116998
If \(n\) be any integer, then \(n(n+1)(2 n+1)\) is:
1 odd number
2 integral multiple of 6
3 perfect square
4 does not necessarily have any of the foregoing proof
Explanation:
B Given that, \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) For \(\mathrm{n}=1 \rightarrow\) \(1 \times 2 \times 3=6\) For \(\mathrm{n}=2 \rightarrow\) \(2 \times 3 \times 5=30\) \(6,30 \neq\) perfect square So, \(n(n+1)(2 n+1)\) always an integral multiple of even numbers. So, it is a integral multiple of \(6(6,30,84\)------)
Manipal UGET-2019
Sets, Relation and Function
116999
If \(\boldsymbol{f}(\mathrm{x})=\left\{\begin{array}{l}3 \mathrm{x}^2+12 x-1,-1 \leq x \leq 2 \\ 37-x, \quad 2 \leq x \leq 3\end{array}\right.\), then
1 \(f(\mathrm{x})\) is decreasing on \([-1,2]\)
2 \(f^{\prime}(2)\) does not exist
3 \(f(\mathrm{x})\) has the maximum value at \(\mathrm{x}=2\)
4 None of the above
Explanation:
B In the interval \([-1,2], \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}+12>0\) hence, \(f(x)\) is increasing in \([-1,2]\) Now, \(\mathrm{f}(\mathrm{x})\) being a polynomial in \(\mathrm{x}_2\) continuous in \(-1 \leq x\lt 2\) and in \(2\lt x \leq 3\) all check at \(x=2\) \(\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} 3(2-h)^2+12(2-h)-1\) \(=12+24-1=35\) \(\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 37-(2+h)=35\) \(f(2)=3(2)^2+12(2)+1=35\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) and hence in the interval [\(1,3]\) Now, \(\operatorname{Lf}^{\prime}(2)=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(2-\mathrm{h})-\mathrm{f}(2)}{-\mathrm{h}}\) \(= \lim _{h \rightarrow 0} \frac{3(2-h)^2+12(2-h)-1-35}{-h}\) \(= \lim _{h \rightarrow 0} \frac{3 h^2-24 h}{-h}=24\) \(R f^{\prime}(2) =\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}\) \(= \lim _{h \rightarrow 0} \frac{37-(2+h)-35}{h}=-1\) Since, \(\operatorname{Lf}(2) \neq R f^{\prime}(2)\), Thus \(f^{\prime}(2)\) does not exist
Manipal UGET-2017
Sets, Relation and Function
117000
If \(\log _3 2, \log _3\left(2^x-5\right)\) and \(\log _3\left(2^x-\frac{7}{2}\right)\) are in \(A P\), the value of \(x\) is
1 2
2 3
3 0
4 \(\frac{1}{3}\)
Explanation:
B Given, \(\log _3 2, \log _3\left(2^x-5\right), \log _3\left(2^x-\frac{7}{2}\right) \text { arein A.P. }\) \(\Rightarrow \quad \log _3\left(2^x-5\right)=\frac{\log _3\left(2^x-\frac{7}{2}\right)+\log _3 2}{2}\) \(\Rightarrow \quad \left(2^x\right)^2+25-10 \cdot 2^x=2 \cdot 2^x-7\) \(\Rightarrow \quad \left(2^x\right)^2-12 \cdot 2^x+25+7=0\) \(\Rightarrow \quad \left(2^x\right)^2-12 \cdot 2^x+32=0\) \(\Rightarrow \quad 2^x=8 \text { or } 2^x=4\) \(\Rightarrow \quad x=3 \text { or } x=2\) \(\Rightarrow \quad x=3(\because x=2 \text { does not satisfy the given series })\)
Manipal UGET-2017
Sets, Relation and Function
117001
If \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \cdot f(\mathrm{y}), f(3)=3, f^{\prime}(0)=11\), then \(f^{\prime}(3)\) is equal to
1 \(11 . \mathrm{e}^{33}\)
2 33
3 11
4 \(\log ^{33}\)
Explanation:
Given, \(f(x+y)=f(x) . f(y)\) So, differentiate with \(\mathrm{x}\). \(f^{\prime}(x+y)=f(y) \cdot f^{\prime}(x)\) Put, \(x=0\) So, \(\quad f^{\prime}(y)=f(y) . f^{\prime}(0)\) \(\mathrm{f}^{\prime}(0)=11\) So, \(\quad \mathrm{f}^{\prime}(\mathrm{y})=11 \mathrm{f}(\mathrm{y})\) Now, \(\quad \mathrm{f}^{\prime}(\mathrm{y})=11 \mathrm{f}(\mathrm{y})\) \(\mathrm{f}^{\prime}(3)=11 \mathrm{f}(3)\) \(\mathrm{f}^{\prime}(3)=33\)
116998
If \(n\) be any integer, then \(n(n+1)(2 n+1)\) is:
1 odd number
2 integral multiple of 6
3 perfect square
4 does not necessarily have any of the foregoing proof
Explanation:
B Given that, \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) For \(\mathrm{n}=1 \rightarrow\) \(1 \times 2 \times 3=6\) For \(\mathrm{n}=2 \rightarrow\) \(2 \times 3 \times 5=30\) \(6,30 \neq\) perfect square So, \(n(n+1)(2 n+1)\) always an integral multiple of even numbers. So, it is a integral multiple of \(6(6,30,84\)------)
Manipal UGET-2019
Sets, Relation and Function
116999
If \(\boldsymbol{f}(\mathrm{x})=\left\{\begin{array}{l}3 \mathrm{x}^2+12 x-1,-1 \leq x \leq 2 \\ 37-x, \quad 2 \leq x \leq 3\end{array}\right.\), then
1 \(f(\mathrm{x})\) is decreasing on \([-1,2]\)
2 \(f^{\prime}(2)\) does not exist
3 \(f(\mathrm{x})\) has the maximum value at \(\mathrm{x}=2\)
4 None of the above
Explanation:
B In the interval \([-1,2], \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}+12>0\) hence, \(f(x)\) is increasing in \([-1,2]\) Now, \(\mathrm{f}(\mathrm{x})\) being a polynomial in \(\mathrm{x}_2\) continuous in \(-1 \leq x\lt 2\) and in \(2\lt x \leq 3\) all check at \(x=2\) \(\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} 3(2-h)^2+12(2-h)-1\) \(=12+24-1=35\) \(\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 37-(2+h)=35\) \(f(2)=3(2)^2+12(2)+1=35\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) and hence in the interval [\(1,3]\) Now, \(\operatorname{Lf}^{\prime}(2)=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(2-\mathrm{h})-\mathrm{f}(2)}{-\mathrm{h}}\) \(= \lim _{h \rightarrow 0} \frac{3(2-h)^2+12(2-h)-1-35}{-h}\) \(= \lim _{h \rightarrow 0} \frac{3 h^2-24 h}{-h}=24\) \(R f^{\prime}(2) =\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}\) \(= \lim _{h \rightarrow 0} \frac{37-(2+h)-35}{h}=-1\) Since, \(\operatorname{Lf}(2) \neq R f^{\prime}(2)\), Thus \(f^{\prime}(2)\) does not exist
Manipal UGET-2017
Sets, Relation and Function
117000
If \(\log _3 2, \log _3\left(2^x-5\right)\) and \(\log _3\left(2^x-\frac{7}{2}\right)\) are in \(A P\), the value of \(x\) is
1 2
2 3
3 0
4 \(\frac{1}{3}\)
Explanation:
B Given, \(\log _3 2, \log _3\left(2^x-5\right), \log _3\left(2^x-\frac{7}{2}\right) \text { arein A.P. }\) \(\Rightarrow \quad \log _3\left(2^x-5\right)=\frac{\log _3\left(2^x-\frac{7}{2}\right)+\log _3 2}{2}\) \(\Rightarrow \quad \left(2^x\right)^2+25-10 \cdot 2^x=2 \cdot 2^x-7\) \(\Rightarrow \quad \left(2^x\right)^2-12 \cdot 2^x+25+7=0\) \(\Rightarrow \quad \left(2^x\right)^2-12 \cdot 2^x+32=0\) \(\Rightarrow \quad 2^x=8 \text { or } 2^x=4\) \(\Rightarrow \quad x=3 \text { or } x=2\) \(\Rightarrow \quad x=3(\because x=2 \text { does not satisfy the given series })\)
Manipal UGET-2017
Sets, Relation and Function
117001
If \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \cdot f(\mathrm{y}), f(3)=3, f^{\prime}(0)=11\), then \(f^{\prime}(3)\) is equal to
1 \(11 . \mathrm{e}^{33}\)
2 33
3 11
4 \(\log ^{33}\)
Explanation:
Given, \(f(x+y)=f(x) . f(y)\) So, differentiate with \(\mathrm{x}\). \(f^{\prime}(x+y)=f(y) \cdot f^{\prime}(x)\) Put, \(x=0\) So, \(\quad f^{\prime}(y)=f(y) . f^{\prime}(0)\) \(\mathrm{f}^{\prime}(0)=11\) So, \(\quad \mathrm{f}^{\prime}(\mathrm{y})=11 \mathrm{f}(\mathrm{y})\) Now, \(\quad \mathrm{f}^{\prime}(\mathrm{y})=11 \mathrm{f}(\mathrm{y})\) \(\mathrm{f}^{\prime}(3)=11 \mathrm{f}(3)\) \(\mathrm{f}^{\prime}(3)=33\)
116998
If \(n\) be any integer, then \(n(n+1)(2 n+1)\) is:
1 odd number
2 integral multiple of 6
3 perfect square
4 does not necessarily have any of the foregoing proof
Explanation:
B Given that, \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) For \(\mathrm{n}=1 \rightarrow\) \(1 \times 2 \times 3=6\) For \(\mathrm{n}=2 \rightarrow\) \(2 \times 3 \times 5=30\) \(6,30 \neq\) perfect square So, \(n(n+1)(2 n+1)\) always an integral multiple of even numbers. So, it is a integral multiple of \(6(6,30,84\)------)
Manipal UGET-2019
Sets, Relation and Function
116999
If \(\boldsymbol{f}(\mathrm{x})=\left\{\begin{array}{l}3 \mathrm{x}^2+12 x-1,-1 \leq x \leq 2 \\ 37-x, \quad 2 \leq x \leq 3\end{array}\right.\), then
1 \(f(\mathrm{x})\) is decreasing on \([-1,2]\)
2 \(f^{\prime}(2)\) does not exist
3 \(f(\mathrm{x})\) has the maximum value at \(\mathrm{x}=2\)
4 None of the above
Explanation:
B In the interval \([-1,2], \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}+12>0\) hence, \(f(x)\) is increasing in \([-1,2]\) Now, \(\mathrm{f}(\mathrm{x})\) being a polynomial in \(\mathrm{x}_2\) continuous in \(-1 \leq x\lt 2\) and in \(2\lt x \leq 3\) all check at \(x=2\) \(\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} 3(2-h)^2+12(2-h)-1\) \(=12+24-1=35\) \(\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 37-(2+h)=35\) \(f(2)=3(2)^2+12(2)+1=35\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) and hence in the interval [\(1,3]\) Now, \(\operatorname{Lf}^{\prime}(2)=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(2-\mathrm{h})-\mathrm{f}(2)}{-\mathrm{h}}\) \(= \lim _{h \rightarrow 0} \frac{3(2-h)^2+12(2-h)-1-35}{-h}\) \(= \lim _{h \rightarrow 0} \frac{3 h^2-24 h}{-h}=24\) \(R f^{\prime}(2) =\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}\) \(= \lim _{h \rightarrow 0} \frac{37-(2+h)-35}{h}=-1\) Since, \(\operatorname{Lf}(2) \neq R f^{\prime}(2)\), Thus \(f^{\prime}(2)\) does not exist
Manipal UGET-2017
Sets, Relation and Function
117000
If \(\log _3 2, \log _3\left(2^x-5\right)\) and \(\log _3\left(2^x-\frac{7}{2}\right)\) are in \(A P\), the value of \(x\) is
1 2
2 3
3 0
4 \(\frac{1}{3}\)
Explanation:
B Given, \(\log _3 2, \log _3\left(2^x-5\right), \log _3\left(2^x-\frac{7}{2}\right) \text { arein A.P. }\) \(\Rightarrow \quad \log _3\left(2^x-5\right)=\frac{\log _3\left(2^x-\frac{7}{2}\right)+\log _3 2}{2}\) \(\Rightarrow \quad \left(2^x\right)^2+25-10 \cdot 2^x=2 \cdot 2^x-7\) \(\Rightarrow \quad \left(2^x\right)^2-12 \cdot 2^x+25+7=0\) \(\Rightarrow \quad \left(2^x\right)^2-12 \cdot 2^x+32=0\) \(\Rightarrow \quad 2^x=8 \text { or } 2^x=4\) \(\Rightarrow \quad x=3 \text { or } x=2\) \(\Rightarrow \quad x=3(\because x=2 \text { does not satisfy the given series })\)
Manipal UGET-2017
Sets, Relation and Function
117001
If \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \cdot f(\mathrm{y}), f(3)=3, f^{\prime}(0)=11\), then \(f^{\prime}(3)\) is equal to
1 \(11 . \mathrm{e}^{33}\)
2 33
3 11
4 \(\log ^{33}\)
Explanation:
Given, \(f(x+y)=f(x) . f(y)\) So, differentiate with \(\mathrm{x}\). \(f^{\prime}(x+y)=f(y) \cdot f^{\prime}(x)\) Put, \(x=0\) So, \(\quad f^{\prime}(y)=f(y) . f^{\prime}(0)\) \(\mathrm{f}^{\prime}(0)=11\) So, \(\quad \mathrm{f}^{\prime}(\mathrm{y})=11 \mathrm{f}(\mathrm{y})\) Now, \(\quad \mathrm{f}^{\prime}(\mathrm{y})=11 \mathrm{f}(\mathrm{y})\) \(\mathrm{f}^{\prime}(3)=11 \mathrm{f}(3)\) \(\mathrm{f}^{\prime}(3)=33\)
116998
If \(n\) be any integer, then \(n(n+1)(2 n+1)\) is:
1 odd number
2 integral multiple of 6
3 perfect square
4 does not necessarily have any of the foregoing proof
Explanation:
B Given that, \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) For \(\mathrm{n}=1 \rightarrow\) \(1 \times 2 \times 3=6\) For \(\mathrm{n}=2 \rightarrow\) \(2 \times 3 \times 5=30\) \(6,30 \neq\) perfect square So, \(n(n+1)(2 n+1)\) always an integral multiple of even numbers. So, it is a integral multiple of \(6(6,30,84\)------)
Manipal UGET-2019
Sets, Relation and Function
116999
If \(\boldsymbol{f}(\mathrm{x})=\left\{\begin{array}{l}3 \mathrm{x}^2+12 x-1,-1 \leq x \leq 2 \\ 37-x, \quad 2 \leq x \leq 3\end{array}\right.\), then
1 \(f(\mathrm{x})\) is decreasing on \([-1,2]\)
2 \(f^{\prime}(2)\) does not exist
3 \(f(\mathrm{x})\) has the maximum value at \(\mathrm{x}=2\)
4 None of the above
Explanation:
B In the interval \([-1,2], \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}+12>0\) hence, \(f(x)\) is increasing in \([-1,2]\) Now, \(\mathrm{f}(\mathrm{x})\) being a polynomial in \(\mathrm{x}_2\) continuous in \(-1 \leq x\lt 2\) and in \(2\lt x \leq 3\) all check at \(x=2\) \(\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} 3(2-h)^2+12(2-h)-1\) \(=12+24-1=35\) \(\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 37-(2+h)=35\) \(f(2)=3(2)^2+12(2)+1=35\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) and hence in the interval [\(1,3]\) Now, \(\operatorname{Lf}^{\prime}(2)=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(2-\mathrm{h})-\mathrm{f}(2)}{-\mathrm{h}}\) \(= \lim _{h \rightarrow 0} \frac{3(2-h)^2+12(2-h)-1-35}{-h}\) \(= \lim _{h \rightarrow 0} \frac{3 h^2-24 h}{-h}=24\) \(R f^{\prime}(2) =\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}\) \(= \lim _{h \rightarrow 0} \frac{37-(2+h)-35}{h}=-1\) Since, \(\operatorname{Lf}(2) \neq R f^{\prime}(2)\), Thus \(f^{\prime}(2)\) does not exist
Manipal UGET-2017
Sets, Relation and Function
117000
If \(\log _3 2, \log _3\left(2^x-5\right)\) and \(\log _3\left(2^x-\frac{7}{2}\right)\) are in \(A P\), the value of \(x\) is
1 2
2 3
3 0
4 \(\frac{1}{3}\)
Explanation:
B Given, \(\log _3 2, \log _3\left(2^x-5\right), \log _3\left(2^x-\frac{7}{2}\right) \text { arein A.P. }\) \(\Rightarrow \quad \log _3\left(2^x-5\right)=\frac{\log _3\left(2^x-\frac{7}{2}\right)+\log _3 2}{2}\) \(\Rightarrow \quad \left(2^x\right)^2+25-10 \cdot 2^x=2 \cdot 2^x-7\) \(\Rightarrow \quad \left(2^x\right)^2-12 \cdot 2^x+25+7=0\) \(\Rightarrow \quad \left(2^x\right)^2-12 \cdot 2^x+32=0\) \(\Rightarrow \quad 2^x=8 \text { or } 2^x=4\) \(\Rightarrow \quad x=3 \text { or } x=2\) \(\Rightarrow \quad x=3(\because x=2 \text { does not satisfy the given series })\)
Manipal UGET-2017
Sets, Relation and Function
117001
If \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \cdot f(\mathrm{y}), f(3)=3, f^{\prime}(0)=11\), then \(f^{\prime}(3)\) is equal to
1 \(11 . \mathrm{e}^{33}\)
2 33
3 11
4 \(\log ^{33}\)
Explanation:
Given, \(f(x+y)=f(x) . f(y)\) So, differentiate with \(\mathrm{x}\). \(f^{\prime}(x+y)=f(y) \cdot f^{\prime}(x)\) Put, \(x=0\) So, \(\quad f^{\prime}(y)=f(y) . f^{\prime}(0)\) \(\mathrm{f}^{\prime}(0)=11\) So, \(\quad \mathrm{f}^{\prime}(\mathrm{y})=11 \mathrm{f}(\mathrm{y})\) Now, \(\quad \mathrm{f}^{\prime}(\mathrm{y})=11 \mathrm{f}(\mathrm{y})\) \(\mathrm{f}^{\prime}(3)=11 \mathrm{f}(3)\) \(\mathrm{f}^{\prime}(3)=33\)
