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

For x $$ \in $$ **R**, x $$ \ne $$ -1,

if (1 + x)^{2016} + x(1 + x)^{2015} + x^{2}(1 + x)^{2014} + . . . . + x^{2016} =

$$\sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,$$ then a_{17} is equal to :

if (1 + x)

$$\sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,$$ then a

A

$${{2017!} \over {17!\,\,\,2000!}}$$

B

$${{2016!} \over {17!\,\,\,1999!}}$$

C

$${{2017!} \over {2000!}}$$

D

$${{2016!} \over {16!}}$$

Assume,

P = (1 + x)^{2016} + x(1 + x)^{2015} + . . . . .+ x^{2015} . (1 + x) + x^{2016} . . . . .(1)

Multiply this with $$\left( {{x \over {1 + x}}} \right),$$

$$\left( {{x \over {1 + x}}} \right)P = $$ x(1 + x)^{2015} + x^{2}(1 + x)^{2014} +

. . . . . . + x^{2016} + $${{{x^{2017}}} \over {1 + x}}$$ . . . . . (2)

Performing (1) $$-$$ (2), we get

$${P \over {1 + x}} = $$ (1 + x)^{2016} $$-$$ $${{{x^{2017}}} \over {1 + x}}$$

$$ \Rightarrow $$ P = (1 + x)^{2017} $$-$$ x^{2017}

$$ \therefore $$ a_{17} = coefficient of x^{17} $$=$$ ^{2017}C_{17} $$=$$ $${{2017!} \over {17!\,\,2000!}}$$

P = (1 + x)

Multiply this with $$\left( {{x \over {1 + x}}} \right),$$

$$\left( {{x \over {1 + x}}} \right)P = $$ x(1 + x)

. . . . . . + x

Performing (1) $$-$$ (2), we get

$${P \over {1 + x}} = $$ (1 + x)

$$ \Rightarrow $$ P = (1 + x)

$$ \therefore $$ a

2

Let x, y, z be positive real numbers such that x + y + z = 12 and x^{3}y^{4}z^{5} = (0.1) (600)^{3}. Then x^{3} + y^{3} + z^{3}is equal to :

A

270

B

258

C

342

D

216

As we know

AM $$ \ge $$ GM

$$ \Rightarrow $$ $${{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}$$ $$ \ge $$ $${\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right)}^4}{{\left( {{z \over 5}} \right)}^5}} \right]^{{1 \over {12}}}}$$

$$ \Rightarrow $$ 1 $$ \ge $$ $${{{x^3}{y^4}{z^5}} \over {{3^3}{4^4}{5^5}}}$$

$$ \Rightarrow $$ x^{3} y^{4} z^{5} $$ \le $$ 3^{3} . 4^{4} . 5^{5}

$$ \Rightarrow $$ x^{3} y^{4} z^{5} $$ \le $$ (0.1)(600)^{3}

but given that,

x^{3} y^{4} z^{5} = (0.1) (600)^{3}

$$ \therefore $$ AM $$=$$ GM

$$ \Rightarrow $$ All the number are equal.

$$ \therefore $$ $${x \over 3} = {y \over 4} = {z \over 5} = k$$

$$ \Rightarrow $$ x $$=$$ 3k, y = 4k, z = 5k

given that,

x + y + z $$=$$ 12

$$ \Rightarrow $$ 3k + 4k + 5k $$=$$ 12

$$ \Rightarrow $$ 12k $$=$$ 12

$$ \Rightarrow $$ k = 1

$$ \therefore $$ x $$=$$ 3, y $$=$$ 4, z $$=$$ 5

So, x^{3} + y^{3} + z^{3}

$$=$$ 3^{3} + 4^{3} + 5^{3}

$$=$$ 216

AM $$ \ge $$ GM

$$ \Rightarrow $$ $${{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}$$ $$ \ge $$ $${\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right)}^4}{{\left( {{z \over 5}} \right)}^5}} \right]^{{1 \over {12}}}}$$

$$ \Rightarrow $$ 1 $$ \ge $$ $${{{x^3}{y^4}{z^5}} \over {{3^3}{4^4}{5^5}}}$$

$$ \Rightarrow $$ x

$$ \Rightarrow $$ x

but given that,

x

$$ \therefore $$ AM $$=$$ GM

$$ \Rightarrow $$ All the number are equal.

$$ \therefore $$ $${x \over 3} = {y \over 4} = {z \over 5} = k$$

$$ \Rightarrow $$ x $$=$$ 3k, y = 4k, z = 5k

given that,

x + y + z $$=$$ 12

$$ \Rightarrow $$ 3k + 4k + 5k $$=$$ 12

$$ \Rightarrow $$ 12k $$=$$ 12

$$ \Rightarrow $$ k = 1

$$ \therefore $$ x $$=$$ 3, y $$=$$ 4, z $$=$$ 5

So, x

$$=$$ 3

$$=$$ 216

3

Let z = 1 + ai be a complex number, a > 0, such that z^{3} is a real number.

Then the sum 1 + z + z^{2} + . . . . .+ z^{11} is equal to :

Then the sum 1 + z + z

A

$$ - 1250\,\sqrt 3 \,i$$

B

$$ 1250\,\sqrt 3 \,i$$

C

$$1365\,\sqrt 3 i$$

D

$$-$$ $$1365\,\sqrt 3 i$$

z = 1 + ai

z^{2} = 1 $$-$$ a^{2} + 2ai

z^{2} . z = {(1 $$-$$ a^{2}) + 2ai} {1 + ai}

= (1 $$-$$ a^{2}) + 2ai + (1 $$-$$ a^{2}) ai $$-$$ 2a^{2}

$$ \because $$ z^{3} is real $$ \Rightarrow $$ 2a + (1 $$-$$ a^{2}) a = 0

a (3 $$-$$ a^{2}) = 0 $$ \Rightarrow $$ a = $$\sqrt 3 $$ (a > 0)

1 + z + z^{2} . . . . . . . z^{11} = $${{{z^{12}} - 1} \over {z - 1}} = {{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {1 + \sqrt 3 i - 1}}$$

= $${{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {\sqrt 3 i}}$$

(1 + $${\sqrt 3 i}$$)^{12} = 2^{12} $${\left( {{1 \over 2} + {{\sqrt 3 } \over 2}i} \right)^{12}}$$

= 2^{12} (cos$${\pi \over 3}$$ + isin$${\pi \over 3}$$)^{12} = 2^{12} (cos4$$\pi $$ + isin4$$\pi $$) = 2^{12}

$$ \Rightarrow $$ $${{{2^{12}} - 1} \over {\sqrt 3 i}} = {{4095} \over {\sqrt 3 i}} = - {{4095} \over 3}\sqrt 3 i = - 1365\sqrt 3 i$$

z

z

= (1 $$-$$ a

$$ \because $$ z

a (3 $$-$$ a

1 + z + z

= $${{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {\sqrt 3 i}}$$

(1 + $${\sqrt 3 i}$$)

= 2

$$ \Rightarrow $$ $${{{2^{12}} - 1} \over {\sqrt 3 i}} = {{4095} \over {\sqrt 3 i}} = - {{4095} \over 3}\sqrt 3 i = - 1365\sqrt 3 i$$

4

Let a_{1}, a_{2}, a_{3}, . . . . . . . , a_{n}, . . . . . be in A.P.

If a_{3} + a_{7} + a_{11} + a_{15} = 72,

then the sum of its first 17 terms is equal to :

If a

then the sum of its first 17 terms is equal to :

A

306

B

153

C

612

D

204

As a_{1} a_{2} . . . . . a_{n} . . . . . are in A.P.

$$ \therefore $$ a_{3} + a_{15} = a_{7} + a_{11} = a_{1} + a_{17}

Given,

a_{3} + a_{7} + a_{11} + a_{15} + a_{15} = 72

$$ \Rightarrow $$ (a_{3} + a_{15}) + (a_{7} + a_{11}) = 72

$$ \Rightarrow $$ 2(a_{1} + a_{17}) = 72

$$ \Rightarrow $$ (a_{1} + a_{17}) = 36

$$ \therefore $$ Sum of first 17 terms

= $${{17} \over 2}$$ (a_{1} + a_{17})

= $${{17} \over 2}$$ $$ \times $$ 36

= 306

$$ \therefore $$ a

Given,

a

$$ \Rightarrow $$ (a

$$ \Rightarrow $$ 2(a

$$ \Rightarrow $$ (a

$$ \therefore $$ Sum of first 17 terms

= $${{17} \over 2}$$ (a

= $${{17} \over 2}$$ $$ \times $$ 36

= 306

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