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Algebra: Basics Test-5

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Question 1
$ \displaystyle f\,\,\,\frac{1}{4}\times \frac{2}{6}\times \frac{3}{8}\times \frac{4}{10}\times \frac{5}{12}\times ..........\times \frac{31}{64}=\frac{1}{{{2}^{t}}},$
the value of t is
A
38
B
40
C
36
D
58
Question 1 Explanation: 
$ \displaystyle \begin{array}{l}=\,\,\,\frac{1}{4}\times \frac{2}{6}\times \frac{3}{8}\times \frac{4}{10}\times \frac{5}{12}\times ..........\times \frac{31}{64}=\frac{1}{{{2}^{p}}}\\\Rightarrow \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times .....to\,\,30\,\,terms\,\,\times \frac{1}{64}=\frac{1}{{{2}^{p}}}\\\Rightarrow \frac{1}{{{2}^{30}}}\times \frac{1}{{{2}^{6}}}=\frac{1}{{{2}^{p}}}\\\Rightarrow \frac{1}{{{2}^{36}}}=\frac{1}{{{2}^{p}}}\\\Rightarrow p=36\end{array}$
Question 2
The value of is $ \displaystyle \frac{{{\left( 243 \right)}^{\frac{k}{5}}}{{.3}^{2k+1}}}{{{9}^{k}}{{.3}^{k-1}}}$
A
11
B
9
C
7
D
3n
Question 2 Explanation: 
$ \displaystyle \begin{array}{l}=\frac{{{\left( 243 \right)}^{\frac{k}{5}}}\times {{3}^{2k+1}}}{{{9}^{k}}\times {{3}^{k-1}}}\\=\frac{{{\left( {{3}^{5}} \right)}^{\frac{k}{5}}}\times {{3}^{2k+1}}}{{{\left( {{3}^{2}} \right)}^{k}}\times {{3}^{k-1}}}\\=\frac{{{3}^{k}}\times {{3}^{2k+1}}}{{{3}^{2k}}\times {{3}^{k-1}}}\\=\frac{{{3}^{k+2k+1}}}{{{3}^{2k+k-1}}}\\=\frac{{{3}^{3k+1}}}{{{3}^{3k-1}}}\\={{3}^{3k+1-3k+1}}\\={{3}^{2}}\\=9\end{array}$
Question 3
$ \displaystyle If\,\,a=0.5\,\,\,and\,\,b=0.2,\,\,\,then\,\,\,\,value\,\,\,of\,\,\,\sqrt{0.6}\,\,\times \,\,{{\left( 3b \right)}^{a}}\,\,\,is\,\,\,equal\,\,\,to$
A
0.9
B
0.6
C
0.5
D
1.3
Question 3 Explanation: 
$ \displaystyle \begin{array}{l}a=0.5\,\,\,and\,\,b=0.2\left( Given \right)\\Therefore\,\,\sqrt{0.6}\times {{\left( 3b \right)}^{a}}\\=\sqrt{0.6}\times {{\left( 3\times 0.2 \right)}^{0.5}}\\=\sqrt{0.6}\times {{\left( 0.6 \right)}^{\frac{1}{2}}}\\=\sqrt{0.6\times .06}=0.6\end{array}$
Question 4
$ \displaystyle if\,\,\,{{s}^{s\sqrt{s}}}={{\left( s\sqrt{s} \right)}^{s}},\,\,then\,\,s\,\,equals$
A
$ \displaystyle \frac{2}{9}$
B
$ \displaystyle \frac{1}{3}$
C
$ \displaystyle \frac{9}{4}$
D
$ \displaystyle \frac{3}{2}$
Question 4 Explanation: 
$ \displaystyle \begin{array}{l}{{s}^{s\sqrt{s}}}={{\left( s\sqrt{s} \right)}^{s}}\\\Rightarrow {{s}^{s.s\frac{1}{2}}}={{\left( s\times {{s}^{\frac{1}{2}}} \right)}^{s}}\\\Rightarrow {{s}^{1+\frac{1}{2}}}={{\left( {{s}^{1+\frac{1}{2}}} \right)}^{s}}\\{{s}^{{{s}^{\frac{3}{2}}}}}={{\left( {{s}^{\frac{3}{2}}} \right)}^{s}}={{s}^{\frac{3s}{2}}}\\\Rightarrow {{s}^{\frac{3}{2}}}=\frac{3s}{2}\Rightarrow {{s}^{\frac{3}{2}}}-\frac{3s}{2}=0\\\Rightarrow s\left( {{s}^{\frac{1}{2}}}-\frac{3}{2} \right)=0\\\Rightarrow s=0\,\,\,or\,\,\,{{s}^{\frac{1}{2}}}=\frac{3}{2}\\\Rightarrow s={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\\s=0\,\,given\,\,\,in\det er\min ate\,\,\,\,value.\\Therefore\,\,\,s=\frac{9}{4}\end{array}$
Question 5
If f =7, g=5 and h=3, then the value of f2+ g2+ h2-fg-gh-hf is:
A
14
B
12
C
-12
D
13
Question 5 Explanation: 
$ \displaystyle \begin{array}{l}{{f}^{2}}+{{g}^{2}}+{{h}^{2}}-fg-gh-hf\\=\frac{1}{2}\left[ {{\left( f-g \right)}^{2}}+{{\left( g-h \right)}^{2}}+{{\left( h-f \right)}^{2}} \right]\\=\frac{1}{2}\left[ {{\left( 7-5 \right)}^{2}}+{{\left( 5-3 \right)}^{2}}+{{\left( 3-7 \right)}^{2}} \right]\\=\frac{1}{2}\left( 4+4+16 \right)\\=\frac{1}{2}\times 24=12\end{array}$
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