Joint Entrance Examination

Graduate Aptitude Test in Engineering

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General Aptitude

1

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

2

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

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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