Application of Derivatives — Page 9 | JEE Mathematics

Q81

The position of a moving car at time t is given by f(t) = at2 + bt + c, t > 0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t1 , t2 ] is attained at the point :

A

{{\left( {{t_1} + {t_2}} \right)} \over 2}

B

{{\left( {{t_2} - {t_1}} \right)} \over 2}

C 2a(t1 + t2) + b

D a(t2 – t1) + b

Correct Answer

Option A

Solution

Vav =

{{f\left( {{t_2}} \right) - f\left( {{t_1}} \right)} \over {{t_2} - {t_1}}}

= f'(t) $\Rightarrow$

{{a\left( {t_2^2 - t_1^2} \right) - b\left( {{t_2} - {t_1}} \right)} \over {{t_2} - {t_1}}}

= 2

a

t + b $\Rightarrow$ a(t2 + t1) + b = 2at + b $\Rightarrow$ t =

{{{t_1} + {t_2}} \over 2}

Q82

The set of all real values of

\lambda

for which the function

f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right),x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

has exactly one maxima and exactly one minima, is :

A

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

B

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

C

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

D

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

Correct Answer

Option A

Solution

f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right)

$\Rightarrow$ f(x) = sin2 x( $\lambda$ + sinx) ....(1) $\therefore$ f'(x) = 2sinx cosx ( $\lambda$ +sinx) + sin2x (cosx) $\Rightarrow$ f'(x) = sin2x(

{{2\lambda + 3\sin x} \over 2}

) For maixma and minima, f'(x) = 0 $\therefore$ sin2x = 0 or 2 $\lambda$ + 3sinx = 0 when sin2x = 0 $\Rightarrow$ x = 0 or when 2 $\lambda$ + 3sinx = 0 $\Rightarrow$ sin x =

- {{2\lambda } \over 3}

As

x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

$\therefore$ -1 < sinx < 1 $\Rightarrow$ -1 <

- {{2\lambda } \over 3}

< 1 $\Rightarrow$

- {3 \over 2}

< $\lambda$ <

{3 \over 2}

$\therefore$ $\lambda$

\in

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

- {0} Note : If $\lambda$ = 0 $\Rightarrow$ f(x) = sin3x [from (1)] Which is monotonic. so no maxima/minima.

Q83

If the tangent to the curve, y = f (x) = xloge x, (x > 0) at a point (c, f(c)) is parallel to the line-segment joining the points (1, 0) and (e, e), then c is equal to :

A

{{e - 1} \over e}

B

{e^{\left( {{1 \over {1 - e}}} \right)}}

C

{e^{\left( {{1 \over {e - 1}}} \right)}}

D

{1 \over {e - 1}}

Correct Answer

Option C

Solution

y = f (x) = xloge x $\Rightarrow$

{{dy} \over {dx}} =

1 + loge x $\Rightarrow$

{\left. {{{dy} \over {dx}}} \right|_{\left( {c,f\left( c \right)} \right)}}

= 1 + loge e = m1 This tangent parallel to the line-segment joining the points (1, 0) and (e, e).

$\therefore$ Slope of line-segment joining the points (1, 0) and (e, e) = m1 $\Rightarrow$

{{e - 0} \over {e - 1}}

= 1 + loge e $\Rightarrow$ loge e =

{{e - 0} \over {e - 1}}

- 1 =

{1 \over {e - 1}}

$\Rightarrow$ c =

{e^{\left( {{1 \over {e - 1}}} \right)}}

Q84

If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2; then p(0) is equal to :

A 6

B 12

C -12

D -24

Correct Answer

Option C

Solution

Since p(x) has relative extreme at x = 1 & 2 so p'(x) = 0 at x = 1 & 2 $\therefore$ Let p'(x) = A(x – 1) (x – 2) $\Rightarrow$ p(x) =

\int {A\left( {{x^2} - 3x + 2} \right)dx}

$\Rightarrow$ p(x) =

{A\left( {{{{x^2}} \over 3} - 3{x^2} + 2x} \right)}

+ C ...(1) As P(1) = 8 From (1)

8 = A\left( {{1 \over 3} - {3 \over 2} + 2} \right) + C

$\Rightarrow$ 8 =

{{5A} \over 6} + C

$\Rightarrow$ 48 = 5A + 5C ...(2) Also P(2) = 4 From (1)

4 = A\left( {{8 \over 3} - 6 + 4} \right) + C

$\Rightarrow$ 4 =

{{2A} \over 3} + C

$\Rightarrow$ 12 = 2A + 3C ......(3) Form (3) & (4), C = –12, A = 24 Now p(0) = C = -12

Q85

If the tangent to the curve y = x + sin y at a point (a, b) is parallel to the line joining

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

and

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

, then :

A b = a

B |b - a| = 1

C

b = {\pi \over 2}

+ a

D |a + b| = 1

Correct Answer

Option B

Solution

Slope of tangent to the curve y = x + siny at (a, b) is =

{{2 - {3 \over 2}} \over {{1 \over 2} - 0}}

= 1 Given curve y = x + sin y $\Rightarrow$

{{dy} \over {dx}} = 1 + \cos y.{{dy} \over {dx}}

$\Rightarrow$ (1 - cos y)

{{dy} \over {dx}}

= 1 $\Rightarrow$

{\left. {{{dy} \over {dx}}} \right|_{\left( {a,b} \right)}}

=

{1 \over {1 + \cos b}}

Now according to question,

{1 \over {1 + \cos b}} = 1

$\Rightarrow$ cos b = 0 $\Rightarrow$ sin b = $\pm$ 1 Point (a, b) lies on curve y = x + sin y $\therefore$ b = a + sin b $\Rightarrow$ |b - a| = |sin b| = 1

Q86

Let P(h, k) be a point on the curve y = x2 + 7x + 2, nearest to the line, y = 3x – 3. Then the equation of the normal to the curve at P is :

A x – 3y – 11 = 0

B x – 3y + 22 = 0

C x + 3y – 62 = 0

D x + 3y + 26 = 0

Correct Answer

Option D

Solution

Let L be the common normal to parabola y = x2 + 7x + 2 and line y = 3x – 3 Slope of tangent of y = x2 + 7x + 2 at P

{{{\left. {{{dy} \over {dx}}} \right|}_{P}}}

= 3 $\Rightarrow$ 2x + 7 = 3 $\Rightarrow$ x = -2 $\therefore$ y = -8 So P(–2, –8) Normal at P : x + 3y + C = 0 $\Rightarrow$ -2 + 3(-8) + C = 0 $\Rightarrow$ C = 26 $\therefore$ Normal : x + 3y + 26 = 0

Q87

A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness the melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which of the thickness of ice decreases, is :

A

{1 \over {18\pi }}

B

{1 \over {36\pi }}

C

{1 \over {54\pi }}

D

{5 \over {6\pi }}

Correct Answer

Option A

Solution

Let the thickness = h cm Volume of ice = v =

{{4\pi } \over 3}\left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)

$\Rightarrow$

{{dv} \over {dt}} = {{4\pi } \over 3}\left( {3{{\left( {10 + h} \right)}^2}} \right).{{dh} \over {dt}}

Given

{{dv} \over {dt}} =

50 cm3/min and h = 5 cm $\therefore$ 50 =

{{4\pi } \over 3}\left( {3{{\left( {10 + 5} \right)}^2}} \right).{{dh} \over {dt}}

$\Rightarrow$

{{dh} \over {dt}} = {{50} \over {4\pi \times {{15}^2}}}

=

{1 \over {18\pi }}

cm/min

Q88

The length of the perpendicular from the origin, on the normal to the curve, x2 + 2xy – 3y2 = 0 at the point (2,2) is

A

\sqrt 2

B

4\sqrt 2

C 2

D

2\sqrt 2

Correct Answer

Option D

Solution

x2 + 2xy – 3y2 = 0 Differentiate the curve 2x + 2y + 2xy' – 6yy' = 0 $\Rightarrow$ $\Rightarrow$ x + y + xy' – 3yy' = 0 $\Rightarrow$ y'(x – 3y) = – (x + y) $\Rightarrow$ y' =

{{x + y} \over {3y - x}}

Slope of normal =

- {{dx} \over {dy}}

=

{{x - 3y} \over {x + y}}

$\therefore$

{\left( { - {{dx} \over {dy}}} \right)_{\left( {2,2} \right)}}

=

{{2 - 6} \over {2 + 2}}

= -1 Normal at (2, 2) y – 2 = – 1 (x – 2) $\Rightarrow$ y + x = 4 $\therefore$ Perpendicular distance from (0,0) =

\left| {{{0 + 0 - 4} \over {\sqrt 2 }}} \right|

=

{2\sqrt 2 }

Q89

Let

\lambda x - 2y = \mu

be a tangent to the hyperbola

{a^2}{x^2} - {y^2} = {b^2}

. Then

{\left( {{\lambda \over a}} \right)^2} - {\left( {{\mu \over b}} \right)^2}

is equal to :

A

-

2

B

-

4

C 2

D 4

Correct Answer

Option D

Solution

$\dfrac{x^{2}}{\left(\dfrac{b^{2}}{a^{2}}\right)}-\dfrac{y^{2}}{b^{2}}=1$ Tangent in slope form $\Rightarrow y=m x \pm \sqrt{\dfrac{b^{2}}{a^{2}} m^{2}-b^{2}}$ i.e., same as $y=\dfrac{\lambda x}{2}-\dfrac{\mu}{2}$ Comparing coefficients,

\begin{aligned} &m=\frac{\lambda}{2}, \frac{b^{2}}{a^{2}} m^{2}-b^{2}=\frac{\mu^{2}}{4} \\\\ &\text{ Eliminating } m, \frac{b^{2}}{a^{2}} \cdot \frac{\lambda^{2}}{4}-b^{2}=\frac{\mu^{2}}{4} \\\\ &\Rightarrow \frac{\lambda^{2}}{a^{2}}-\frac{\mu^{2}}{b^{2}}=4 \end{aligned}

Q90

Let ƒ(x) = xcos–1(–sin|x|),

x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]

, then which of the following is true?

A ƒ' is decreasing in

\left( { - {\pi \over 2},0} \right)

and increasing in

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

B ƒ '(0) =

{ - {\pi \over 2}}

C ƒ is not differentiable at x = 0

D ƒ' is increasing in

\left( { - {\pi \over 2},0} \right)

and decreasing in

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

Correct Answer

Option A

Solution

We know, cos-1(-x) = $\pi$ - cos-1x $\therefore$ ƒ(x) = x( $\pi$ - cos–1(sin|x|)) = x( $\pi$ -

{\pi \over 2}

+ sin–1(sin|x|)) = x( $\pi$ -

{\pi \over 2}

+ sin–1(sin|x|)) = x

{\pi \over 2}

+ x|x| $\therefore$ f(x) =

\left\{ \begin{array}{ll}{x{\pi \over 2} - {x^2},} & {x < 0} \\ {x{\pi \over 2} + {x^2},} & {x \ge 0} \end{array} \right.

Now f'(x) =

\left\{ \begin{array}{ll}{{\pi \over 2} - 2x,} & {x < 0} \\ {{\pi \over 2} + 2x,} & {x \ge 0} \end{array} \right.

and f''(x) =

\left\{ \begin{array}{ll}{ - 2,} & {x < 0} \\ {2,} & {x \ge 0} \end{array} \right.

$\therefore$ ƒ' is decreasing in

\left( { - {\pi \over 2},0} \right)

and increasing in

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