Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

The coefficient of x^{2} in the expansion of the product

(2$$-$$x^{2}) .((1 + 2x + 3x^{2})^{6} + (1 $$-$$ 4x^{2})^{6}) is :

(2$$-$$x

A

107

B

106

C

108

D

155

Given,

(2 $$-$$ x^{2}) . (1 + 2x + 3x^{2}) ^{6} + (1 $$-$$ 4x^{2})^{6})

Let, a = ((1 + 2x + 3x^{2})^{6} + (1 $$-$$ 4x^{2})^{6})

$$\therefore\,\,\,\,$$ Given statement becomes,

(2 $$-$$ x^{2}) . (a)

= 2a $$-$$ x^{2} (a)

Here coefficients of x^{2} is

= 2 (coefficient of x^{2} in a ) $$-$$ 1 (constant in a)

(1 + 2x + 3x^{2}) ^{6} = ^{6}C_{0} + ^{6}C_{1} (2x + 3x^{2}) + ^{6}C_{2} (2x + 3x^{2})^{2}

+ . . . . . .+ (2x + 3x^{2})^{6}

(1 $$-$$ 4x^{2})^{6} = ^{6}C_{0} $$-$$ ^{6}C_{1} (4x^{2}) + ^{6}C_{2} (4x^{2})^{2}

+ . . . . .+ (4x^{2})^{6}

Coefficient of x^{2} in (1 + 2x + 3x^{2})^{6}

=^{6}C_{1} $$ \times $$ 3 + ^{6}C_{2} $$ \times $$ 4

= 18 + 60

Coefficient of x^{2} in (1 $$-$$ 4x^{2})^{6}

= $$-$$^{6}C_{1} $$ \times $$ 4

= $$-$$ 24

Coefficient of x^{2} in ((1 + 2x + 3x^{2})^{6} + (1 $$-$$ 4x^{2})^{6})

= 60 + 18 $$-$$ 24

= 54

Constant term in (1 + 2x + 3x^{2})^{6} = ^{6}C_{0} = 1

Constant term in (1 $$-$$ 4x)^{6} = ^{6}C_{0} = 1

$$\therefore\,\,\,\,$$ Constant term in ((1 + 2x + 3x^{2})^{6} + (1 $$-$$ 4x)^{6})

= 1 + 1 = 2

$$\therefore\,\,\,\,$$ Coefficient of x^{2} in (2 $$-$$ x^{2}) ((1 + 2x + 3x^{2})^{6} + (1 $$-$$ 4x^{2})^{6})

= 2 (54) $$-$$ 1 (2)

= 108 $$-$$ 2

= 106

(2 $$-$$ x

Let, a = ((1 + 2x + 3x

$$\therefore\,\,\,\,$$ Given statement becomes,

(2 $$-$$ x

= 2a $$-$$ x

Here coefficients of x

= 2 (coefficient of x

(1 + 2x + 3x

+ . . . . . .+ (2x + 3x

(1 $$-$$ 4x

+ . . . . .+ (4x

Coefficient of x

=

= 18 + 60

Coefficient of x

= $$-$$

= $$-$$ 24

Coefficient of x

= 60 + 18 $$-$$ 24

= 54

Constant term in (1 + 2x + 3x

Constant term in (1 $$-$$ 4x)

$$\therefore\,\,\,\,$$ Constant term in ((1 + 2x + 3x

= 1 + 1 = 2

$$\therefore\,\,\,\,$$ Coefficient of x

= 2 (54) $$-$$ 1 (2)

= 108 $$-$$ 2

= 106

2

If the fractional part of the number $$\left\{ {{{{2^{403}}} \over {15}}} \right\} is \, {k \over {15}}$$, then k is equal to :

A

8

B

14

C

6

D

1

$${{{2^{403}}} \over {15}}$$

$$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$$

$$ = {8 \over {15}}{\left( {16} \right)^{100}}$$

$$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$$

$$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$$

$$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$$

$$ = {8 \over {15}} + 8$$ (integer)

$$ \therefore $$ Fractional part $$ = {8 \over {15}}$$

According to the question,

$${k \over {15}} = {8 \over {15}}$$

$$ \Rightarrow $$ K $$=$$ 8

$$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$$

$$ = {8 \over {15}}{\left( {16} \right)^{100}}$$

$$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$$

$$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$$

$$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$$

$$ = {8 \over {15}} + 8$$ (integer)

$$ \therefore $$ Fractional part $$ = {8 \over {15}}$$

According to the question,

$${k \over {15}} = {8 \over {15}}$$

$$ \Rightarrow $$ K $$=$$ 8

3

The coefficient of t^{4} in the expansion of $${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$ is :

A

14

B

15

C

10

D

12

$${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$

= (1 $$-$$ t^{6})^{3} (1 $$-$$ t)^{$$-$$3}

= (1 $$-$$^{3}C_{1}t^{6} + ^{3}C_{2}t^{12} $$-$$ ^{3}C_{3}t^{18}) $$ \times $$ (1 $$-$$ t)^{$$-$$3}

coefficient of t^{4} is 1 $$ \times $$ coefficient of t^{4} in (1 $$-$$ t)^{$$-$$3}

= 1 $$ \times $$^{3+4$$-$$1}C_{4} (By multinomial theorem)

=^{6}C_{4} = 15

= (1 $$-$$ t

= (1 $$-$$

coefficient of t

= 1 $$ \times $$

=

4

If $${\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$ then k is equal to

A

100

B

200

C

50

D

400

$${\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$$

$$ \Rightarrow 100 = k$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$$

$$ \Rightarrow 100 = k$$

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