Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

In a triangle $$ABC$$, medians $$AD$$ and $$BE$$ are drawn. If $$AD=4$$,

$$\angle DAB = {\pi \over 6}$$ and $$\angle ABE = {\pi \over 3}$$, then the area of the $$\angle \Delta ABC$$ is

$$\angle DAB = {\pi \over 6}$$ and $$\angle ABE = {\pi \over 3}$$, then the area of the $$\angle \Delta ABC$$ is

A

$${{64} \over 3}$$

B

$${8 \over 3}$$

C

$${{16} \over 3}$$

D

$${{32} \over {3\sqrt 3 }}$$

$$AP = {2 \over 3}AD = {8 \over 3};\,\,PD = {4 \over 3};\,\,$$

Let $$PB=x$$

$$\tan {60^ \circ } = {{8/3} \over x}$$

or $$x = {8 \over {3\sqrt 3 }}$$

Area of $$\Delta ABD$$

$$ = {1 \over 2} \times 4 \times {8 \over {3\sqrt 3 }} = {{16} \over {3\sqrt 3 }}$$

$$\therefore$$ Area of $$\Delta ABC$$

$$ = 2 \times {{16} \over {3\sqrt 3 }} = {{32} \over {3\sqrt 3 }}$$

$$\left[ \, \right.$$ As median of a $$\Delta $$ divides it into two $$\Delta 's$$ of equal area. $$\left. \, \right]$$

2

MCQ (Single Correct Answer)

The sum of the radii of inscribed and circumscribed circles for an $$n$$ sided regular polygon of side $$a, $$ is

A

$${a \over 4}\cot \left( {{\pi \over {2n}}} \right)$$

B

$$a\cot \left( {{\pi \over {n}}} \right)$$

C

$${a \over 2}\cot \left( {{\pi \over {2n}}} \right)$$

D

$$a\cot \left( {{\pi \over {2n}}} \right)$$

$$\tan \left( {{\pi \over n}} \right) = {a \over {2r}};\,\,\sin \left( {{\pi \over n}} \right) = {a \over {2R}}$$

$$r + R = {a \over 2}\left[ {\cot {\pi \over n} + \cos ec{\pi \over n}} \right]$$

$$ = {a \over 2}\left[ {{{\cos {\pi \over n} + 1} \over {\sin {\pi \over n}}}} \right]$$

$$ = {a \over 2}\left[ {{{2{{\cos }^2}{\pi \over {2n}}} \over {2\sin {\pi \over {2n}}\cos {\pi \over {2n}}}}} \right]$$

$$ = {a \over 2}\cot {\pi \over {2\pi }}$$

$$r + R = {a \over 2}\left[ {\cot {\pi \over n} + \cos ec{\pi \over n}} \right]$$

$$ = {a \over 2}\left[ {{{\cos {\pi \over n} + 1} \over {\sin {\pi \over n}}}} \right]$$

$$ = {a \over 2}\left[ {{{2{{\cos }^2}{\pi \over {2n}}} \over {2\sin {\pi \over {2n}}\cos {\pi \over {2n}}}}} \right]$$

$$ = {a \over 2}\cot {\pi \over {2\pi }}$$

3

MCQ (Single Correct Answer)

In a triangle with sides $$a, b, c,$$ $${r_1} > {r_2} > {r_3}$$ (which are the ex-radii) then

A

$$a>b>c$$

B

$$a < b < c$$

C

$$a > b$$ and $$b < c$$

D

$$a < b$$ and $$b > c$$

$${r_1} > {r_2} > {r_3}$$

$$ \Rightarrow {\Delta \over {s - a}} > {\Delta \over {s - b}} > {\Delta \over {s - c}};$$

$$ \Rightarrow s - a < s - b < s - c$$

$$ \Rightarrow - a < - b < - c$$

$$ \Rightarrow a > b > c$$

$$ \Rightarrow {\Delta \over {s - a}} > {\Delta \over {s - b}} > {\Delta \over {s - c}};$$

$$ \Rightarrow s - a < s - b < s - c$$

$$ \Rightarrow - a < - b < - c$$

$$ \Rightarrow a > b > c$$

4

MCQ (Single Correct Answer)

The sides of a triangle are $$3x + 4y,$$ $$4x + 3y$$ and $$5x + 5y$$ where $$x$$, $$y>0$$ then the triangle is

A

right angled

B

obtuse angled

C

equilateral

D

none of these

Let $$\,\,\,\,a = 3x + 4y,b = 4x + 3y$$

and $$c = 5x + 5y$$

as $$\,\,\,\,x,y > 0,c = 5x + 5y$$ is the largest side

$$\therefore$$ $$C$$ is the largest angle. Now

$$\cos \,C = {{{{\left( {3x + 4y} \right)}^2} + {{\left( {4x + 3y} \right)}^3} - {{\left( {5x + 5y} \right)}^2}} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}}$$

$$ = {{ - 2xy} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}} < 0$$

$$\therefore$$ $$C$$ is obtuse angle $$ \Rightarrow \Delta ABC$$ is obtuse angled

and $$c = 5x + 5y$$

as $$\,\,\,\,x,y > 0,c = 5x + 5y$$ is the largest side

$$\therefore$$ $$C$$ is the largest angle. Now

$$\cos \,C = {{{{\left( {3x + 4y} \right)}^2} + {{\left( {4x + 3y} \right)}^3} - {{\left( {5x + 5y} \right)}^2}} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}}$$

$$ = {{ - 2xy} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}} < 0$$

$$\therefore$$ $$C$$ is obtuse angle $$ \Rightarrow \Delta ABC$$ is obtuse angled

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Complex Numbers

Quadratic Equation and Inequalities

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