Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The sum of series $${1 \over {2\,!}} + {1 \over {4\,!}} + {1 \over {6\,!}} + ........$$ is

A

$${{\left( {{e^2} - 2} \right)} \over e}\,$$

B

$${{{{\left( {e - 1} \right)}^2}} \over {2e}}$$

C

$${{\left( {{e^2} - 1} \right)} \over {2e}}\,$$

D

$${{\left( {{e^2} - 1} \right)} \over 2}$$

We know that

$$e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........$$

and

$${e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........$$

$$\therefore$$ $$e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \right]$$

$$\therefore$$ $${1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + .......$$

$$ = {{e + {e^{ - 1}}} \over 2} - 1$$

$$ = {{{e^2} + 1 - 2e} \over {2e}}$$

$$ = {{{{\left( {e - 1} \right)}^2}} \over {2e}}$$

$$e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........$$

and

$${e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........$$

$$\therefore$$ $$e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \right]$$

$$\therefore$$ $${1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + .......$$

$$ = {{e + {e^{ - 1}}} \over 2} - 1$$

$$ = {{{e^2} + 1 - 2e} \over {2e}}$$

$$ = {{{{\left( {e - 1} \right)}^2}} \over {2e}}$$

2

MCQ (Single Correct Answer)

The sum of the first n terms of the series $${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....\,is\,{{n{{(n + 1)}^2}} \over 2}$$ when n is even. When n is odd the sum is

A

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

B

$${{{n^2}(n + 1)} \over 2}$$

C

$${{n{{(n + 1)}^2}} \over 4}$$

D

$$\,{{3n(n + 1)} \over 2}$$

If $$n$$ is odd, the required sum is

$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$

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

[ As $$\left( {n - 1} \right)$$ is even

$$\therefore$$ using given formula for the sum of $$\left( {n - 1} \right)$$ terms.]

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

$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$

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

[ As $$\left( {n - 1} \right)$$ is even

$$\therefore$$ using given formula for the sum of $$\left( {n - 1} \right)$$ terms.]

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

3

MCQ (Single Correct Answer)

Let $${{T_r}}$$ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, $$m \ne n,\,\,{T_m} = {1 \over n}\,\,and\,{T_n} = {1 \over m},\,$$ then a - d equals

A

$${1 \over m} + {1 \over n}$$

B

1

C

$${1 \over {m\,n}}$$

D

0

$${T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)$$

$${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$$

$$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$$

$$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$$

From $$\left( 1 \right)$$ $$a = {1 \over {mn}} \Rightarrow a - d = 0$$

$${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$$

$$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$$

$$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$$

From $$\left( 1 \right)$$ $$a = {1 \over {mn}} \Rightarrow a - d = 0$$

4

MCQ (Single Correct Answer)

The sum of the serier $${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..............$$ up to $$\infty $$ is equal to

A

$$\log {\,_e}\left( {{4 \over e}} \right)\,\,$$

B

$$2\,\log {\,_e}2$$

C

$$\log {\,_e}2 - 1\,$$

D

$$\log {\,_e}2$$

$${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty $$

$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$

$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $$

$$ = \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$$

$$ = 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$$

$$ = 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]$$

$$ = 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$$

$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$

$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $$

$$ = \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$$

$$ = 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$$

$$ = 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]$$

$$ = 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$$

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