mathrm{n}+1}+1\) \(=1+2^{\mathrm{n}}[\mathrm{n}+1-2]\) \(=1+2^{\mathrm{n}}(\mathrm{n}-1) \text { always positive. }\)\(\therefore \quad \mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n})>0\) So, \(\quad \mathrm{g}(\mathrm{n})>\mathrm{f}(\mathrm{n})\)
117036
Let \(T\) \& \(U\) be the set of all orthogonal matrices of order 3 over \(R\) \& the set of all non-singular matrices of order 3 over \(R\) respectively. Let \(A=\{-1,0,1\}\), then
1 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\).
2 There does not exist bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\)
3 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}\) but no between \(\mathrm{A} \& \mathrm{U}\).
4 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{U}\) but not between \(\mathrm{A} \& \mathrm{~T}\).
Explanation:
: \(\mathrm{T}=\text { Orthogonal Matrices }\) \(\mathrm{T}=\{+1,-1\}\) \(\mathrm{U}=\{\mathrm{R}-0\}\) \(\mathrm{A}=\{-1,0,1\}\) Ans: b: \(\mathrm{T}=\) Orthogonal Matrices \(\mathrm{T}=\{+1,-1\}\) Bijective \(\rightarrow\) On to and one-one \(\mathrm{A} \rightarrow \mathrm{T}\) Bijective mapping is not possible \(\mathrm{A} \rightarrow \mathrm{U}\) Bijective mapping is not possible
117031
The distinct linear functions which map \([-1,1\) ] onto [0,2 ] are
1 \(f(x)=x+1, g(x)=-x+1\)
2 \(f(x)=x-1, g(x)=x+1\)
3 \(f(x)=-x-1, g(x)=x-1\)
4 none of these
Explanation:
A Let, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) be the required linear function, then \(f(x)\) is either strictly increasing or strictly decreasing. \(\therefore \mathrm{f}^{\prime}(\mathrm{x})>0\) or \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[-1,1]\) So, \(\mathrm{a}>0\) or \(\mathrm{a}\lt 0\) Case-I When \(\mathrm{a}>0\) \(\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}\) is strictly increasing and maps \([-1,1]\) onto \([0,2]\). \(\therefore \mathrm{f}(-1)=0 \text { and } \mathrm{f}(1)=2\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(0=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(2=\mathrm{a}+\mathrm{b}\) On solving equation (i) and (ii), we get - \(\mathrm{a}=1, \mathrm{~b}=1\) \(\therefore \mathrm{f}(\mathrm{x})=\mathrm{x}+1\) Case-II When \(\mathrm{a}\lt 0\) In this case \(f(x)\) is strictly decreasing and maps \([-1,1]\) onto [0,2]. Therefore, \(\mathrm{f}(-1)=2\) and \(\mathrm{f}(1)=0\) \(\mathrm{f}(-1)=-\mathrm{a}+\mathrm{b}\) \(2=-\mathrm{a}+\mathrm{b}\) \(\mathrm{f}(1)=\mathrm{a}+\mathrm{b}\) \(0=\mathrm{a}+\mathrm{b}\) On solving equation (iii) and (iv), we get - \(\mathrm{a}=-1, \mathrm{~b}=1\) \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+1\)Hence, the distinct function are \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) and \(\mathrm{f}(\mathrm{x})=-\) \(\mathrm{x}+1\).
SRMJEEE-2007
Sets, Relation and Function
117032
Let \(f(x)=2 x^n+\lambda, \lambda \in R, n \in N\), and \(f(4)=133\), \(f(5)=255\). Then the sum of all the positive integer divisors of \(\{(\mathbf{f}(3)-\mathbf{f}(2)\}\) is
1 59
2 60
3 61
4 58
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{\mathrm{n}}+\lambda, \lambda \in \mathrm{R} \text { and } \mathrm{n} \in \mathrm{N}\) \(\mathrm{f}(4)=133\) \(\mathrm{f}(5)=255\) \(\mathrm{f}(4)=133=2 \times(4)^{\mathrm{n}}+\lambda\) \(\mathrm{f}(5)=255=2 \times(5)^{\mathrm{n}}+\lambda\) Now, subtracting the equation- \(2\left\{(5)^{\mathrm{n}}-(4)^{\mathrm{n}}\right\}=255-133\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=\frac{122}{2}\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=61\) \((5)^{\mathrm{n}}-(4)^{\mathrm{n}}=(5)^3-(4)^3\) \(\mathrm{n}=3\) From equation (i) - \(2 \times(4)^3+\lambda=133\) \(\lambda=133-128\) \(\lambda=5\) Now, \(\mathrm{f}(3)-\mathrm{f}(2)\) \(=\left\{2(3)^3+\lambda\right)-\left(2(2)^3+\lambda\right\}\) \(=2\left(3^3-2^3\right)=2(27-8)=38\) The number of divisor is \(1,2,19,38\) Sum of divisor \(1+2+19+38=60\)
Shift-II
Sets, Relation and Function
117033
Let \(f(n)=2^{n+1}, g(n)=1+(n+1) 2^n\) for all \(n \in N\). Then
1 \(f(n)>g(n)\)
2 f(n) \(\lt \) g(n)
3 \(f(n)\) and \(g(n)\) are not comparable
4 \(f(n)>g(n)\) if \(n\) be even and \(f(n)\lt g(n)\) if \(n\) be odd.
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=2^{\mathrm{n}+1}\) and \(\mathrm{g}(\mathrm{n})=1+(\mathrm{n}+1) 2^{\mathrm{n}} \forall \mathrm{n} \in \mathrm{N}\) Now, \(\mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n}) =(\mathrm{n}+1) 2^{\mathrm{n}}-2^{\mathrm{n}+1}+1\) \(=1+2^{\mathrm{n}}[\mathrm{n}+1-2]\) \(=1+2^{\mathrm{n}}(\mathrm{n}-1) \text { always positive. }\)\(\therefore \quad \mathrm{g}(\mathrm{n})-\mathrm{f}(\mathrm{n})>0\) So, \(\quad \mathrm{g}(\mathrm{n})>\mathrm{f}(\mathrm{n})\)
117036
Let \(T\) \& \(U\) be the set of all orthogonal matrices of order 3 over \(R\) \& the set of all non-singular matrices of order 3 over \(R\) respectively. Let \(A=\{-1,0,1\}\), then
1 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\).
2 There does not exist bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}, \mathrm{U}\)
3 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{T}\) but no between \(\mathrm{A} \& \mathrm{U}\).
4 There exists bijective mapping between \(\mathrm{A}\) and \(\mathrm{U}\) but not between \(\mathrm{A} \& \mathrm{~T}\).
Explanation:
: \(\mathrm{T}=\text { Orthogonal Matrices }\) \(\mathrm{T}=\{+1,-1\}\) \(\mathrm{U}=\{\mathrm{R}-0\}\) \(\mathrm{A}=\{-1,0,1\}\) Ans: b: \(\mathrm{T}=\) Orthogonal Matrices \(\mathrm{T}=\{+1,-1\}\) Bijective \(\rightarrow\) On to and one-one \(\mathrm{A} \rightarrow \mathrm{T}\) Bijective mapping is not possible \(\mathrm{A} \rightarrow \mathrm{U}\) Bijective mapping is not possible
