Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A tower stands at the centre of a circular park. $$A$$ and $$B$$ are two points on the boundary of the park such that $$AB(=a)$$ subtends an angle of $${60^ \circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $$A$$ or $$B$$ is $${30^ \circ }$$. The height of the tower is

A

$$a/\sqrt 3 $$

B

$$a\sqrt 3 $$

C

$$2a/\sqrt 3 $$

D

$$2a\sqrt 3 $$

In the $$\Delta AOB,\,\,\angle AOB = {60^ \circ },$$ and

$$\angle OBA = \angle OAB$$

(since $$OA=OB=AB$$ radius of same circle).

$$\therefore$$ $$\Delta AOB$$ is a equilateral triangle.

Let the height of tower is $$h$$

$$m.$$ Given distance between two points $$A$$ & $$B$$ lie on boundary of

circular park, subtends an angle of $${60^ \circ }$$ at the foot of the tower

is $$AB$$ i.e. $$AB$$$$=a.$$ A tower $$OC$$ stands at the center of a circular

park. Angle of elevation of the top of the tower from $$A$$ and $$B$$ is $${30^ \circ }$$ .

In $$\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$$

$$ \Rightarrow {1 \over {\sqrt 3 }} = {h \over a} \Rightarrow h = {a \over {\sqrt 3 }}$$

$$\angle OBA = \angle OAB$$

(since $$OA=OB=AB$$ radius of same circle).

$$\therefore$$ $$\Delta AOB$$ is a equilateral triangle.

Let the height of tower is $$h$$

$$m.$$ Given distance between two points $$A$$ & $$B$$ lie on boundary of

circular park, subtends an angle of $${60^ \circ }$$ at the foot of the tower

is $$AB$$ i.e. $$AB$$$$=a.$$ A tower $$OC$$ stands at the center of a circular

park. Angle of elevation of the top of the tower from $$A$$ and $$B$$ is $${30^ \circ }$$ .

In $$\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$$

$$ \Rightarrow {1 \over {\sqrt 3 }} = {h \over a} \Rightarrow h = {a \over {\sqrt 3 }}$$

2

MCQ (Single Correct Answer)

In a triangle $$ABC$$, let $$\angle C = {\pi \over 2}$$. If $$r$$ is the inradius and $$R$$ is the circumradius of the triangle $$ABC$$, then $$2(r+R)$$ equals

A

$$b+c$$

B

$$a+b$$

C

$$a+b+c$$

D

$$c+a$$

We know by sin c rule

$${c \over {\sin C}} = 2R \Rightarrow c = 2R\sin C$$

$$ \Rightarrow c = 2R$$

$$\left( \, \right.$$as $$\,\,\,\angle C = {90^ \circ }$$ $$\left. \, \right)$$

Also $$\tan {C \over 2} = {r \over {s - c}}$$

$$ \Rightarrow \tan {\pi \over 4} = {r \over {s - c}}$$

$$\left( \, \right.$$as $$\,\,\,\angle C = {90^ \circ }$$ $$\left. \, \right)$$

$$ \Rightarrow r = s - c = {{a + b - c} \over 2}$$

$$ \Rightarrow 2r + c = a + b$$

$$ \Rightarrow 2r + 2R = a + b$$

(using $$c=2R$$)

$${c \over {\sin C}} = 2R \Rightarrow c = 2R\sin C$$

$$ \Rightarrow c = 2R$$

$$\left( \, \right.$$as $$\,\,\,\angle C = {90^ \circ }$$ $$\left. \, \right)$$

Also $$\tan {C \over 2} = {r \over {s - c}}$$

$$ \Rightarrow \tan {\pi \over 4} = {r \over {s - c}}$$

$$\left( \, \right.$$as $$\,\,\,\angle C = {90^ \circ }$$ $$\left. \, \right)$$

$$ \Rightarrow r = s - c = {{a + b - c} \over 2}$$

$$ \Rightarrow 2r + c = a + b$$

$$ \Rightarrow 2r + 2R = a + b$$

(using $$c=2R$$)

3

MCQ (Single Correct Answer)

If in a $$\Delta ABC$$, the altitudes from the vertices $$A, B, C$$ on opposite sides are in H.P, then $$\sin A,\sin B,\sin C$$ are in

A

G. P.

B

A. P.

C

A.P-G.P.

D

H. P

$$\Delta = {1 \over 2}{p_1}a = {1 \over 2}{p_2}b = {1 \over 2}{p_3}b$$

$${p_1},{p_2},{p_3},$$ are in $$H.P.$$

$$ \Rightarrow {{2\Delta } \over a},{{2\Delta } \over b},{{2\Delta } \over c}$$ are in $$H.P.$$

$$ \Rightarrow {1 \over a},{1 \over b},{1 \over c},$$ are in $$H.P.$$

$$ \Rightarrow a,b,c$$ are in $$A.P.$$

$$ \Rightarrow $$ $$K\sin A,K\sin B,K\sin C$$ are in $$A.P.$$

$$ \Rightarrow $$ $$\sin A,\sin B,\sin C$$ are in $$A.P.$$

$${p_1},{p_2},{p_3},$$ are in $$H.P.$$

$$ \Rightarrow {{2\Delta } \over a},{{2\Delta } \over b},{{2\Delta } \over c}$$ are in $$H.P.$$

$$ \Rightarrow {1 \over a},{1 \over b},{1 \over c},$$ are in $$H.P.$$

$$ \Rightarrow a,b,c$$ are in $$A.P.$$

$$ \Rightarrow $$ $$K\sin A,K\sin B,K\sin C$$ are in $$A.P.$$

$$ \Rightarrow $$ $$\sin A,\sin B,\sin C$$ are in $$A.P.$$

4

MCQ (Single Correct Answer)

The sides of a triangle are $$\sin \alpha ,\,\cos \alpha $$ and $$\sqrt {1 + \sin \alpha \cos \alpha } $$ for some $$0 < \alpha < {\pi \over 2}$$. Then the greatest angle of the triangle is

A

$${150^ \circ }$$

B

$${90^ \circ }$$

C

$${120^ \circ }$$

D

$${60^ \circ }$$

Let $$a = \sin \alpha ,b = \cos \alpha $$

and $$c = \sqrt {1 + \sin \alpha \cos \alpha } $$

Clearly $$a$$ and $$b < 1$$ but $$c > 1$$

as $$\,\,\,\sin \alpha > 0$$ and $$\cos \alpha > 0$$

$$\therefore$$ $$c$$ is the greatest side and greatest angle is $$C$$

$$\therefore$$ $$\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}$$

$$ = {{{{\sin }^2}\alpha + {{\cos }^2}\alpha - 1 - \sin \alpha \cos \alpha } \over {2\,\sin \alpha \cos \alpha }}$$

$$ = - {1 \over 2}$$

$$\therefore$$ $$C = {120^ \circ }$$

and $$c = \sqrt {1 + \sin \alpha \cos \alpha } $$

Clearly $$a$$ and $$b < 1$$ but $$c > 1$$

as $$\,\,\,\sin \alpha > 0$$ and $$\cos \alpha > 0$$

$$\therefore$$ $$c$$ is the greatest side and greatest angle is $$C$$

$$\therefore$$ $$\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}$$

$$ = {{{{\sin }^2}\alpha + {{\cos }^2}\alpha - 1 - \sin \alpha \cos \alpha } \over {2\,\sin \alpha \cos \alpha }}$$

$$ = - {1 \over 2}$$

$$\therefore$$ $$C = {120^ \circ }$$

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Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations