Let $$f:R \to R$$ be defined as $$f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x &

$$f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr } } \right.$$ Now, $$f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6({x^2} - x - 20)} & ; & { - 5 < x < 4} \cr {6({x^2} - x - 6)} & ; & {x > 4} \cr } } \right.$$ $$f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6(x - 5)(x + 4)} & ; & { - 5 < x &

Application of Derivatives — Page 10 | JEE Mathematics

Q91

If c is a point at which Rolle's theorem holds for the function, f(x) =

{\log _e}\left( {{{{x^2} + \alpha } \over {7x}}} \right)

in the interval [3, 4], where a

\in

R, then ƒ''(c) is equal to

A

{1 \over {12}}

B

{{\sqrt 3 } \over 7}

C

-{1 \over {12}}

D

-{1 \over {24}}

Correct Answer

Option A

Solution

For Rolle’s theorem to be applicable in [3, 4] ƒ(3) = ƒ(4) $\Rightarrow$

{\log _e}\left( {{{9 + \alpha } \over {21}}} \right) = {\log _e}\left( {{{16 + \alpha } \over {28}}} \right)

$\Rightarrow$

\left( {{{9 + \alpha } \over {21}}} \right) = \left( {{{16 + \alpha } \over {28}}} \right)

$\Rightarrow$ 36 + 4 $\alpha$ = 48 + 3 $\alpha$ $\Rightarrow$ $\alpha$ = 12 According to Rolle’s theorem, f'(c) = 0 where c

\in

(3, 4) f'(x) =

{{7x} \over {{x^2} + 12}}

$\times$

\left( {{{2x\left( {7x} \right) - \left( {{x^2} + 12} \right)7} \over {{7^2}{x^2}}}} \right)

$\Rightarrow$ f'(x) =

{{{x^2} - 12} \over {\left( {{x^2} + 12} \right)x}}

$\therefore$ f'(c) =

{{{c^2} - 12} \over {\left( {{c^2} + 12} \right)c}}

From Rolle's theorem

{{{c^2} - 12} \over {\left( {{c^2} + 12} \right)c}}

= 0 $\Rightarrow$ c2 = 12 Now f''(c) =

{{2c\left( {{c^3} + 12c} \right) - \left( {{c^2} - 12} \right)\left( {3{c^2} + 12} \right)} \over {{{\left( {{c^2} + 12} \right)}^2}{c^2}}}

=

{{2\left( {12} \right)\left( {24} \right) - 0} \over {{{\left( {24} \right)}^2} \times 12}}

=

{1 \over {12}}

Q92

Let ƒ(x) be a polynomial of degree 5 such that x = ±1 are its critical points. If

\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4

, then which one of the following is not true?

A ƒ(1) - 4ƒ(-1) = 4.

B x = 1 is a point of minima and x = -1 is a point of maxima of ƒ.

C x = 1 is a point of maxima and x = -1 is a point of minimum of ƒ.

D ƒ is an odd function.

Correct Answer

Option B

Solution

let f(x) = ax5 + bx4 + cx3 + dx2 + ex + f Given

\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4

$\Rightarrow$

\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f} \over {{x^3}}}} \right)

= 4 As this limit exists so d = e = f = 0 $\therefore$ f(x) = ax5 + bx4 + cx3

\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3}} \over {{x^3}}}} \right)

= 4 $\Rightarrow$ 2 + c = 4 $\Rightarrow$ c = 2 $\therefore$ f(x) = ax5 + bx4 + 2x3 $\Rightarrow$ f'(x) = 5ax4 + 4bx3 + 6x2 As x = ±1 are its critical points so f'(x) = 0 at x = ±1. f'(1) = 5a + 4b + 6 = 0 ....(1) and f'(-1) = 5a - 4b + 6 = 0 .....(

2) Solving (1) and (2), a =

- {6 \over 5}

and b = 0 $\therefore$ f(x) =

- {6 \over 5}{x^5} + 2{x^3}

So f'(x) = -6x4 + 6x2 = 6x2(1 + x)(1 - x) Sign scheme for f'(x) It is clear that maxima at x = 1 and minima at x = –1.

Q93

The value of c in the Lagrange's mean value theorem for the function ƒ(x) = x3 - 4x2 + 8x + 11, when x

\in

[0, 1] is:

A

{2 \over 3}

B

{{\sqrt 7 - 2} \over 3}

C

{{4 - \sqrt 5 } \over 3}

D

{{4 - \sqrt 7 } \over 3}

Correct Answer

Option D

Solution

ƒ(x) = x3 - 4x2 + 8x + 11 f(0) = 11 f(1) = 16 Using LMVT f'(c) =

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

$\Rightarrow$ 3c2 – 8c + 8 =

{{16 - 11} \over {1 - 0}}

$\Rightarrow$ 3c2 – 8c + 3 = 0 $\therefore$ c =

{{8 \pm 2\sqrt 7 } \over 6}

$\therefore$ c =

{{4 - \sqrt 7 } \over 3}

as c

\in

[0, 1]

Q94

Let f : (–1,

\infty

)

\to

R be defined by f(0) = 1 and f(x) =

{1 \over x}{\log _e}\left( {1 + x} \right)

, x

\ne

0. Then the function f :

A decreases in (–1,

\infty

)

B decreases in (–1, 0) and increases in (0,

\infty

)

C increases in (–1,

\infty

)

D increases in (–1, 0) and decreases in (0,

\infty

)

Correct Answer

Option A

Solution

f(x) = {1 \over x}{\log _e}\left( {1 + x} \right)

\Rightarrow f'(x) = {{x{1 \over {1 + x}} - 1{{\log }_e}\left( {1 + x} \right)} \over {{x^2}}}

\Rightarrow f'(x) = {{x - \left( {1 + x} \right){{\log }_e}\left( {1 + x} \right)} \over {{x^2}\left( {1 + x} \right)}}

Let

g(x) = x - \left( {1 + x} \right){\log _e}\left( {1 + x} \right)

\Rightarrow g'(x) = 1 - \left( {1 + x} \right){1 \over {1 + x}} - \left( 1 \right) \times {\log _e}\left( {1 + x} \right)

= 1 - 1 - {\log _e}\left( {1 + x} \right)

= - {\log _e}\left( {1 + x} \right)

For

x \in \left( { - 1,0} \right),g'(x) > 0

and for

x \in \left( {0,\infty } \right),g'(x) < 0

Also,

g(0) = 0 - \left( {1 + 0} \right){\log _e}\left( {1 + 0} \right) = 0

\therefore g'(x) < 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)

\Rightarrow f'(x) < 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)

$\Rightarrow$ f(x) is a decreasing function for all

x \in \left( { - 1,\infty } \right)

Q95

Let the function, ƒ:[-7, 0]

\to

R be continuous on [-7,0] and differentiable on (-7, 0). If ƒ(-7) = - 3 and ƒ'(x)

\le

2, for all x

\in

(-7,0), then for all such functions ƒ, ƒ(-1) + ƒ(0) lies in the interval:

A

\left[ { - 6,20} \right]

B

\left( { - \infty ,\left. {20} \right]} \right.

C

\left[ { - 3,11} \right]

D

\left( { - \infty ,\left. {11} \right]} \right.

Correct Answer

Option B

Solution

Using Lagrange’s Mean Value Theorem in [–7, –1]

{{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}

= f'(c1) As ƒ'(x) $\le$ 2 then f'(c1) $\le$ 2 $\therefore$

{{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}

$\le$ 2 $\Rightarrow$

{{f\left( { - 1} \right) + 3} \over 6}

$\le$ 2 $\Rightarrow$ f(-1) $\le$ 9 Using Lagrange’s Mean Value Theorem in [–7, 0] $\Rightarrow$

{{f\left( 0 \right) - f\left( { - 7} \right)} \over {0 - \left( { - 7} \right)}}

$\le$ 2 $\Rightarrow$ f(0) $\le$ 11 $\therefore$ ƒ(-1) + ƒ(0) $\le$ 11 + 9 $\Rightarrow$ ƒ(-1) + ƒ(0) $\le$ 20

Q96

The function f(x) =

{{4{x^3} - 3{x^2}} \over 6} - 2\sin x + \left( {2x - 1} \right)\cos x

:

A increases in

\left( { - \infty ,{1 \over 2}} \right]

B decreases in

\left( { - \infty ,{1 \over 2}} \right]

C increases in

\left[ {{1 \over 2},\infty } \right)

D decreases in

\left[ {{1 \over 2},\infty } \right)

Correct Answer

Option C

Solution

Given,

f(x) = {{4{x^3} - 3{x^2}} \over 6} - 2\sin x + (2x - 1)\cos x

f'(x) = {{12{x^2} - 6x} \over 6} - 2\cos x + (2x - 1)( - \sin x) + \cos x(2)

= (2{x^2} - x) - 2\cos x - 2x\sin x + \sin x + 2\cos x

= 2{x^2} - x - 2x\sin x + \sin x

= 2x(x - \sin x) - 1(x - \sin x)

f'(x) = (2x - 1)(x - \sin x)

for

x > 0,x - \sin x > 0

x < 0,x - \sin x < 0

for

x \in ( - \infty ,0] \cup \left[ {{1 \over 2},\infty } \right),f'(x) \ge 0

for

x \in \left[ {0,{1 \over 2}} \right],f'(x) \le 0

Hence, f(x) increases in

\left[ {{1 \over 2},\infty } \right)

.

Q97

If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1 : 2 is :

A 0

B 2t3

C -2t3

D -t3

Correct Answer

Option C

Solution

Given

P(t,{t^3})

Let

Q = ({t_1},t_1^3)

Slope of tangent at point p,

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

\Rightarrow {\left. {{{dy} \over {dx}}} \right|_{(t,{t^3})}} = 3{t^2}

This slope is same as slope of line PQ. Slope of

PQ = {{t_1^3 - {t_3}} \over {{t_1} - t}}

$\therefore$

3{t^2} = {{t_1^3 - {t_3}} \over {{t_1} - t}}

\Rightarrow 3{t^2} = {{({t_1} - t)(t_1^2 + t{t_1} + {t^2})} \over {({t_1} - t)}}

\Rightarrow 3{t^2} = t_1^2 + t{t_1} + {t^2}

\Rightarrow {t_1} = - 2t

$\therefore$

Q = \left( { - 2t, - 8{t^3}} \right)

$\therefore$

h = {{2t - 2t} \over 3} = 0

k = {{2{t^3} - 8{t^3}} \over 3} = - 2{t^3}

$\therefore$ Point

M = (0, - 2{t^3})

Q98

For which of the following curves, the line

x + \sqrt 3 y = 2\sqrt 3

is the tangent at the point

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

?

A

2{x^2} - 18{y^2} = 9

B

{y^2} = {1 \over {6\sqrt 3 }}x

C

{x^2} + 9{y^2} = 9

D

{x^2} + {y^2} = 7

Correct Answer

Option C

Solution

Tangent to

{x^2} + 9{y^2} = 9

at point

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

is

x\left( {{{3\sqrt 3 } \over 2}} \right) + 9y\left( {{1 \over 2}} \right) = 9

3\sqrt 3 x + 9y = 18 \Rightarrow x + \sqrt 3 y = 2\sqrt 3

$\Rightarrow$ option (1) is true.

Q99

Let

f:R \to R

be defined as

f(x) = \left\{ \begin{array}{ll}{ - 55x,} & {if\,x < - 5} \\ {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \\ {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \end{array} \right.

Let A = {x

\in

R : f is increasing}. Then A is equal to :

A

( - 5,\infty )

B

( - \infty , - 5) \cup (4,\infty )

C

( - 5, - 4) \cup (4,\infty )

D

( - \infty , - 5) \cup ( - 4,\infty )

Correct Answer

Option C

Solution

f(x) = \left\{ \begin{array}{ll}{ - 55x,} & {if\,x < - 5} \\ {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \\ {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \end{array} \right.

Now,

f'(x) = \left\{ \begin{array}{lll}{ - 55} & ; & {x < - 5} \\ {6({x^2} - x - 20)} & ; & { - 5 < x < 4} \\ {6({x^2} - x - 6)} & ; & {x > 4} \end{array} \right.

f'(x) = \left\{ \begin{array}{lll}{ - 55} & ; & {x < - 5} \\ {6(x - 5)(x + 4)} & ; & { - 5 < x < 4} \\ {6(x - 3)(x + 2)} & ; & {x > 4} \end{array} \right.

Hence, f(x) is monotonically increasing in interval

( - 5, - 4) \cup (4,\infty )

Q100

If the curve y = ax2 + bx + c, x

\in

R, passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are :

A a =

-

1, b = 1, c = 1

B a = 1, b = 1, c = 0

C a =

{1 \over 2}

, b =

{1 \over 2}

, c = 1

D a = 1, b = 0, c = 1

Correct Answer

Option B

Solution

Given curve y = ax2 + bx + c, x

\in

R This curve passes through the point (1, 2) $\therefore$

2 = a + b + c

..... (i) Given, slope of tangent at origin is 1 $\therefore$

{{dy} \over {dx}} = 2ax + b \Rightarrow {\left. {{{dy} \over {dx}}} \right|_{(0,0)}} = 1

\Rightarrow b = 1 \Rightarrow a + c = 1

(0, 0) lie on curve $\therefore$ c = 0, a = 1