Application of Derivatives — Page 7 | JEE Mathematics

Q61

If the curves y2 = 6x, 9x2 + by2 = 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

Correct Answer

Option A

Solution

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

-1

. Let m1, and m2 are the tangent of the slope of the two curves respectively

\therefore\,\,\,

m1 m2 = $-$ 1. Now let they intersect at point (x1, y1)

\therefore\,\,\,

y_1^2 = 6x,

and

9x_1^2 + b\,y_1^2 = 16

y2 = 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}

9x2 + by2 = 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 m1 m2 = $-$ 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}

Q62

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

A

2\sqrt 2

B 3

C -3

D

-2\sqrt 2

Correct Answer

Option A

Solution

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

Q63

If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :

A

6\sqrt 2 \pi

B

6\sqrt 3 \pi

C

8\sqrt 2 \pi

D

8\sqrt 3 \pi

Correct Answer

Option D

Solution

Sphere of radius r = 3 cm Let b, h be base radius and height of cone respectively. So, volume of cone =

{1 \over 2}

$\pi$ b2h In right

\Delta

ABC by Pythagoras theorem (h $-$ r)2 + b2 = r2 $\Rightarrow$ b2 = r2 $-$ (h $-$ r)2 = r2 $-$ (h2 $-$ 2hr + r2) = 2hr $-$ h2

\therefore\,\,\,

Volume (v) =

{1 \over 3}

$\pi$ h[2hr $-$ h2] =

{1 \over 3}

[ 2h2r $-$ h3]

{{dv} \over {dh}}

=

{1 \over 3}

[4hr $-$ 3h2] = 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 + b2 = r2 $\Rightarrow$ b2 = 2hr $-$ h2 = 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

$\pi$ cm2

Q64

Let

f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10

,

x \in [ - 1,1]

. If [a, b] is the range of the function f, then 4a

-

b is equal to :

A 11

B 11

-

\pi

C 11 +

\pi

D 15

-

\pi

Correct Answer

Option B

Solution

f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10\,\forall x \in [ - 1,1]

\Rightarrow f'(x) = - {2 \over {\sqrt {1 - {x^2}} }} - {4 \over {1 + {x^2}}} - 6x - 2 So f(x) is decreasing function and range of f(x) is [f(1), f(

-

1)], which is [

\pi

+ 5, 5

\pi

+ 9] Now

4a - b = 4(\pi + 5) - (5\pi + 9)

= 11 - \pi $$

Q65

If the tangent to the curve

y = {x \over {{x^2} - 3}}

,

x \in \rho ,\left( {x \ne \pm \sqrt 3 } \right)

, at a point (

\alpha

,

\beta

)

\ne

(0, 0) on it is parallel to the line 2x + 6y – 11 = 0, then :

A | 6

\alpha

+ 2

\beta

| = 9

B | 2

\alpha

+ 6

\beta

| = 11

C | 2

\alpha

+ 6

\beta

| = 19

D | 6

\alpha

+ 2

\beta

| = 19

Correct Answer

Option D

Solution

{{dy} \over {dx}}{|_{(\alpha ,\beta )}} = {{ - {\alpha ^2} - 3} \over {{{\left( {{\alpha ^2} - 3} \right)}^2}}}

Given that

{{ - {\alpha ^2} - 3} \over {{{\left( {{\alpha ^2} - 3} \right)}^2}}} = - {1 \over 3}

$\Rightarrow$ $\alpha$ = 0, $\pm$ 3 (

\alpha \ne

0)

Q66

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is

{\tan ^{ - 1}}\left( {{1 \over 2}} \right)

. Water is poured into it at a constant rate of 5 cubic meter per minute. The the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10m; is :-

A

{1 \over {15\pi }}

B

{1 \over {5\pi }}

C

{1 \over {10\pi }}

D

{2 \over \pi }

Correct Answer

Option B

Solution

Given

{{dv} \over {dt}}

= 5 cm3/min and $\theta$ =

{\tan ^{ - 1}}\left( {{1 \over 2}} \right)

$\Rightarrow$ tan $\theta$ =

{1 \over 2}

=

{r \over h}

$\Rightarrow$ h = 2r Volume of the cone, v =

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

=

{1 \over 3}\pi {\left( {{h \over 2}} \right)^2}h

=

{{\pi {h^3}} \over {12}}

$\therefore$

{{dv} \over {dt}} = {\pi \over {12}}\left( {3{h^2}} \right){{dh} \over {dt}}

$\Rightarrow$ 5 =

{\pi \over 4}{h^2}{{dh} \over {dt}}

$\Rightarrow$ 5 =

{\pi \over 4}{\left( {10} \right)^2}{{dh} \over {dt}}

$\Rightarrow$

{{dh} \over {dt}}

=

{1 \over {5\pi }}

Q67

If the tangent to the curve, y = x3 + ax – b at the point (1, –5) is perpendicular to the line, –x + y + 4 = 0, then which one of the following points lies on the curve ?

A (2, –2)

B (2, –1)

C (–2, 2)

D (–2, 1)

Correct Answer

Option A

Solution

Slope of the tangent to the curve y = x3 + ax – b at point (1, –5) m1 =

{\left. {{{dy} \over {dx}}} \right|_{\left( {1, - 5} \right)}}

= 3x2 + a = 3 + a Slope of the line –x + y + 4 = 0, m2 = 1 As line and tangent to the curve are perpendicular to each other, $\therefore$ m1 $\times$ m2 = -1 $\Rightarrow$ (3 + a) $\times$ 1 = -1 $\Rightarrow$ a = - 4 $\therefore$ Curve becomes y = x3 - 4x – b This curve goes through (1, –5) $\therefore$ -5 = 1 - 4 - b $\Rightarrow$ b = 2 So curve is y = x3 - 4x – 2 By checking each options you can see, (2, –2) lies on the curve.

Q68

Let S be the set of all values of x for which the tangent to the curve y = ƒ(x) = x3 – x2 – 2x at (x, y) is parallel to the line segment joining the points (1, ƒ(1)) and (–1, ƒ(–1)), then S is equal to :

A

\left\{ { {1 \over 3}, - 1} \right\}

B

\left\{ { - {1 \over 3}, 1} \right\}

C

\left\{ { - {1 \over 3}, - 1} \right\}

D

\left\{ { {1 \over 3}, 1} \right\}

Correct Answer

Option B

Solution

Given ƒ(x) = x3 – x2 – 2x $\therefore$ ƒ(1) = 1 – 1 – 2 = - 2 and ƒ(-1) = -1 – 1 + 2 = 0 So point A(1, ƒ(1)) = (1, -2) and point B(–1, ƒ(–1)) = (-1, 0) Slope of tangent at point (x, y) to the curve y = ƒ(x) = x3 – x2 – 2x is

{{dy} \over {dx}}

= 3x2 - 2x - 2 Slope of line segment joining the points (1, -2) and (–1, 0) is =

{{ - 2 - 0} \over {1 - \left( { - 1} \right)}}

As tangent to the curve and line segment both are parallel then slope of them are same. $\therefore$ 3x2 - 2x - 2 =

{{ - 2 - 0} \over {1 - \left( { - 1} \right)}}

$\Rightarrow$ 3x2 - 2x - 1 = 0 $\Rightarrow$ x =

{{2 \pm \sqrt {4 - 4.3\left( { - 1} \right)} } \over {2.3}}

=

{{2 \pm 4} \over 6}

$\therefore$ x = 1,

- {1 \over 3}

So, S =

\left\{ { - {1 \over 3}, 1} \right\}

Q69

If ƒ(x) is a non-zero polynomial of degree four, having local extreme points at x = –1, 0, 1; then the set S = {x

\in

R : ƒ(x) = ƒ(0)} Contains exactly :

A four rational numbers.

B four irrational numbers.

C two irrational and one rational number.

D two irrational and two rational numbes.

Correct Answer

Option C

Solution

Local extreme points of f(x) is at x = –1, 0, 1. $\therefore$ f'(x) = 0 has three solutions x = –1, 0, 1. $\therefore$ f'(x) = k(x + 1)x(x - 1)

\int {f'(x)dx} = \int {k(x + 1)x(x - 1)dx}

f(x) =

k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C

Also given that ƒ(x) = ƒ(0) $\therefore$

k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C

= C $\Rightarrow$

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

= 0 $\Rightarrow$ x = 0,

- \sqrt 2

,

\sqrt 2

$\therefore$ x has two irrational and one rational value.

Q70

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is

A

\sqrt 3

B

2\sqrt 3

C

\sqrt 6

D

{2 \over 3} {\sqrt 3}

Correct Answer

Option B

Solution

Here r = 3 cos $\theta$ and

{h \over 2}

= 3 sin $\theta$ $\Rightarrow$ h = 6 sin $\theta$ We know, Volume of cylinder V = $\pi$ r2 h $\Rightarrow$ V = $\pi$ (9cos2 $\theta$ )(6 sin $\theta$ ) $\Rightarrow$ V = 54 $\pi$ (sin $\theta$ - sin3 $\theta$ ) For maxima/minima of volume,

{{dV} \over {dh}} = 0

$\Rightarrow$ cos $\theta$ - 3sin2 $\theta$ cos $\theta$ = 0 $\Rightarrow$ cos $\theta$ (1 - 3sin2 $\theta$ ) = 0 $\Rightarrow$ cos $\theta$ = 0 $\Rightarrow$ $\theta$ =

{\pi \over 2}

(not possible) or 1 - 3sin2 $\theta$ = 0 $\Rightarrow$ sin $\theta$ =

{1 \over {\sqrt 3 }}

$\therefore$ h = 6 sin $\theta$ =

{6 \over {\sqrt 3 }}

=

2\sqrt 3