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 function f defined by

f(x) = x^{3} $$-$$ 3x^{2} + 5x + 7 , is :

f(x) = x

A

increasing in **R**.

B

decreasing in **R**.

C

decreasing in (0, $$\infty $$) and increasing in ($$-$$ $$\infty $$, 0)

D

increasing in (0, $$\infty $$) and decreasing in ($$-$$ $$\infty $$, 0)

2

If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm^{2}) of this cone is :

A

$$6\sqrt 2 \pi $$

B

$$6\sqrt 3 \pi $$

C

$$8\sqrt 2 \pi $$

D

$$8\sqrt 3 \pi $$

Sphere of radius r = 3 cm

Let b, h be base radius and height of cone respectively.

So, volume of cone = $${1 \over 2}$$ $$\pi $$b^{2}h

In right $$\Delta $$ABC by Pythagoras theorem

(h $$-$$ r)^{2} + b^{2} = r^{2}

$$ \Rightarrow $$ b^{2} = r^{2} $$-$$ (h $$-$$ r)^{2} = r^{2} $$-$$ (h^{2} $$-$$ 2hr + r^{2}) = 2hr $$-$$ h^{2}

$$\therefore\,\,\,$$ Volume (v) = $${1 \over 3}$$ $$\pi $$h[2hr $$-$$ h^{2}] = $${1 \over 3}$$ [ 2h^{2}r $$-$$ h^{3}]

$${{dv} \over {dh}}$$ = $${1 \over 3}$$ [4hr $$-$$ 3h^{2}] = 0

$$ \Rightarrow $$ h (4r $$-$$ 3h) = 0

$${{{d^2}v} \over {d{h^2}}}$$ = $${1 \over 3}$$ [4r $$-$$ 6h]

At h = $${{4r} \over 3}$$, $${{{d^2}y} \over {d{h^2}}}$$ = $${1 \over 3}$$ $$\left[ {4r - {{4r} \over 3} \times 6} \right] = {1 \over 3}\left[ {4r - 8r} \right] < 0$$

$$ \Rightarrow $$ maximum volume cours at h = $${{4r} \over 3}$$ = $${4 \over 3}$$ $$ \times $$ 3 = 4 cm

As from (1),

(h $$-$$ r)^{2} + b^{2} = r^{2}

$$ \Rightarrow $$ b^{2} = 2hr $$-$$ h^{2} = 2.$${{4r} \over 3}$$ r $$-$$ $${{16{r^2}} \over 9}$$ = $${{8{r^2}} \over 3}$$ $$-$$ $${{16{r^2}} \over 9}$$

= $${{\left( {24 - 16} \right){r^2}} \over 9}$$ = $${{8{r^2}} \over 9}$$

$$ \Rightarrow $$ b = $${{2\sqrt 2 } \over 3}$$ r = 2 $$\sqrt 2 \,\,m$$

Therefore curved surface area = $$\pi bl$$

= $$\pi $$b$$\sqrt {{h^2} + {r^2}} $$ = $$\pi $$2$$\sqrt 2 $$ $$\sqrt {{4^2} + 8} $$ = 8$$\sqrt 3 $$$$\pi $$ cm^{2}

Let b, h be base radius and height of cone respectively.

So, volume of cone = $${1 \over 2}$$ $$\pi $$b

In right $$\Delta $$ABC by Pythagoras theorem

(h $$-$$ r)

$$ \Rightarrow $$ b

$$\therefore\,\,\,$$ Volume (v) = $${1 \over 3}$$ $$\pi $$h[2hr $$-$$ h

$${{dv} \over {dh}}$$ = $${1 \over 3}$$ [4hr $$-$$ 3h

$$ \Rightarrow $$ h (4r $$-$$ 3h) = 0

$${{{d^2}v} \over {d{h^2}}}$$ = $${1 \over 3}$$ [4r $$-$$ 6h]

At h = $${{4r} \over 3}$$, $${{{d^2}y} \over {d{h^2}}}$$ = $${1 \over 3}$$ $$\left[ {4r - {{4r} \over 3} \times 6} \right] = {1 \over 3}\left[ {4r - 8r} \right] < 0$$

$$ \Rightarrow $$ maximum volume cours at h = $${{4r} \over 3}$$ = $${4 \over 3}$$ $$ \times $$ 3 = 4 cm

As from (1),

(h $$-$$ r)

$$ \Rightarrow $$ b

= $${{\left( {24 - 16} \right){r^2}} \over 9}$$ = $${{8{r^2}} \over 9}$$

$$ \Rightarrow $$ b = $${{2\sqrt 2 } \over 3}$$ r = 2 $$\sqrt 2 \,\,m$$

Therefore curved surface area = $$\pi bl$$

= $$\pi $$b$$\sqrt {{h^2} + {r^2}} $$ = $$\pi $$2$$\sqrt 2 $$ $$\sqrt {{4^2} + 8} $$ = 8$$\sqrt 3 $$$$\pi $$ cm

3

If $$\beta $$ is one of the angles between the normals to the ellipse, x^{2} + 3y^{2} = 9 at the points (3 cos $$\theta $$, $$\sqrt 3 \sin \theta $$) and ($$-$$ 3 sin $$\theta $$, $$\sqrt 3 \,\cos \theta $$); $$\theta \in \left( {0,{\pi \over 2}} \right);$$ then $${{2\,\cot \beta } \over {\sin 2\theta }}$$ is equal to :

A

$${2 \over {\sqrt 3 }}$$

B

$${1 \over {\sqrt 3 }}$$

C

$$\sqrt 2 $$

D

$${{\sqrt 3 } \over 4}$$

Since, x^{2} + 3y^{2} = 9

$$ \Rightarrow $$ 2x + 6y $${{dy} \over {dx}}$$ = 0

$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{ - x} \over {3y}}$$

Slope of normal is $$-$$ $${{dx} \over {dy}}$$ = $${{3y} \over x}$$

$$ \Rightarrow $$ $${\left( { - {{dx} \over {dy}}} \right)_{\left( {3\cos \theta ,\sqrt 3 \sin \theta } \right)}}$$

= $${{3\sqrt 3 \sin \theta } \over {3\cos \theta }}$$ = $$\sqrt 3 \tan \theta $$ = m_{1}

& $${\left( { - {{dx} \over {dy}}} \right)_{\left( { - 3\sin \theta ,\sqrt 3 \cos \theta } \right)}}$$

= $${{3\sqrt 3 \cos \theta } \over { - 3\sin \theta }}$$ = $$ - \sqrt 3 \cot \theta $$ = m_{2}

As, $$\beta $$ is the angle between the normals to the given ellipse then

tan$$\beta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$

= $$\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3\tan \theta \cot \theta }}} \right|$$ = $$\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3}}} \right|$$

So, tan $$\beta $$ = $${{\sqrt 3 } \over 2}$$ $$\left| {\tan \theta + \cot \theta } \right|$$

$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}\left| {{{\sin \theta } \over {\cos \theta }} + {{\cos \theta } \over {\sin \theta }}} \right|$$

$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}$$ $$\left| {{1 \over {\sin \theta \cos \theta }}} \right|$$

$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over {\sin 2\theta }}$$

$$ \Rightarrow $$ $${{2\cot \beta } \over {\sin 2\theta }}$$ = $${2 \over {\sqrt 3 }}$$

$$ \Rightarrow $$ 2x + 6y $${{dy} \over {dx}}$$ = 0

$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{ - x} \over {3y}}$$

Slope of normal is $$-$$ $${{dx} \over {dy}}$$ = $${{3y} \over x}$$

$$ \Rightarrow $$ $${\left( { - {{dx} \over {dy}}} \right)_{\left( {3\cos \theta ,\sqrt 3 \sin \theta } \right)}}$$

= $${{3\sqrt 3 \sin \theta } \over {3\cos \theta }}$$ = $$\sqrt 3 \tan \theta $$ = m

& $${\left( { - {{dx} \over {dy}}} \right)_{\left( { - 3\sin \theta ,\sqrt 3 \cos \theta } \right)}}$$

= $${{3\sqrt 3 \cos \theta } \over { - 3\sin \theta }}$$ = $$ - \sqrt 3 \cot \theta $$ = m

As, $$\beta $$ is the angle between the normals to the given ellipse then

tan$$\beta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$

= $$\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3\tan \theta \cot \theta }}} \right|$$ = $$\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3}}} \right|$$

So, tan $$\beta $$ = $${{\sqrt 3 } \over 2}$$ $$\left| {\tan \theta + \cot \theta } \right|$$

$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}\left| {{{\sin \theta } \over {\cos \theta }} + {{\cos \theta } \over {\sin \theta }}} \right|$$

$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}$$ $$\left| {{1 \over {\sin \theta \cos \theta }}} \right|$$

$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over {\sin 2\theta }}$$

$$ \Rightarrow $$ $${{2\cot \beta } \over {\sin 2\theta }}$$ = $${2 \over {\sqrt 3 }}$$

4

Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x^{3} $$-$$ 9x^{2} + 12x + 5 in the interval [0, 3]. Then M $$-$$m is equal to :

A

5

B

9

C

4

D

1

Here, f(x) = 2x^{3} $$-$$ 9x^{2} + 12x + 5

$$ \Rightarrow $$ f'(x) = 6x^{2} $$-$$ 18x + 12 = 0

For maxima or minima put f'(x) = 0

$$ \Rightarrow $$ x^{2} $$-$$ 3x + 2 = 0

$$ \Rightarrow $$ x = 1 or x = 2

Now, f''(x) = 12x $$-$$ 18

$$ \Rightarrow $$ f''(1) = 12(1) $$-$$ 18 = $$-$$ 6 < 0

Hence, f(x) has maxima at x = 1

$$ \therefore $$ maximum value = M = f(1) = 2 $$-$$ 9 + 12 + 5 = 10

And, f''(2) = 12(2) $$-$$ 18 = 6 > 0.

Hence, f(x) has minima at x = 2.

$$ \therefore $$ minimum value = m = f(2) = 2(8) $$-$$ 9(4) + 12(2) + 5 = 9

$$ \therefore $$ M $$-$$ m = 10 $$-$$ 9 = 1

$$ \Rightarrow $$ f'(x) = 6x

For maxima or minima put f'(x) = 0

$$ \Rightarrow $$ x

$$ \Rightarrow $$ x = 1 or x = 2

Now, f''(x) = 12x $$-$$ 18

$$ \Rightarrow $$ f''(1) = 12(1) $$-$$ 18 = $$-$$ 6 < 0

Hence, f(x) has maxima at x = 1

$$ \therefore $$ maximum value = M = f(1) = 2 $$-$$ 9 + 12 + 5 = 10

And, f''(2) = 12(2) $$-$$ 18 = 6 > 0.

Hence, f(x) has minima at x = 2.

$$ \therefore $$ minimum value = m = f(2) = 2(8) $$-$$ 9(4) + 12(2) + 5 = 9

$$ \therefore $$ M $$-$$ m = 10 $$-$$ 9 = 1

Number in Brackets after Paper Name Indicates No of Questions

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JEE Main 2016 (Online) 9th April Morning Slot (2) *keyboard_arrow_right*

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*