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General Aptitude

1

If (27)^{999} is divided by 7, then the remainder is :

A

1

B

2

C

3

D

6

We have,

$${{{{\left( {27} \right)}^{999}}} \over 7}$$

= $${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$$

= $${{28\,\lambda - 1} \over 7}$$

= $${{28\,\lambda - 7 + 7 - 1} \over \lambda }$$

= $${{7\left( {4\lambda - 1} \right) + 6} \over 7}$$

$$\therefore\,\,\,$$ Remainder = 6

$${{{{\left( {27} \right)}^{999}}} \over 7}$$

= $${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$$

= $${{28\,\lambda - 1} \over 7}$$

= $${{28\,\lambda - 7 + 7 - 1} \over \lambda }$$

= $${{7\left( {4\lambda - 1} \right) + 6} \over 7}$$

$$\therefore\,\,\,$$ Remainder = 6

2

The coefficient of x^{−5} in the binomial expansion of

$${\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}},$$ where x $$ \ne $$ 0, 1, is :

$${\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}},$$ where x $$ \ne $$ 0, 1, is :

A

1

B

4

C

$$-$$ 4

D

$$-$$ 1

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

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

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

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

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

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

[**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 = -5

then $$r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}$$ = $${{25} \over 3} \times {6 \over 5}$$ = 10

$$\therefore$$ T_{11} is the term with x^{-5}.

$$\therefore$$ T_{11} = $${}^{10}{C_{10}}$$ = 1

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

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

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

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

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

[

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 = -5

then $$r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}$$ = $${{25} \over 3} \times {6 \over 5}$$ = 10

$$\therefore$$ T

$$\therefore$$ T

3

The sum of the co-efficients of all odd degree terms in the expansion of

$${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$$, $$\left( {x > 1} \right)$$ is

$${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$$, $$\left( {x > 1} \right)$$ is

A

2

B

-1

C

0

D

1

4

If n is the degree of the polynomial,

$${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8} + $$ $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$

and m is the coefficient of x^{n} in it, then the ordered pair (n, m) is equal to :

$${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8} + $$ $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$

and m is the coefficient of x

A

(24, (10)^{8})

B

(8, 5(10)^{4})

C

(12, (20)^{4})

D

(12, 8(10)^{4})

Given,

$${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$$ + $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$

= $${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$

+ $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$$

= $${\left[ {{{2(\sqrt {5x^3 + 1} + \sqrt {5x^3 - 1} } \over 2}} \right]^8}$$ + $${\left[ {{{2(\sqrt {5x^3 - 1} - \sqrt {5x^3 - 1} } \over 2}} \right]^8}$$

= ($${\sqrt {5{x^3} + 1} }$$ + $${\sqrt {5{x^3} - 1} }$$)^{8} + ($${\sqrt {5{x^3} - 1} }$$ - $${\sqrt {5{x^3} - 1} }$$)^{8}

= 2 $$\left[ {{}^8{C_0}} \right.{(\sqrt {5{x^3} + 1} )^8}$$ +

+ $${}^8{C_2}{(\sqrt {5{x^3} + 1} )^6}\sqrt {5{x^3} - 1} $$

+ $${}^8{C_4}{(\sqrt {5{x^3} + 1} )^4}{(5{x^3} - 1)^2}$$

+ $${}^8{C_6}{(\sqrt {5{x^3} + 1} )^2}{(5{x^3} - 1)^3}$$

+ $$\left. {{}^8{C_8}{{(5{x^3} - 1)}^4}} \right]$$

= 2$$[{}^8{C_0}{(5{x^3} + 1)^4}$$

+ $${}^8{C_2}{(5{x^3} + 1)^3}(5{x^3} - 1)$$

+ $${}^8{C_4}(5{x^3} + 1)^2{(5{x^3} - 1)^2}$$

+ $${}^8{C_6}(5{x^3} + 1){(5{x^3} - 1)^3}$$

+ $${}^8{C_8}{(5{x^3} - 1)^4}]$$

Here maximum power of x is 12

$$ \therefore $$ Degree of polynomial = 12

Coefficient of x^{12}

= 2 [^{8}C_{6} 5^{4} + ^{8}C_{2}
$$\times$$ 5^{3} $$\times$$ 5 + ^{8}C_{4}
$$\times$$ 5^{2} $$\times$$ 5^{2} + ^{8}C_{6 } $$\times$$ 5 $$\times$$ 5^{3} + ^{8}C_{8} $$\times$$ 5^{4}]

= 160000 = (20)^{4}

$${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$$ + $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$

= $${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$$

+ $${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$$

= $${\left[ {{{2(\sqrt {5x^3 + 1} + \sqrt {5x^3 - 1} } \over 2}} \right]^8}$$ + $${\left[ {{{2(\sqrt {5x^3 - 1} - \sqrt {5x^3 - 1} } \over 2}} \right]^8}$$

= ($${\sqrt {5{x^3} + 1} }$$ + $${\sqrt {5{x^3} - 1} }$$)

= 2 $$\left[ {{}^8{C_0}} \right.{(\sqrt {5{x^3} + 1} )^8}$$ +

+ $${}^8{C_2}{(\sqrt {5{x^3} + 1} )^6}\sqrt {5{x^3} - 1} $$

+ $${}^8{C_4}{(\sqrt {5{x^3} + 1} )^4}{(5{x^3} - 1)^2}$$

+ $${}^8{C_6}{(\sqrt {5{x^3} + 1} )^2}{(5{x^3} - 1)^3}$$

+ $$\left. {{}^8{C_8}{{(5{x^3} - 1)}^4}} \right]$$

= 2$$[{}^8{C_0}{(5{x^3} + 1)^4}$$

+ $${}^8{C_2}{(5{x^3} + 1)^3}(5{x^3} - 1)$$

+ $${}^8{C_4}(5{x^3} + 1)^2{(5{x^3} - 1)^2}$$

+ $${}^8{C_6}(5{x^3} + 1){(5{x^3} - 1)^3}$$

+ $${}^8{C_8}{(5{x^3} - 1)^4}]$$

Here maximum power of x is 12

$$ \therefore $$ Degree of polynomial = 12

Coefficient of x

= 2 [

= 160000 = (20)

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