Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If m is the A.M. of two distinct real numbers l and n $$(l,n > 1)$$ and $${G_1},{G_2}$$ and $${G_3}$$ are three geometric means between $$l$$ and n, then $$G_1^4\, + 2G_2^4\, + G_3^4$$ equals:

A

$$4\,lm{n^2}$$

B

$$4\,{l^2}{m^2}{n^2}$$

C

$$4\,{l^2}m\,n$$

D

$$4\,l\,{m^2}n$$

$$m = {{l + n} \over 2}$$ and common ratio of

$$G.P.$$ $$ = r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$$

$$\therefore$$ $${G_1} = {l^{3/4}}\,{n^{1/4}},$$ $${G_2} = {l^{1/2}}{n^{1/2}},\,$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$$

$$G_1^4 + 2G_2^4 + G_3^4$$

$$ = {l^3}n + 2{l^2}{n^2} + {\ln ^3}$$

$$ = \ln {\left( {1 + n} \right)^2}$$

$$ = \ln \times 2{m^2}$$

$$ = 4l{m^2}n$$

$$G.P.$$ $$ = r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$$

$$\therefore$$ $${G_1} = {l^{3/4}}\,{n^{1/4}},$$ $${G_2} = {l^{1/2}}{n^{1/2}},\,$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$$

$$G_1^4 + 2G_2^4 + G_3^4$$

$$ = {l^3}n + 2{l^2}{n^2} + {\ln ^3}$$

$$ = \ln {\left( {1 + n} \right)^2}$$

$$ = \ln \times 2{m^2}$$

$$ = 4l{m^2}n$$

2

MCQ (Single Correct Answer)

The sum of first 9 terms of the series.

$${{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + ......$$

$${{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + ......$$

A

142

B

192

C

71

D

96

$${n^{th}}$$ term of series

$$ = {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$$

Sum of $$n$$ term $$ = \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$$

$$ = {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \right]$$

$$ = {1 \over 4}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6} + {{2n\left( {n + 1} \right)} \over 2} + n} \right]$$

Sum of $$9$$ terms

$$ = {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96$$

$$ = {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$$

Sum of $$n$$ term $$ = \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$$

$$ = {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \right]$$

$$ = {1 \over 4}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6} + {{2n\left( {n + 1} \right)} \over 2} + n} \right]$$

Sum of $$9$$ terms

$$ = {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96$$

3

MCQ (Single Correct Answer)

If $${(10)^9} + 2{(11)^1}\,({10^8}) + 3{(11)^2}\,{(10)^7} + ......... + 10{(11)^9} = k{(10)^9},$$, then k is equal to :

A

100

B

110

C

$${{121} \over {10}}$$

D

$${{441} \over {100}}$$

Let $${10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$

Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}$$

Multiplied by $${{11} \over {10}}$$ on both the sides

$${{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}$$

$$x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}$$

$$ \Rightarrow x = {10^{11}} = k{.10^9}$$

Given $$ \Rightarrow k = 100$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$

Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}$$

Multiplied by $${{11} \over {10}}$$ on both the sides

$${{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}$$

$$x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}$$

$$ \Rightarrow x = {10^{11}} = k{.10^9}$$

Given $$ \Rightarrow k = 100$$

4

MCQ (Single Correct Answer)

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :

A

$$2 - \sqrt 3 $$

B

$$2 + \sqrt 3 $$

C

$$\sqrt 2 + \sqrt 3 $$

D

$$3 + \sqrt 2 $$

Let $$a,ar,a{r^2}$$ are in $$G.P.$$

According to the question

$$a,2ar,a{r^2}$$ are in $$A.P.$$

$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$

$$ \Rightarrow 4r = 1 + {r^2}$$

$$ \Rightarrow {r^2} - 4r + 1 = 0$$

$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$

Since $$r > 1$$

$$\therefore$$ $$\pi = 2 - \sqrt 3 $$ is rejected

Hence, $$r = 2 + \sqrt 3 $$

According to the question

$$a,2ar,a{r^2}$$ are in $$A.P.$$

$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$

$$ \Rightarrow 4r = 1 + {r^2}$$

$$ \Rightarrow {r^2} - 4r + 1 = 0$$

$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$

Since $$r > 1$$

$$\therefore$$ $$\pi = 2 - \sqrt 3 $$ is rejected

Hence, $$r = 2 + \sqrt 3 $$

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Complex Numbers

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