Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If the coefficints of $${x^3}$$ and $${x^4}$$ in the expansion of $$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$ in powers of $$x$$ are both zero, then $$\left( {a,\,b} \right)$$ is equal to:

A

$$\left( {14,{{272} \over 3}} \right)$$

B

$$\left( {16,{{272} \over 3}} \right)$$

C

$$\left( {16,{{251} \over 3}} \right)$$

D

$$\left( {14,{{251} \over 3}} \right)$$

$$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$

= $${\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}$$

= $$\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)}^2} - {}^{18}{C_3}{{\left( {2x} \right)}^3} + ....} \right]$$

Coefficient of x^{3} is

$${\left( { - 2} \right)^3}.{}^{18}{C_3} + a{\left( { - 2} \right)^2}.{}^{18}{C_2} + b\left( { - 2} \right).{}^{18}{C_1}$$ = 0

$$ \Rightarrow 153a - 9b = 1632$$ ....... (1)

Coefficient of x^{4} is

$${\left( { - 2} \right)^4}.{}^{18}{C_4} + a{\left( { - 2} \right)^3}.{}^{18}{C_3} + b{\left( { - 2} \right)^2}.{}^{18}{C_2}$$ = 0

$$ \Rightarrow 3b - 32a = -240$$ ....... (2)

Solving (1) and (2), we get $$a$$ = 16, b = $${{172} \over 3}$$

= $${\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}$$

= $$\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)}^2} - {}^{18}{C_3}{{\left( {2x} \right)}^3} + ....} \right]$$

Coefficient of x

$${\left( { - 2} \right)^3}.{}^{18}{C_3} + a{\left( { - 2} \right)^2}.{}^{18}{C_2} + b\left( { - 2} \right).{}^{18}{C_1}$$ = 0

$$ \Rightarrow 153a - 9b = 1632$$ ....... (1)

Coefficient of x

$${\left( { - 2} \right)^4}.{}^{18}{C_4} + a{\left( { - 2} \right)^3}.{}^{18}{C_3} + b{\left( { - 2} \right)^2}.{}^{18}{C_2}$$ = 0

$$ \Rightarrow 3b - 32a = -240$$ ....... (2)

Solving (1) and (2), we get $$a$$ = 16, b = $${{172} \over 3}$$

2

MCQ (Single Correct Answer)

The term independent of $$x$$ in expansion of

$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$ is

$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$ is

A

4

B

120

C

210

D

310

$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$

= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$$

= $${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$$

= $${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$$

= $${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$$

= $${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$$

[**Note:**

For $${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$$ the $$\left( {r + 1} \right)$$^{th} term with power m of x is

$$r = {{n\alpha - m} \over {\alpha + \beta }}$$]

Here $$\alpha = {1 \over 3}$$, $$\beta = {1 \over 2}$$ and m = 0

then $$r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$$ = $${{10} \over 3} \times {6 \over 5}$$ = 4

$$\therefore$$ T_{5} is the term independent of x.

$$\therefore$$ T_{5} = $${}^{10}{C_4}$$ = 210

= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$$

= $${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$$

= $${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$$

= $${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$$

= $${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$$

[

For $${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$$ the $$\left( {r + 1} \right)$$

$$r = {{n\alpha - m} \over {\alpha + \beta }}$$]

Here $$\alpha = {1 \over 3}$$, $$\beta = {1 \over 2}$$ and m = 0

then $$r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$$ = $${{10} \over 3} \times {6 \over 5}$$ = 4

$$\therefore$$ T

$$\therefore$$ T

3

MCQ (Single Correct Answer)

If $$n$$ is a positive integer, then $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ is :

A

an irrational number

B

an odd positive integer

C

an even positive integer

D

a rational number other than positive integers

Let $${\left( {a + x} \right)^n}$$ = Odd trems(A) + Even terms(B)

So $${\left( {a - x} \right)^n}$$ = Odd terms(A) - Even terms(B)

$$\therefore$$ $${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$$

= (A + B) - (A - B)

= 2B

= 2[even terms]

= 2[ T_{2} + T_{4} + T_{6} + ....... ]

So in case of $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$

= 2[ T_{2} + T_{4} + T_{6} + ....... ]

= 2[ $${}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...$$

Here in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$, 2n - 1 is odd number. So there will be always $${\sqrt 3 }$$ in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$.

So $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ will be always irrational number.

So $${\left( {a - x} \right)^n}$$ = Odd terms(A) - Even terms(B)

$$\therefore$$ $${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$$

= (A + B) - (A - B)

= 2B

= 2[even terms]

= 2[ T

So in case of $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$

= 2[ T

= 2[ $${}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...$$

Here in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$, 2n - 1 is odd number. So there will be always $${\sqrt 3 }$$ in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$.

So $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ will be always irrational number.

4

MCQ (Single Correct Answer)

The coefficient of $${x^7}$$ in the expansion of $${\left( {1 - x - {x^2} + {x^3}} \right)^6}$$ is

A

$$-132$$

B

$$-144$$

C

$$132$$

D

$$144$$

Given,

$${\left( {1 - x - {x^2} + {x^3}} \right)^6}$$

= $${\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6}$$

= $${\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}$$

= $$\left( {1 + {}^6{C_1}( - x) + {}^6{C_2}{{( - x)}^2} + {}^6{C_3}{{( - x)}^3} + .......} \right)\times$$

$$\left( {1 + {}^6{C_1}( - {x^2}) + {}^6{C_2}{{( - {x^2})}^2} + {}^6{C_3}{{( - {x^2})}^3} + .......} \right)$$

$$\therefore$$ Coefficient of x^{7} = $$ - {}^6{C_1} \times - {}^6{C_3} + \left( { - {}^6{C_3}} \right) \times {}^6{C_2} + \left( { - {}^6{C_5}} \right) \times - {}^6{C_1}$$

= 120 - 300 + 36 = - 144

$${\left( {1 - x - {x^2} + {x^3}} \right)^6}$$

= $${\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6}$$

= $${\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}$$

= $$\left( {1 + {}^6{C_1}( - x) + {}^6{C_2}{{( - x)}^2} + {}^6{C_3}{{( - x)}^3} + .......} \right)\times$$

$$\left( {1 + {}^6{C_1}( - {x^2}) + {}^6{C_2}{{( - {x^2})}^2} + {}^6{C_3}{{( - {x^2})}^3} + .......} \right)$$

$$\therefore$$ Coefficient of x

= 120 - 300 + 36 = - 144

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