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 curve satisfying the differential equation, ydx $$-$$(x + 3y^{2})dy = 0 and passing through the point (1, 1), also passes through the point :

A

$$\left( {{1 \over 4}, - {1 \over 2}} \right)$$

B

$$\left( { - {1 \over 3},{1 \over 3}} \right)$$

C

$$\left( {{1 \over 3}, - {1 \over 3}} \right)$$

D

$$\left( {{1 \over 4}, {1 \over 2}} \right)$$

Given,

y dx = $$\left( {x + 3{y^2}} \right)dy$$

$$ \Rightarrow $$$$\,\,\,$$ y $${{dx} \over {dy}}$$ = x + 3y^{2}

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

If = $${e^{ - \int {{1 \over y}dy} }}$$ = $${e^{ - \ln y}}$$ = $${1 \over y}$$

$$\therefore\,\,\,$$ Soluation is ,

x . $${1 \over y}$$ = $$\int {3y.{1 \over y}dy} $$

$$ \Rightarrow $$$$\,\,\,$$ $${x \over y}$$ = 3y + c

This curve passing through (1, 1)

$$\therefore\,\,\,$$ 1 = 3 + c

$$ \Rightarrow $$$$\,\,\,$$ c = $$-$$ 2

$$\therefore\,\,\,$$ Curve is, x = 3y^{2} $$-$$ 2y

Now put every point in this equation, and see which point satisfy this equation.

Following this method you can see ($$-$$ $${1 \over 3}$$, $${1 \over 3}$$) point satisfy this equation.

y dx = $$\left( {x + 3{y^2}} \right)dy$$

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

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

If = $${e^{ - \int {{1 \over y}dy} }}$$ = $${e^{ - \ln y}}$$ = $${1 \over y}$$

$$\therefore\,\,\,$$ Soluation is ,

x . $${1 \over y}$$ = $$\int {3y.{1 \over y}dy} $$

$$ \Rightarrow $$$$\,\,\,$$ $${x \over y}$$ = 3y + c

This curve passing through (1, 1)

$$\therefore\,\,\,$$ 1 = 3 + c

$$ \Rightarrow $$$$\,\,\,$$ c = $$-$$ 2

$$\therefore\,\,\,$$ Curve is, x = 3y

Now put every point in this equation, and see which point satisfy this equation.

Following this method you can see ($$-$$ $${1 \over 3}$$, $${1 \over 3}$$) point satisfy this equation.

2

If the curves y^{2} = 6x, 9x^{2} + by^{2} = 16 intersect each other at right angles, then the value of b is :

A

$${9 \over 2}$$

B

6

C

$${7 \over 2}$$

D

4

When two curves intersect each other at right angle, then at the point of intersection the product of tangent of slopes = $$-1$$.

Let m_{1}, and m_{2} are the tangent of the slope of the two curves respectively

$$\therefore\,\,\,$$ m_{1} m_{2} = $$-$$ 1.

Now let they intersect at point (x_{1}, y_{1})

$$\therefore\,\,\,$$ $$y_1^2 = 6x,$$ and $$9x_1^2 + b\,y_1^2 = 16$$

y^{2} = 6x

$$ \Rightarrow \,\,\,\,2y{{dy} \over {dx}} = 6$$

$$ \Rightarrow \,\,\,\,{{dy} \over {dx}} = {3 \over y}$$

$$\therefore\,\,\,$$ $${\left( {{{dy} \over {dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} = {3 \over {{y_1}}} = {m_1}$$

9x^{2} + by^{2} = 16

$$ = 18x + 2by{{dy} \over {dx}} = O$$

$$ \Rightarrow \,\,\,\,{\left( {{{dy} \over {dx}}} \right)_{({x_1},{y_1})}} = - {{9{x_1}} \over {b{y_1}}} = {m_2}$$

As m_{1} m_{2} = $$-$$1

$$\therefore\,\,\,$$ $${3 \over {y{}_1}} \times - {{9{x_1}} \over {b{y_1}}} = - 1$$

$$ \Rightarrow \,\,\,\,27{x_1} = by_1^2$$

$$ \Rightarrow \,\,\,\,\,27{x_1} = b.6{x_1}$$ $$\,\,\,$$ [as $$y_1^2 = 6{x_1}\,]$$

$$ \Rightarrow \,\,\,\,b = {{27} \over 6} = {9 \over 2}$$

Let m

$$\therefore\,\,\,$$ m

Now let they intersect at point (x

$$\therefore\,\,\,$$ $$y_1^2 = 6x,$$ and $$9x_1^2 + b\,y_1^2 = 16$$

y

$$ \Rightarrow \,\,\,\,2y{{dy} \over {dx}} = 6$$

$$ \Rightarrow \,\,\,\,{{dy} \over {dx}} = {3 \over y}$$

$$\therefore\,\,\,$$ $${\left( {{{dy} \over {dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} = {3 \over {{y_1}}} = {m_1}$$

9x

$$ = 18x + 2by{{dy} \over {dx}} = O$$

$$ \Rightarrow \,\,\,\,{\left( {{{dy} \over {dx}}} \right)_{({x_1},{y_1})}} = - {{9{x_1}} \over {b{y_1}}} = {m_2}$$

As m

$$\therefore\,\,\,$$ $${3 \over {y{}_1}} \times - {{9{x_1}} \over {b{y_1}}} = - 1$$

$$ \Rightarrow \,\,\,\,27{x_1} = by_1^2$$

$$ \Rightarrow \,\,\,\,\,27{x_1} = b.6{x_1}$$ $$\,\,\,$$ [as $$y_1^2 = 6{x_1}\,]$$

$$ \Rightarrow \,\,\,\,b = {{27} \over 6} = {9 \over 2}$$

3

Let $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$,

$$x \in R - \left\{ { - 1,0,1} \right\}$$.

If $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$, then the local minimum value of h(x) is

$$x \in R - \left\{ { - 1,0,1} \right\}$$.

If $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$, then the local minimum value of h(x) is

A

$$2\sqrt 2 $$

B

3

C

-3

D

$$-2\sqrt 2 $$

Given $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$

As $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$

= $${{{x^2} + {1 \over {{x^2}}}} \over {x - {1 \over x}}}$$

= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2.x.{1 \over x}} \over {x - {1 \over x}}}$$

= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2} \over {x - {1 \over x}}}$$

Let $${x - {1 \over x} = t}$$

So $$f\left( t \right) = {{{t^2} + 2} \over t}$$ = $$t + {2 \over t}$$

then $$f'\left( t \right) = 1 - {2 \over {{t^2}}}$$

At maximum or minimum $$f'\left( t \right) = 0$$.

$$\therefore$$ $$1 - {2 \over {{t^2}}} = 0$$

$$ \Rightarrow t = \pm \sqrt 2 $$

Now we need to find $$f''\left( t \right)$$ which will tell at which point among $$ + \sqrt 2 $$ and $$ - \sqrt 2 $$ will be maximum and minimum.

$$f''\left( t \right) = + {4 \over {{t^3}}}$$

So when $$t = + \sqrt 2 $$, then $$f''\left( t \right) = + {4 \over {{{\left( {\sqrt 2 } \right)}^3}}}$$ = $$ + \sqrt 2 $$

As $$f''\left( t \right) > 0$$ when $$t = + \sqrt 2 $$ then at $$t = + \sqrt 2 $$ the function $$f\left( t \right)$$ will be minimum.

Min of $$f\left( t \right) = {{{{\left( {\sqrt 2 } \right)}^2} + 2} \over {\sqrt 2 }}$$ = $${4 \over {\sqrt 2 }}$$ = $$2\sqrt 2 $$

So local minimum of $$f\left( t \right)$$ = $$2\sqrt 2 $$

As $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$

= $${{{x^2} + {1 \over {{x^2}}}} \over {x - {1 \over x}}}$$

= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2.x.{1 \over x}} \over {x - {1 \over x}}}$$

= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2} \over {x - {1 \over x}}}$$

Let $${x - {1 \over x} = t}$$

So $$f\left( t \right) = {{{t^2} + 2} \over t}$$ = $$t + {2 \over t}$$

then $$f'\left( t \right) = 1 - {2 \over {{t^2}}}$$

At maximum or minimum $$f'\left( t \right) = 0$$.

$$\therefore$$ $$1 - {2 \over {{t^2}}} = 0$$

$$ \Rightarrow t = \pm \sqrt 2 $$

Now we need to find $$f''\left( t \right)$$ which will tell at which point among $$ + \sqrt 2 $$ and $$ - \sqrt 2 $$ will be maximum and minimum.

$$f''\left( t \right) = + {4 \over {{t^3}}}$$

So when $$t = + \sqrt 2 $$, then $$f''\left( t \right) = + {4 \over {{{\left( {\sqrt 2 } \right)}^3}}}$$ = $$ + \sqrt 2 $$

As $$f''\left( t \right) > 0$$ when $$t = + \sqrt 2 $$ then at $$t = + \sqrt 2 $$ the function $$f\left( t \right)$$ will be minimum.

Min of $$f\left( t \right) = {{{{\left( {\sqrt 2 } \right)}^2} + 2} \over {\sqrt 2 }}$$ = $${4 \over {\sqrt 2 }}$$ = $$2\sqrt 2 $$

So local minimum of $$f\left( t \right)$$ = $$2\sqrt 2 $$

4

Let S = { t $$ \in R:f(x) = \left| {x - \pi } \right|.\left( {{e^{\left| x \right|}} - 1} \right)$$$$\sin \left| x \right|$$ is not differentiable at t}, then the set S is equal to

A

{0, $$\pi $$}

B

$$\phi $$ (an empty set)

C

{0}

D

{$$\pi $$}

Check differtiability at x = $$\pi $$ and x = 0

**at x = 0 : **

We have L. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {0 - h} \right) - f\left( 0 \right)} \over { - h}}$$

= $$\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| h \right| - 0} \over { - h}} = 0$$

R. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {0 + h} \right) - f\left( 0 \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - o} \over h}$$

= 0

$$\therefore,\,\,$$ LHD = RHD

Therefore, function is differentiable at x = $$\pi $$.

**at x = $$\pi $$ : **

L. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi - h} \right) - f\left( \pi \right)} \over { - h}}$$

= $$\mathop {\lim }\limits_{h \to 0} {{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {\pi - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0.} \over { - h}}$$

= 0

RH. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi + h} \right) - f\left( \pi \right)} \over h}$$

= $$\mathop {\lim }\limits_{h \to 0} {{\left| {\pi + h - \pi } \right|\left( {{e^{\left| {\pi + h} \right|}} - 1} \right)\sin \left| {\pi + h} \right| - 0} \over h}$$

= 0

$$\therefore\,\,\,$$ L. H. D = R H D

Therefore, function is differentiable at x = $$\pi $$,

Since, the function f(x) is differentiable at all the points including 0 and $$\pi $$.

So, f(x) is differentiable everywhere.

Therefore, set S is empty set.

S = $$\phi $$

We have L. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {0 - h} \right) - f\left( 0 \right)} \over { - h}}$$

= $$\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| h \right| - 0} \over { - h}} = 0$$

R. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {0 + h} \right) - f\left( 0 \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - o} \over h}$$

= 0

$$\therefore,\,\,$$ LHD = RHD

Therefore, function is differentiable at x = $$\pi $$.

L. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi - h} \right) - f\left( \pi \right)} \over { - h}}$$

= $$\mathop {\lim }\limits_{h \to 0} {{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {\pi - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0.} \over { - h}}$$

= 0

RH. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi + h} \right) - f\left( \pi \right)} \over h}$$

= $$\mathop {\lim }\limits_{h \to 0} {{\left| {\pi + h - \pi } \right|\left( {{e^{\left| {\pi + h} \right|}} - 1} \right)\sin \left| {\pi + h} \right| - 0} \over h}$$

= 0

$$\therefore\,\,\,$$ L. H. D = R H D

Therefore, function is differentiable at x = $$\pi $$,

Since, the function f(x) is differentiable at all the points including 0 and $$\pi $$.

So, f(x) is differentiable everywhere.

Therefore, set S is empty set.

S = $$\phi $$

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Trigonometric Functions & Equations *keyboard_arrow_right*

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Complex Numbers *keyboard_arrow_right*

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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*

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*