116919
The value of \([\sin x]+[1+\sin x]+[2+\sin x]\) in \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) can be ([.] is the greatest integer function) can be
1 0
2 1
3 2
4 3
Explanation:
A Given, \(\mathrm{x} \in\left[\pi, \frac{3 \pi}{2}\right]\) Then, the value of \([\sin x]+[1+\sin x]+[2+\sin x]=\) ? Where, \([\cdot]\) is the greatest integer function. Then, from \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) - \(-1 \leq \sin x \leq 0\) So, \([1+\sin x]=0\) and \([\sin x]=-1\) Hence, \({[\sin x]+[1+\sin x]+[2+\sin x]=-1+0+2+[\sin x]}\) \(=-1+0+2-1\) \(=-2+2\) \({[\sin \mathrm{x}]+[1+\sin \mathrm{x}]+[2+\sin \mathrm{x}]=0}\)
BCECE-2018
Sets, Relation and Function
116920
The function \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is
1 a periodic function with period \(2 \pi\)
2 a periodic function with period \(\frac{2 \pi}{5}\)
3 a periodic function with period \(\frac{2 \pi}{\sqrt{5}}\)
4 not a periodic function
Explanation:
D Given, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) Differentiate on both side with respect to \(x\), we get- \(\mathrm{f}^{\prime}(\mathrm{x})=-2 \sin 5 \mathrm{x} \times 5+3 \cos \sqrt{5} \mathrm{x} \times \sqrt{5}\) \(\mathrm{f}^{\prime}(\mathrm{x})=-10 \sin 5 \mathrm{x}+3 \sqrt{5} \cos \sqrt{5} \mathrm{x}\) Then, \(2 \cos 5 \mathrm{x}, 3 \sin \sqrt{5} \mathrm{x}\) are periodic function with periods \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\). But \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\) have no common multiple. So, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is not periodic function.
BCECE-2018
Sets, Relation and Function
116921
If \(a, b, c\) are positive real numbers, then \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}=\)
1 0
2 1
3 2
4 3
Explanation:
C Given, \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are positive real number, Then, \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}\) \(\frac{1}{\frac{\log a b c}{\log a b}} +\frac{1}{\frac{\log a b c}{\log b c}}+\frac{1}{\frac{\log a b c}{\log c a}} \quad\left(\because \log _b a=\frac{\log a}{\log b}\right)\) \(=\frac{\log a b}{\log a b c}+\frac{\log b c}{\log a b c}+\frac{\log c a}{\log a b c}\) \(=\frac{\log a b+\log b c+\log c a}{\log a b c}\) \(=\frac{\log (a b \times b c \times c a)}{\log a b c}\) \(=\frac{\log \left(a^2 b^2 c^2\right)}{\log a b c}\) \(=\frac{\log (a b c)^2}{\log (a b c)}\) \(=\frac{2 \log (a b c)}{\log (a b c)}\) \(=2\)
BCECE-2017
Sets, Relation and Function
116922
The number of functions \(f:\{1,2,3,4\} \rightarrow\{a \in\) \(\mathrm{Z}:|\mathrm{a}| \leq 8\}\) satisfying \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1, \forall \mathrm{n} \in\) \(\{1,2,3\}\) is
1 2
2 1
3 4
4 3
Explanation:
A Given, \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1\) \(\mathrm{n} \cdot \mathrm{f}(\mathrm{n})+\mathrm{f}(\mathrm{n}+1)=1\) When \(\mathrm{n}=1\) \(\mathrm{f}(1)+\mathrm{f}(2)=1\) When \(\mathrm{n}=2\) \(2 \mathrm{f}(2)+\mathrm{f}(3)=2\) When \(\mathrm{n}=3\) \(3 f(3)+f(4)=3\) Now, multiple by 2 in equation (i), we get - \(2 \mathrm{f}(1)+2 \mathrm{f}(2)=2\) On subtracting equation (iv) from (ii), we get - \(\mathrm{f}(3)-2 \mathrm{f}(1)=0\) \(\mathrm{f}(3)=2 \mathrm{f}(1)\) Now, putting the value in equation (iii), we get- \(3[2 f(1)]+f(4)=3\) \(6 f(1)+f(4)=3\) \(f(4)=3-6 f(1)\) Therefore, \(-8 \leq \mathrm{f}(4) \leq 8\) \(-8 \leq 3-6 \mathrm{f}(1) \leq 8\) \(-11 \leq-6 \mathrm{f}(1) \leq 5\) \(-\frac{5}{6} \leq \mathrm{f}(1) \leq \frac{11}{6}\) \(\mathrm{f}(1)=0,1\) \(\mathrm{f}(1)=0, \quad \mathrm{f}(2)=1\) \(\mathrm{f}(3)=0, \mathrm{f}(4)=3\) \(\mathrm{f}(1)=1, \mathrm{f}(2)=0\) \(\mathrm{f}(3)=2, \mathrm{f}(4)=-3\) Case - I : There can be 2 function such that like this.
Shift-II
Sets, Relation and Function
116924
If the periods of the functions \(\sin (a x+b)\) and \(\tan (c x+d)\) are respectively \(\frac{4}{7}\) and \(\frac{2}{5}\), then \(\boldsymbol{\operatorname { s i n }}(|\mathbf{a}|+|\mathbf{c}|)+\cos (|\mathbf{a}|-|\mathbf{c}|)=\)
1 -1
2 0
3 1
4 2
Explanation:
A We know that, Period of \(\sin (a x+b)=\left|\frac{2 \pi}{a}\right|\) Period of \(\tan (a x+b)=\frac{\pi}{|a|}\) \(\therefore \frac{2 \pi}{|\mathrm{a}|}=\frac{4}{7}\) \(\frac{\pi}{|\mathrm{c}|}=\frac{2}{5}\) \(|\mathrm{a}|=\frac{7 \pi}{2}\) and \(|\mathrm{c}|=\frac{5 \pi}{2}\) \(\therefore \sin (|\mathrm{a}|+|\mathrm{c}|)+\cos (|\mathrm{a}|-|\mathrm{c}|)=\sin \left(\frac{7 \pi}{2}\right. \left.+\frac{5 \pi}{2}\right)\) \(+\cos \left(\frac{7 \pi}{2}-\frac{5 \pi}{2}\right)\) \(=\sin (6 \pi)+\cos (\pi)=0+(-1)=-1\)
116919
The value of \([\sin x]+[1+\sin x]+[2+\sin x]\) in \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) can be ([.] is the greatest integer function) can be
1 0
2 1
3 2
4 3
Explanation:
A Given, \(\mathrm{x} \in\left[\pi, \frac{3 \pi}{2}\right]\) Then, the value of \([\sin x]+[1+\sin x]+[2+\sin x]=\) ? Where, \([\cdot]\) is the greatest integer function. Then, from \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) - \(-1 \leq \sin x \leq 0\) So, \([1+\sin x]=0\) and \([\sin x]=-1\) Hence, \({[\sin x]+[1+\sin x]+[2+\sin x]=-1+0+2+[\sin x]}\) \(=-1+0+2-1\) \(=-2+2\) \({[\sin \mathrm{x}]+[1+\sin \mathrm{x}]+[2+\sin \mathrm{x}]=0}\)
BCECE-2018
Sets, Relation and Function
116920
The function \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is
1 a periodic function with period \(2 \pi\)
2 a periodic function with period \(\frac{2 \pi}{5}\)
3 a periodic function with period \(\frac{2 \pi}{\sqrt{5}}\)
4 not a periodic function
Explanation:
D Given, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) Differentiate on both side with respect to \(x\), we get- \(\mathrm{f}^{\prime}(\mathrm{x})=-2 \sin 5 \mathrm{x} \times 5+3 \cos \sqrt{5} \mathrm{x} \times \sqrt{5}\) \(\mathrm{f}^{\prime}(\mathrm{x})=-10 \sin 5 \mathrm{x}+3 \sqrt{5} \cos \sqrt{5} \mathrm{x}\) Then, \(2 \cos 5 \mathrm{x}, 3 \sin \sqrt{5} \mathrm{x}\) are periodic function with periods \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\). But \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\) have no common multiple. So, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is not periodic function.
BCECE-2018
Sets, Relation and Function
116921
If \(a, b, c\) are positive real numbers, then \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}=\)
1 0
2 1
3 2
4 3
Explanation:
C Given, \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are positive real number, Then, \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}\) \(\frac{1}{\frac{\log a b c}{\log a b}} +\frac{1}{\frac{\log a b c}{\log b c}}+\frac{1}{\frac{\log a b c}{\log c a}} \quad\left(\because \log _b a=\frac{\log a}{\log b}\right)\) \(=\frac{\log a b}{\log a b c}+\frac{\log b c}{\log a b c}+\frac{\log c a}{\log a b c}\) \(=\frac{\log a b+\log b c+\log c a}{\log a b c}\) \(=\frac{\log (a b \times b c \times c a)}{\log a b c}\) \(=\frac{\log \left(a^2 b^2 c^2\right)}{\log a b c}\) \(=\frac{\log (a b c)^2}{\log (a b c)}\) \(=\frac{2 \log (a b c)}{\log (a b c)}\) \(=2\)
BCECE-2017
Sets, Relation and Function
116922
The number of functions \(f:\{1,2,3,4\} \rightarrow\{a \in\) \(\mathrm{Z}:|\mathrm{a}| \leq 8\}\) satisfying \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1, \forall \mathrm{n} \in\) \(\{1,2,3\}\) is
1 2
2 1
3 4
4 3
Explanation:
A Given, \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1\) \(\mathrm{n} \cdot \mathrm{f}(\mathrm{n})+\mathrm{f}(\mathrm{n}+1)=1\) When \(\mathrm{n}=1\) \(\mathrm{f}(1)+\mathrm{f}(2)=1\) When \(\mathrm{n}=2\) \(2 \mathrm{f}(2)+\mathrm{f}(3)=2\) When \(\mathrm{n}=3\) \(3 f(3)+f(4)=3\) Now, multiple by 2 in equation (i), we get - \(2 \mathrm{f}(1)+2 \mathrm{f}(2)=2\) On subtracting equation (iv) from (ii), we get - \(\mathrm{f}(3)-2 \mathrm{f}(1)=0\) \(\mathrm{f}(3)=2 \mathrm{f}(1)\) Now, putting the value in equation (iii), we get- \(3[2 f(1)]+f(4)=3\) \(6 f(1)+f(4)=3\) \(f(4)=3-6 f(1)\) Therefore, \(-8 \leq \mathrm{f}(4) \leq 8\) \(-8 \leq 3-6 \mathrm{f}(1) \leq 8\) \(-11 \leq-6 \mathrm{f}(1) \leq 5\) \(-\frac{5}{6} \leq \mathrm{f}(1) \leq \frac{11}{6}\) \(\mathrm{f}(1)=0,1\) \(\mathrm{f}(1)=0, \quad \mathrm{f}(2)=1\) \(\mathrm{f}(3)=0, \mathrm{f}(4)=3\) \(\mathrm{f}(1)=1, \mathrm{f}(2)=0\) \(\mathrm{f}(3)=2, \mathrm{f}(4)=-3\) Case - I : There can be 2 function such that like this.
Shift-II
Sets, Relation and Function
116924
If the periods of the functions \(\sin (a x+b)\) and \(\tan (c x+d)\) are respectively \(\frac{4}{7}\) and \(\frac{2}{5}\), then \(\boldsymbol{\operatorname { s i n }}(|\mathbf{a}|+|\mathbf{c}|)+\cos (|\mathbf{a}|-|\mathbf{c}|)=\)
1 -1
2 0
3 1
4 2
Explanation:
A We know that, Period of \(\sin (a x+b)=\left|\frac{2 \pi}{a}\right|\) Period of \(\tan (a x+b)=\frac{\pi}{|a|}\) \(\therefore \frac{2 \pi}{|\mathrm{a}|}=\frac{4}{7}\) \(\frac{\pi}{|\mathrm{c}|}=\frac{2}{5}\) \(|\mathrm{a}|=\frac{7 \pi}{2}\) and \(|\mathrm{c}|=\frac{5 \pi}{2}\) \(\therefore \sin (|\mathrm{a}|+|\mathrm{c}|)+\cos (|\mathrm{a}|-|\mathrm{c}|)=\sin \left(\frac{7 \pi}{2}\right. \left.+\frac{5 \pi}{2}\right)\) \(+\cos \left(\frac{7 \pi}{2}-\frac{5 \pi}{2}\right)\) \(=\sin (6 \pi)+\cos (\pi)=0+(-1)=-1\)
116919
The value of \([\sin x]+[1+\sin x]+[2+\sin x]\) in \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) can be ([.] is the greatest integer function) can be
1 0
2 1
3 2
4 3
Explanation:
A Given, \(\mathrm{x} \in\left[\pi, \frac{3 \pi}{2}\right]\) Then, the value of \([\sin x]+[1+\sin x]+[2+\sin x]=\) ? Where, \([\cdot]\) is the greatest integer function. Then, from \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) - \(-1 \leq \sin x \leq 0\) So, \([1+\sin x]=0\) and \([\sin x]=-1\) Hence, \({[\sin x]+[1+\sin x]+[2+\sin x]=-1+0+2+[\sin x]}\) \(=-1+0+2-1\) \(=-2+2\) \({[\sin \mathrm{x}]+[1+\sin \mathrm{x}]+[2+\sin \mathrm{x}]=0}\)
BCECE-2018
Sets, Relation and Function
116920
The function \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is
1 a periodic function with period \(2 \pi\)
2 a periodic function with period \(\frac{2 \pi}{5}\)
3 a periodic function with period \(\frac{2 \pi}{\sqrt{5}}\)
4 not a periodic function
Explanation:
D Given, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) Differentiate on both side with respect to \(x\), we get- \(\mathrm{f}^{\prime}(\mathrm{x})=-2 \sin 5 \mathrm{x} \times 5+3 \cos \sqrt{5} \mathrm{x} \times \sqrt{5}\) \(\mathrm{f}^{\prime}(\mathrm{x})=-10 \sin 5 \mathrm{x}+3 \sqrt{5} \cos \sqrt{5} \mathrm{x}\) Then, \(2 \cos 5 \mathrm{x}, 3 \sin \sqrt{5} \mathrm{x}\) are periodic function with periods \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\). But \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\) have no common multiple. So, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is not periodic function.
BCECE-2018
Sets, Relation and Function
116921
If \(a, b, c\) are positive real numbers, then \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}=\)
1 0
2 1
3 2
4 3
Explanation:
C Given, \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are positive real number, Then, \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}\) \(\frac{1}{\frac{\log a b c}{\log a b}} +\frac{1}{\frac{\log a b c}{\log b c}}+\frac{1}{\frac{\log a b c}{\log c a}} \quad\left(\because \log _b a=\frac{\log a}{\log b}\right)\) \(=\frac{\log a b}{\log a b c}+\frac{\log b c}{\log a b c}+\frac{\log c a}{\log a b c}\) \(=\frac{\log a b+\log b c+\log c a}{\log a b c}\) \(=\frac{\log (a b \times b c \times c a)}{\log a b c}\) \(=\frac{\log \left(a^2 b^2 c^2\right)}{\log a b c}\) \(=\frac{\log (a b c)^2}{\log (a b c)}\) \(=\frac{2 \log (a b c)}{\log (a b c)}\) \(=2\)
BCECE-2017
Sets, Relation and Function
116922
The number of functions \(f:\{1,2,3,4\} \rightarrow\{a \in\) \(\mathrm{Z}:|\mathrm{a}| \leq 8\}\) satisfying \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1, \forall \mathrm{n} \in\) \(\{1,2,3\}\) is
1 2
2 1
3 4
4 3
Explanation:
A Given, \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1\) \(\mathrm{n} \cdot \mathrm{f}(\mathrm{n})+\mathrm{f}(\mathrm{n}+1)=1\) When \(\mathrm{n}=1\) \(\mathrm{f}(1)+\mathrm{f}(2)=1\) When \(\mathrm{n}=2\) \(2 \mathrm{f}(2)+\mathrm{f}(3)=2\) When \(\mathrm{n}=3\) \(3 f(3)+f(4)=3\) Now, multiple by 2 in equation (i), we get - \(2 \mathrm{f}(1)+2 \mathrm{f}(2)=2\) On subtracting equation (iv) from (ii), we get - \(\mathrm{f}(3)-2 \mathrm{f}(1)=0\) \(\mathrm{f}(3)=2 \mathrm{f}(1)\) Now, putting the value in equation (iii), we get- \(3[2 f(1)]+f(4)=3\) \(6 f(1)+f(4)=3\) \(f(4)=3-6 f(1)\) Therefore, \(-8 \leq \mathrm{f}(4) \leq 8\) \(-8 \leq 3-6 \mathrm{f}(1) \leq 8\) \(-11 \leq-6 \mathrm{f}(1) \leq 5\) \(-\frac{5}{6} \leq \mathrm{f}(1) \leq \frac{11}{6}\) \(\mathrm{f}(1)=0,1\) \(\mathrm{f}(1)=0, \quad \mathrm{f}(2)=1\) \(\mathrm{f}(3)=0, \mathrm{f}(4)=3\) \(\mathrm{f}(1)=1, \mathrm{f}(2)=0\) \(\mathrm{f}(3)=2, \mathrm{f}(4)=-3\) Case - I : There can be 2 function such that like this.
Shift-II
Sets, Relation and Function
116924
If the periods of the functions \(\sin (a x+b)\) and \(\tan (c x+d)\) are respectively \(\frac{4}{7}\) and \(\frac{2}{5}\), then \(\boldsymbol{\operatorname { s i n }}(|\mathbf{a}|+|\mathbf{c}|)+\cos (|\mathbf{a}|-|\mathbf{c}|)=\)
1 -1
2 0
3 1
4 2
Explanation:
A We know that, Period of \(\sin (a x+b)=\left|\frac{2 \pi}{a}\right|\) Period of \(\tan (a x+b)=\frac{\pi}{|a|}\) \(\therefore \frac{2 \pi}{|\mathrm{a}|}=\frac{4}{7}\) \(\frac{\pi}{|\mathrm{c}|}=\frac{2}{5}\) \(|\mathrm{a}|=\frac{7 \pi}{2}\) and \(|\mathrm{c}|=\frac{5 \pi}{2}\) \(\therefore \sin (|\mathrm{a}|+|\mathrm{c}|)+\cos (|\mathrm{a}|-|\mathrm{c}|)=\sin \left(\frac{7 \pi}{2}\right. \left.+\frac{5 \pi}{2}\right)\) \(+\cos \left(\frac{7 \pi}{2}-\frac{5 \pi}{2}\right)\) \(=\sin (6 \pi)+\cos (\pi)=0+(-1)=-1\)
116919
The value of \([\sin x]+[1+\sin x]+[2+\sin x]\) in \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) can be ([.] is the greatest integer function) can be
1 0
2 1
3 2
4 3
Explanation:
A Given, \(\mathrm{x} \in\left[\pi, \frac{3 \pi}{2}\right]\) Then, the value of \([\sin x]+[1+\sin x]+[2+\sin x]=\) ? Where, \([\cdot]\) is the greatest integer function. Then, from \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) - \(-1 \leq \sin x \leq 0\) So, \([1+\sin x]=0\) and \([\sin x]=-1\) Hence, \({[\sin x]+[1+\sin x]+[2+\sin x]=-1+0+2+[\sin x]}\) \(=-1+0+2-1\) \(=-2+2\) \({[\sin \mathrm{x}]+[1+\sin \mathrm{x}]+[2+\sin \mathrm{x}]=0}\)
BCECE-2018
Sets, Relation and Function
116920
The function \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is
1 a periodic function with period \(2 \pi\)
2 a periodic function with period \(\frac{2 \pi}{5}\)
3 a periodic function with period \(\frac{2 \pi}{\sqrt{5}}\)
4 not a periodic function
Explanation:
D Given, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) Differentiate on both side with respect to \(x\), we get- \(\mathrm{f}^{\prime}(\mathrm{x})=-2 \sin 5 \mathrm{x} \times 5+3 \cos \sqrt{5} \mathrm{x} \times \sqrt{5}\) \(\mathrm{f}^{\prime}(\mathrm{x})=-10 \sin 5 \mathrm{x}+3 \sqrt{5} \cos \sqrt{5} \mathrm{x}\) Then, \(2 \cos 5 \mathrm{x}, 3 \sin \sqrt{5} \mathrm{x}\) are periodic function with periods \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\). But \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\) have no common multiple. So, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is not periodic function.
BCECE-2018
Sets, Relation and Function
116921
If \(a, b, c\) are positive real numbers, then \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}=\)
1 0
2 1
3 2
4 3
Explanation:
C Given, \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are positive real number, Then, \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}\) \(\frac{1}{\frac{\log a b c}{\log a b}} +\frac{1}{\frac{\log a b c}{\log b c}}+\frac{1}{\frac{\log a b c}{\log c a}} \quad\left(\because \log _b a=\frac{\log a}{\log b}\right)\) \(=\frac{\log a b}{\log a b c}+\frac{\log b c}{\log a b c}+\frac{\log c a}{\log a b c}\) \(=\frac{\log a b+\log b c+\log c a}{\log a b c}\) \(=\frac{\log (a b \times b c \times c a)}{\log a b c}\) \(=\frac{\log \left(a^2 b^2 c^2\right)}{\log a b c}\) \(=\frac{\log (a b c)^2}{\log (a b c)}\) \(=\frac{2 \log (a b c)}{\log (a b c)}\) \(=2\)
BCECE-2017
Sets, Relation and Function
116922
The number of functions \(f:\{1,2,3,4\} \rightarrow\{a \in\) \(\mathrm{Z}:|\mathrm{a}| \leq 8\}\) satisfying \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1, \forall \mathrm{n} \in\) \(\{1,2,3\}\) is
1 2
2 1
3 4
4 3
Explanation:
A Given, \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1\) \(\mathrm{n} \cdot \mathrm{f}(\mathrm{n})+\mathrm{f}(\mathrm{n}+1)=1\) When \(\mathrm{n}=1\) \(\mathrm{f}(1)+\mathrm{f}(2)=1\) When \(\mathrm{n}=2\) \(2 \mathrm{f}(2)+\mathrm{f}(3)=2\) When \(\mathrm{n}=3\) \(3 f(3)+f(4)=3\) Now, multiple by 2 in equation (i), we get - \(2 \mathrm{f}(1)+2 \mathrm{f}(2)=2\) On subtracting equation (iv) from (ii), we get - \(\mathrm{f}(3)-2 \mathrm{f}(1)=0\) \(\mathrm{f}(3)=2 \mathrm{f}(1)\) Now, putting the value in equation (iii), we get- \(3[2 f(1)]+f(4)=3\) \(6 f(1)+f(4)=3\) \(f(4)=3-6 f(1)\) Therefore, \(-8 \leq \mathrm{f}(4) \leq 8\) \(-8 \leq 3-6 \mathrm{f}(1) \leq 8\) \(-11 \leq-6 \mathrm{f}(1) \leq 5\) \(-\frac{5}{6} \leq \mathrm{f}(1) \leq \frac{11}{6}\) \(\mathrm{f}(1)=0,1\) \(\mathrm{f}(1)=0, \quad \mathrm{f}(2)=1\) \(\mathrm{f}(3)=0, \mathrm{f}(4)=3\) \(\mathrm{f}(1)=1, \mathrm{f}(2)=0\) \(\mathrm{f}(3)=2, \mathrm{f}(4)=-3\) Case - I : There can be 2 function such that like this.
Shift-II
Sets, Relation and Function
116924
If the periods of the functions \(\sin (a x+b)\) and \(\tan (c x+d)\) are respectively \(\frac{4}{7}\) and \(\frac{2}{5}\), then \(\boldsymbol{\operatorname { s i n }}(|\mathbf{a}|+|\mathbf{c}|)+\cos (|\mathbf{a}|-|\mathbf{c}|)=\)
1 -1
2 0
3 1
4 2
Explanation:
A We know that, Period of \(\sin (a x+b)=\left|\frac{2 \pi}{a}\right|\) Period of \(\tan (a x+b)=\frac{\pi}{|a|}\) \(\therefore \frac{2 \pi}{|\mathrm{a}|}=\frac{4}{7}\) \(\frac{\pi}{|\mathrm{c}|}=\frac{2}{5}\) \(|\mathrm{a}|=\frac{7 \pi}{2}\) and \(|\mathrm{c}|=\frac{5 \pi}{2}\) \(\therefore \sin (|\mathrm{a}|+|\mathrm{c}|)+\cos (|\mathrm{a}|-|\mathrm{c}|)=\sin \left(\frac{7 \pi}{2}\right. \left.+\frac{5 \pi}{2}\right)\) \(+\cos \left(\frac{7 \pi}{2}-\frac{5 \pi}{2}\right)\) \(=\sin (6 \pi)+\cos (\pi)=0+(-1)=-1\)
116919
The value of \([\sin x]+[1+\sin x]+[2+\sin x]\) in \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) can be ([.] is the greatest integer function) can be
1 0
2 1
3 2
4 3
Explanation:
A Given, \(\mathrm{x} \in\left[\pi, \frac{3 \pi}{2}\right]\) Then, the value of \([\sin x]+[1+\sin x]+[2+\sin x]=\) ? Where, \([\cdot]\) is the greatest integer function. Then, from \(x \in\left[\pi, \frac{3 \pi}{2}\right]\) - \(-1 \leq \sin x \leq 0\) So, \([1+\sin x]=0\) and \([\sin x]=-1\) Hence, \({[\sin x]+[1+\sin x]+[2+\sin x]=-1+0+2+[\sin x]}\) \(=-1+0+2-1\) \(=-2+2\) \({[\sin \mathrm{x}]+[1+\sin \mathrm{x}]+[2+\sin \mathrm{x}]=0}\)
BCECE-2018
Sets, Relation and Function
116920
The function \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is
1 a periodic function with period \(2 \pi\)
2 a periodic function with period \(\frac{2 \pi}{5}\)
3 a periodic function with period \(\frac{2 \pi}{\sqrt{5}}\)
4 not a periodic function
Explanation:
D Given, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) Differentiate on both side with respect to \(x\), we get- \(\mathrm{f}^{\prime}(\mathrm{x})=-2 \sin 5 \mathrm{x} \times 5+3 \cos \sqrt{5} \mathrm{x} \times \sqrt{5}\) \(\mathrm{f}^{\prime}(\mathrm{x})=-10 \sin 5 \mathrm{x}+3 \sqrt{5} \cos \sqrt{5} \mathrm{x}\) Then, \(2 \cos 5 \mathrm{x}, 3 \sin \sqrt{5} \mathrm{x}\) are periodic function with periods \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\). But \(\frac{2 \pi}{5}\) and \(\frac{2 \pi}{\sqrt{5}}\) have no common multiple. So, \(f(x)=2 \cos 5 x+3 \sin \sqrt{5} x\) is not periodic function.
BCECE-2018
Sets, Relation and Function
116921
If \(a, b, c\) are positive real numbers, then \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}=\)
1 0
2 1
3 2
4 3
Explanation:
C Given, \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are positive real number, Then, \(\frac{1}{\log _{a b} a b c}+\frac{1}{\log _{b c} a b c}+\frac{1}{\log _{c a} a b c}\) \(\frac{1}{\frac{\log a b c}{\log a b}} +\frac{1}{\frac{\log a b c}{\log b c}}+\frac{1}{\frac{\log a b c}{\log c a}} \quad\left(\because \log _b a=\frac{\log a}{\log b}\right)\) \(=\frac{\log a b}{\log a b c}+\frac{\log b c}{\log a b c}+\frac{\log c a}{\log a b c}\) \(=\frac{\log a b+\log b c+\log c a}{\log a b c}\) \(=\frac{\log (a b \times b c \times c a)}{\log a b c}\) \(=\frac{\log \left(a^2 b^2 c^2\right)}{\log a b c}\) \(=\frac{\log (a b c)^2}{\log (a b c)}\) \(=\frac{2 \log (a b c)}{\log (a b c)}\) \(=2\)
BCECE-2017
Sets, Relation and Function
116922
The number of functions \(f:\{1,2,3,4\} \rightarrow\{a \in\) \(\mathrm{Z}:|\mathrm{a}| \leq 8\}\) satisfying \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1, \forall \mathrm{n} \in\) \(\{1,2,3\}\) is
1 2
2 1
3 4
4 3
Explanation:
A Given, \(\mathrm{f}(\mathrm{n})+\frac{1}{\mathrm{n}} \mathrm{f}(\mathrm{n}+1)=1\) \(\mathrm{n} \cdot \mathrm{f}(\mathrm{n})+\mathrm{f}(\mathrm{n}+1)=1\) When \(\mathrm{n}=1\) \(\mathrm{f}(1)+\mathrm{f}(2)=1\) When \(\mathrm{n}=2\) \(2 \mathrm{f}(2)+\mathrm{f}(3)=2\) When \(\mathrm{n}=3\) \(3 f(3)+f(4)=3\) Now, multiple by 2 in equation (i), we get - \(2 \mathrm{f}(1)+2 \mathrm{f}(2)=2\) On subtracting equation (iv) from (ii), we get - \(\mathrm{f}(3)-2 \mathrm{f}(1)=0\) \(\mathrm{f}(3)=2 \mathrm{f}(1)\) Now, putting the value in equation (iii), we get- \(3[2 f(1)]+f(4)=3\) \(6 f(1)+f(4)=3\) \(f(4)=3-6 f(1)\) Therefore, \(-8 \leq \mathrm{f}(4) \leq 8\) \(-8 \leq 3-6 \mathrm{f}(1) \leq 8\) \(-11 \leq-6 \mathrm{f}(1) \leq 5\) \(-\frac{5}{6} \leq \mathrm{f}(1) \leq \frac{11}{6}\) \(\mathrm{f}(1)=0,1\) \(\mathrm{f}(1)=0, \quad \mathrm{f}(2)=1\) \(\mathrm{f}(3)=0, \mathrm{f}(4)=3\) \(\mathrm{f}(1)=1, \mathrm{f}(2)=0\) \(\mathrm{f}(3)=2, \mathrm{f}(4)=-3\) Case - I : There can be 2 function such that like this.
Shift-II
Sets, Relation and Function
116924
If the periods of the functions \(\sin (a x+b)\) and \(\tan (c x+d)\) are respectively \(\frac{4}{7}\) and \(\frac{2}{5}\), then \(\boldsymbol{\operatorname { s i n }}(|\mathbf{a}|+|\mathbf{c}|)+\cos (|\mathbf{a}|-|\mathbf{c}|)=\)
1 -1
2 0
3 1
4 2
Explanation:
A We know that, Period of \(\sin (a x+b)=\left|\frac{2 \pi}{a}\right|\) Period of \(\tan (a x+b)=\frac{\pi}{|a|}\) \(\therefore \frac{2 \pi}{|\mathrm{a}|}=\frac{4}{7}\) \(\frac{\pi}{|\mathrm{c}|}=\frac{2}{5}\) \(|\mathrm{a}|=\frac{7 \pi}{2}\) and \(|\mathrm{c}|=\frac{5 \pi}{2}\) \(\therefore \sin (|\mathrm{a}|+|\mathrm{c}|)+\cos (|\mathrm{a}|-|\mathrm{c}|)=\sin \left(\frac{7 \pi}{2}\right. \left.+\frac{5 \pi}{2}\right)\) \(+\cos \left(\frac{7 \pi}{2}-\frac{5 \pi}{2}\right)\) \(=\sin (6 \pi)+\cos (\pi)=0+(-1)=-1\)
