Application of Derivatives — Page 19 | JEE Mathematics

Q181

If

\beta

is one of the angles between the normals to the ellipse, x2 + 3y2 = 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}

Correct Answer

Option A

Solution

Since, x2 + 3y2 = 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

= m1 &

{\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

= m2 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 }}

Q182

If m is the minimum value of k for which the function f(x) = x

\sqrt {kx - {x^2}}

is increasing in the interval [0,3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to :

A

\left( {5,3\sqrt 6 } \right)

B

\left( {4,3\sqrt 3 } \right)

C

\left( {4,3\sqrt 2 } \right)

D

\left( {3,3\sqrt 3 } \right)

Correct Answer

Option B

Solution

f\left( x \right) = x\sqrt {kx - {x^2}}

$\Rightarrow$

f'\left( x \right) = \sqrt {kx - {x^2}} + {{(k - 2x)x} \over {2\sqrt {kx - {x^2}} }}

\Rightarrow {{2\left( {kx - {x^2}} \right) + kx - 2{x^2}} \over {2\sqrt {kx - {x^2}} }} = {{3kx - 4{x^2}} \over {2\sqrt {kx - {x^2}} }}

\Rightarrow {{x(3k - 4x)} \over {2\sqrt {kx - {x^2}} }}

Now for increasing function for f'(x) $\ge$ 0,

\forall x \in [0,3]

\Rightarrow kx - {x^2} \ge 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(3k - 4x) \ge 0,\,\forall x \in \left[ {0,3} \right]\,

\Rightarrow x(x - k) \le 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(4x - 3k) \le 0,\,\forall x \in \left[ {0,3} \right]\,

k $\ge$ 3 and k $\ge$ 4 $\Rightarrow$ k $\ge$ 4 $\Rightarrow$ m = 4 So f(x) is maximum when k = 4 then

3\sqrt {4 \times 3 - {3^2}} = 3\sqrt 3 = M

$\therefore$ (m, M) = (4,

3\sqrt 3

)

Q183

The maximum volume (in cu.m) of the right circular cone having slant height 3 m is :

A 2

\sqrt3

\pi

B 3

\sqrt3

\pi

C 6

\pi

D

{4 \over 3}\pi

Correct Answer

Option A

Solution

$\therefore$ h = 3 cos $\theta$ r = 3 sin $\theta$ We know volume of right circular cone, V =

{1 \over 3}\pi {r^2}h

=

{1 \over 3}\pi

(3 sin $\theta$ )2 3 cos $\theta$ = 9 $\pi$ sin2 $\theta$ cos $\theta$ For maximum or minimum value of volume,

{{dv} \over {d\theta }}

= 0 $\therefore$ (2sin $\theta$ cos $\theta$ ) cos $\theta$ + 3sin2 $\theta$ .( $-$ sin $\theta$ ) = 0 $\Rightarrow$ 2 sin $\theta$ cos2 $\theta$ $-$ sin3 $\theta$ = 0 $\Rightarrow$ 2 sin $\theta$ (1 $-$ sin2 $\theta$ ) $-$ sin3 $\theta$ = 0 $\Rightarrow$ 2 sin $\theta$ $-$ 2 sin3 $\theta$ $-$ sin3 $\theta$ = 0 $\Rightarrow$ 3 sin3 $\theta$ = 2 sin $\theta$ $\Rightarrow$ sin2 $\theta$ =

{2 \over 3}

$\Rightarrow$ sin $\theta$ =

\sqrt {{2 \over 3}}

{{{d^2}v} \over {d{\theta ^2}}}

= 2cos $\theta$ $-$ 3(3sin $\theta$ cos $\theta$ ) = 2 cos $\theta$ $-$ 9 sin $\theta$ cos $\theta$ = 2 $\times$

{1 \over {\sqrt 3 }}

$-$ 9 $\times$

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

$\times$

{1 \over {\sqrt 3 }}

=

{2 \over {\sqrt 3 }} - 3\sqrt 2 \, < 0

$\therefore$ Volume is maximum when sin $\theta$ =

\sqrt {{2 \over 3}}

$\therefore$ Maximum volume is = 9 $\pi$

{\left( {\sqrt {{2 \over 3}} } \right)^2} \times {1 \over {\sqrt 3 }}

= 9 $\pi$ $\times$

{2 \over 3} \times {1 \over {\sqrt 3 }}

=

2\sqrt 3 \,\pi

Q184

A helicopter is flying along the curve given by y – x3/2 = 7, (x

\ge

0). A soldier positioned at the point

\left( {{1 \over 2},7} \right)

wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is -

A

{1 \over 6}\sqrt {{7 \over 3}}

B

{{\sqrt 5 } \over 6}

C

{1 \over 2}

D

{1 \over 3}

\sqrt {{7 \over 3}}

Correct Answer

Option A

Solution

y - {x^{3/2}} = 7\left( {x \ge 0} \right)

{{dy} \over {dx}} = {3 \over 2}{x^{1/2}}

\left( {{3 \over 2}\sqrt x } \right)\left( {{{7 - y} \over {{1 \over 2} - x}}} \right) = - 1

\left( {{3 \over 2}\sqrt x } \right)\left( {{{ - {x^{3/2}}} \over {{1 \over 2} - x}}} \right) = - 1

{3 \over 2}.{x^2} = {1 \over 2} - x

3{x^2} = 1 - 2x

3{x^2} + 2x - 1 = 0

3{x^2} + 3x - x - 1 = 0

\left( {x + 1} \right)\left( {3x - 1} \right) = 0

$\therefore$

x = - 1

(rejected)

x = {1 \over 3}

y = 7 + {x^{3/2}} = 7 + {\left( {{1 \over 3}} \right)^{3/2}}

{\ell _{AB}} = \sqrt {{{\left( {{1 \over 2} - {1 \over 3}} \right)}^2} + {{\left( {{1 \over 3}} \right)}^3}} = \sqrt {{1 \over {36}} + {1 \over {27}}}

= \sqrt {{{3 + 4} \over {9 \times 12}}}

= \sqrt {{7 \over {108}}} = {1 \over 6}\sqrt {{7 \over 3}}

Q185

If the function f given by f(x) = x3 – 3(a – 2)x2 + 3ax + 7, for some a

\in

R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation,

{{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)

is :

A

-

7

B 5

C 7

D 6

Correct Answer

Option C

Solution

f '(x) = 3x2 $-$ 6(a $-$ 2)x + 3a f '(x) $\ge$ 0 $\forall$ x

\in

(0, 1] f '(x) $\le$ 0 $\forall$ x

\in

[1, 5) $\Rightarrow$ f '(x) = 0 at x = 1 $\Rightarrow$ a = 5 f(x) $-$ 14 = (x $-$ 1)2 (x $-$ 7)

{{f(x) - 14} \over {{{\left( {x - 1} \right)}^2}}} = x - 7

Q186

The tangent to the curve y = x2 – 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point :

A

\left\{ {{1 \over 4},{7 \over 2}} \right\}

B

\left( { - {1 \over 8},7} \right)

C

\left( {{7 \over 2},{1 \over 4}} \right)

D

\left( {{1 \over 8}, - 7} \right)

Correct Answer

Option D

Solution

y = x2 $-$ 5x + 5

{{dy} \over {dx}} = 2x - 5 = 2 \Rightarrow x = {7 \over 2}

at x =

{7 \over 2}

, y =

{{ - 1} \over 4}

Equation of tangent at

\left( {{7 \over 2},{{ - 1} \over 4}} \right)

is 2x $-$ y $-$

{{29} \over 4}

= 0 Now check options x =

{1 \over 8}

, y = $-$ 7

Q187

The shortest distance between the point

\left( {{3 \over 2},0} \right)

and the curve y =

\sqrt x

, (x > 0), is -

A

{{\sqrt 3 } \over 2}

B

{5 \over 4}

C

{3 \over 2}

D

{{\sqrt 5 } \over 2}

Correct Answer

Option D

Solution

Let points

\left( {{3 \over 2},0} \right),\left( {{t^2},t} \right),t > 0

Distance =

\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}}

=

\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}}

So minimum distance is

\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}

Q188

Let C be a curve given by y(x) = 1 +

\sqrt {4x - 3} ,x > {3 \over 4}.

If P is a point on C, such that the tangent at P has slope

{2 \over 3}

, then a point through which the normal at P passes, is :

A (2, 3)

B (4,

-

3)

C (1, 7)

D (3,

-

4),

Correct Answer

Option C

Solution

Given, y = 1 +

\sqrt {4x - 3}

$\therefore$

{{dy} \over {dx}}

=

{1 \over {2\sqrt {4x - 3} }} \times 4 = {2 \over 3}

$\Rightarrow$ 4x $-$ 3 = 9 $\Rightarrow$ x = 3 $\therefore$ y = 1 +

\sqrt {12 - 3}

= 4 $\therefore$ Equation of normal at point P(3,4) y $-$ 4 = $-$

{3 \over 2}

(x $-$ 3) $\Rightarrow$ 2y $-$ 8 = $-$ 3x + 9 $\Rightarrow$ 3x + 2y $-$ 17 = 0