The maximum value of $$f(x) = \left| {\matrix{ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}

$$f(x) = \left| {\matrix{ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|$$ $${C_1} \to {C_1} + {C_2}$$ = $$\left| {\matrix{ 2 & {1 + {{\cos }^2}x} & {\cos 2x} \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|$$ $${R_1} \to {R_1} - {R_2}$$ = $$\left| {\matrix{ 0 & 1 & 0 \cr 2 & {{{\cos }

Consider the function f : R $$ \to $$ R defined by $$f(x) = \left\{ \matrix{ \left( {2 - \sin \left( {{1 \over x}} \right)} \right)|x|,x \ne 0 \hfill \cr 0,\,\,x = 0 \hfill \cr} \right.$$. Then f is :

$$f(x) = \left\{ {\matrix{ { - \left( {2 - \sin {1 \over x}} \right)x} & , & {x < 0} \cr 0 & , & {x = 0} \cr {\left( {2 - \sin {1 \over x}} \right)x} & , & {x > 0} \cr } } \right.$$ $$f'(x) = \left\{ \matrix{ - x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) - \left( {2 - \sin {1 \over x}} \right),x < 0 \hfill \cr x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) + \left( {2 - \sin {1 \over x}} \right),x > 0

Application of Derivatives — Page 11 | JEE Mathematics

Q101

If the curves,

{{{x^2}} \over a} + {{{y^2}} \over b} = 1

and

{{{x^2}} \over c} + {{{y^2}} \over d} = 1

intersect each other at an angle of 90

^\circ

, then which of the following relations is TRUE?

A a

-

c = b + d

B a + b = c + d

C

ab = {{c + d} \over {a + b}}

D a

-

b = c

-

d

Correct Answer

Option D

Solution

{{{x^2}} \over a} + {{{y^2}} \over b} = 1

..........(1) Differentiating both sides :

{{2x} \over a} + {{2y} \over b}{{dy} \over {dx}} = 0 \Rightarrow {y \over b}{{dy} \over {dx}} = {{ - x} \over a}

$\Rightarrow$

{{dy} \over {dx}} = {{ - bx} \over {ay}}

............(2)

{{{x^2}} \over c} + {{{y^2}} \over d} = 1

........(3) Differentiating both sides :

{{dy} \over {dx}} = {{ - dx} \over {cy}}

...........(4)

{m_1}{m_2} = - 1 \Rightarrow {{ - bx} \over {ay}} \times {{ - dx} \over {cy}} = - 1

\Rightarrow bd{x^2} = - ac{y^2}

.............(5)

(1) - (3) \Rightarrow \left( {{1 \over a} - {1 \over c}} \right){x^2} + \left( {{1 \over b} - {1 \over d}} \right){y^2} = 0

\Rightarrow {{c - a} \over {ac}}{x^2} + {{d - b} \over {bd}} \times \left( {{{ - bd} \over {ac}}} \right){x^2} = 0

(using 5)

\Rightarrow (c - a) - (d - b) = 0

\Rightarrow c - a = d - b

\Rightarrow c - d = a - b

Q102

If Rolle's theorem holds for the function

f(x) = {x^3} - a{x^2} + bx - 4

,

x \in [1,2]

with

f'\left( {{4 \over 3}} \right) = 0

, then ordered pair (a, b) is equal to :

A (

-

5,

-

8)

B (5,

-

8)

C (

-

5, 8)

D (5, 8)

Correct Answer

Option D

Solution

f(1) = f(2)

\Rightarrow 1 - a + b - 4 = 8 - 4a + 2b - 4

3a - b = 7

..... (1)

f'(x) = 3{x^2} - 2ax + b

\Rightarrow f'\left( {{4 \over 3}} \right) = 0 \Rightarrow 3 \times {{16} \over 9} - {8 \over 3}a + b = 0

\Rightarrow - 8a + 3b = - 16

..... (2) $\therefore$

a = 5,b = 8

Q103

The maximum slope of the curve

y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x

occurs at the point :

A

\left( {3,{{21} \over 2}} \right)

B (0, 0)

C (2, 9)

D (2, 2)

Correct Answer

Option D

Solution

Given,

y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x

{{dy} \over {dx}} = {1 \over 2} \times 4{x^3} - 15{x^2} + 36x - 19

$\Rightarrow$ Slope M =

2{x^3} - 15{x^2} + 36x - 19

At max of slope

{{dM} \over {dx}} = 0

$\therefore$

{{dM} \over {dx}} = 6{x^2} - 30x + 36 = 0

\Rightarrow 6({x^2} - 5x + 6) = 0

\Rightarrow 6(x - 2)(x - 3) = 0

$\therefore$

x = 2,3

Now,

{{{d^2}M} \over {d{x^2}}} = 6(2x - 5)

at

x = 2,{{{d^2}M} \over {d{x^2}}} = 6(4 - 5) = - 6 < 0

$\therefore$ at x = 2 slope is maximum. At x = 2, y = 8 - 40 + 72 - 38 = 2 $\therefore$ Required point = (2, 2)

Q104

Let slope of the tangent line to a curve at any point P(x, y) be given by

{{x{y^2} + y} \over x}

. If the curve intersects the line x + 2y = 4 at x =

-

2, then the value of y, for which the point (3, y) lies on the curve, is :

A

- {{18} \over {19}}

B

- {{4} \over {3}}

C

{{18} \over {35}}

D

- {{18} \over {11}}

Correct Answer

Option A

Solution

{{dy} \over {dx}} = {{x{y^2} + y} \over x}

\Rightarrow {{xdy - ydx} \over {{y^2}}} = xdx

\Rightarrow - d\left( {{x \over y}} \right) = d\left( {{{{x^2}} \over 2}} \right)

\Rightarrow {{ - x} \over y} = {{{x^2}} \over 2} + C

Curve intersect the line x + 2y = 4 at x = $-$ 2 So, $-$ 2 + 2y = 4 $\Rightarrow$ y = 3 So the curve passes through ( $-$ 2, 3)

\Rightarrow {2 \over 3} = 2 + C

\Rightarrow C = {{ - 4} \over 3}

$\therefore$ curve is

{{ - x} \over y} = {{{x^2}} \over 2} - {4 \over 3}

It also passes through (3, y)

{{ - 3} \over y} = {9 \over 2} - {4 \over 3}

\Rightarrow {{ - 3} \over y} = {{19} \over 6}

\Rightarrow y = - {{18} \over {19}}

Q105

Let f be a real valued function, defined on R

-

{

-

1, 1} and given by f(x) = 3 loge

\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}

. Then in which of the following intervals, function f(x) is increasing?

A (

-

\infty

,

-

1)

\cup

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

B (

-

\infty

,

\infty

)

-

{

-

1, 1)

C (

-

\infty

,

{{1 \over 2}}

]

-

{

-

1}

D (

-

1,

{{1 \over 2}}

]

Correct Answer

Option A

Solution

f(x) = 3 loge

\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}

f'(x) = {{3(x + 1)} \over {x - 1}} \times {{(x + 1) - (x - 1)} \over {{{(x + 1)}^2}}} + {2 \over {{{(x - 1)}^2}}} > 0

= {6 \over {{x^2} - 1}} + {2 \over {{{(x - 1)}^2}}} > 0

= {{2(3(x - 1) + (x + 1))} \over {{{(x - 1)}^2}(x + 1)}} = {{4(2x - 1)} \over {{{(x - 1)}^2}(x + 1)}} > 0

\Rightarrow x \in ( - \infty , - 1) \cup \left[ {{1 \over 2},\infty ) - \{ 1\} } \right.

Q106

The maximum value of

f(x) = \left| \begin{array}{lll}{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \\ {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \\ {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \end{array} \right|,x \in R

is :

A

\sqrt 5

B

{3 \over 4}

C 5

D

\sqrt 7

Correct Answer

Option A

Solution

f(x) = \left| \begin{array}{lll}{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \\ {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \\ {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \end{array} \right|

{C_1} \to {C_1} + {C_2}

=

\left| \begin{array}{lll}2 & {1 + {{\cos }^2}x} & {\cos 2x} \\ 2 & {{{\cos }^2}x} & {\cos 2x} \\ 1 & {{{\cos }^2}x} & {\sin 2x} \end{array} \right|

{R_1} \to {R_1} - {R_2}

=

\left| \begin{array}{lll}0 & 1 & 0 \\ 2 & {{{\cos }^2}x} & {\cos 2x} \\ 1 & {{{\cos }^2}x} & {\sin 2x} \end{array} \right|

= ( - 1)[2\sin 2x - \cos 2x] = \cos 2x - 2\sin 2x

We know, maximum value of acosx $\pm$ bsinx =

\sqrt {{a^2} + {b^2}}

$\therefore$ Here maximum value =

\sqrt {{1^2} + {{\left( { - 2} \right)}^2}}

= \sqrt 5

Q107

Let

x=2

be a local minima of the function

f(x)=2x^4-18x^2+8x+12,x\in(-4,4)

. If M is local maximum value of the function

f

in (

-4,4)

, then M =

A

18\sqrt6-\dfrac{33}{2}

B

18\sqrt6-\dfrac{31}{2}

C

12\sqrt6-\dfrac{33}{2}

D

12\sqrt6-\dfrac{31}{2}

Correct Answer

Option C

Solution

\begin{aligned} & f(x)=8 x^3-36 x+8 \\\\ & =4\left(2 x^3-9 x+2\right) \\\\ & =4(x-2)\left(2 x^2+4 x-1\right) \\\\ & =4(x-2)\left(x-\frac{-2+\sqrt{6}}{2}\right)\left(x-\frac{-2 \sqrt{6}}{2}\right) \end{aligned}

Local maxima occurs at $x=\dfrac{-2+\sqrt{6}}{2}=x_0$

f\left(x_0\right)=12 \sqrt{6}-\frac{33}{2}

Q108

Consider the function f : R

\to

R defined by

f(x) = \left\{ \begin{array}{ll}\left( {2 - \sin \left( {{1 \over x}} \right)} \right)|x|,x \ne 0 \\ 0,\,\,x = 0 \end{array} \right.

. Then f is :

A not monotonic on (

-

\infty

, 0) and (0,

\infty

)

B monotonic on (0,

\infty

) only

C monotonic on (

-

\infty

, 0) only

D monotonic on (

-

\infty

, 0)

\cup

(0,

\infty

)

Correct Answer

Option A

Solution

f(x) = \left\{ \begin{array}{lll}{ - \left( {2 - \sin {1 \over x}} \right)x} & , & {x < 0} \\ 0 & , & {x = 0} \\ {\left( {2 - \sin {1 \over x}} \right)x} & , & {x > 0} \end{array} \right.

f'(x) = \left\{ \begin{array}{ll}- x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) - \left( {2 - \sin {1 \over x}} \right),x < 0 \\ x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) + \left( {2 - \sin {1 \over x}} \right),x > 0 \end{array} \right.

=

\left\{ \begin{array}{ll}- {1 \over x}\cos {1 \over x} + \sin {1 \over x} - 2,x < 0 \\ {1 \over x}\cos {1 \over x} - \sin {1 \over x} + 2,x > 0 \end{array} \right.

$\therefore$ f'(x) is an oscillating function which is non-monotonic on ( $-$ $\infty$ , 0) and (0, $\infty$ ).

Q109

Let

A = [{a_{ij}}]

be a 3

\times

3 matrix, where

{a_{ij}} = \left\{ \begin{array}{lll}1 & , & {if\,i = j} \\ { - x} & , & {if\,\left| {i - j} \right| = 1} \\ {2x + 1} & , & {otherwise.} \end{array} \right.

Let a function f : R

\to

R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to:

A

- {{20} \over {27}}

B

{{88} \over {27}}

C

{{20} \over {27}}

D

- {{88} \over {27}}

Correct Answer

Option D

Solution

A = \left[ \begin{array}{lll}1 & { - x} & {2x + 1} \\ { - x} & 1 & { - x} \\ {2x + 1} & { - x} & 1 \end{array} \right]

\left| A \right| = 4{x^3} - 4{x^2} - 4x = f(x)

f'(x) = 4(3{x^2} - 2x - 1) = 0

\Rightarrow x = 1;x = {{ - 1} \over 3}

$\therefore$

f(1) = - 4;f\left( { - {1 \over 3}} \right) = {{20} \over {27}}

Sum

= - 4 + {{20} \over 7} = - {{88} \over {27}}

Q110

Let 'a' be a real number such that the function f(x) = ax2 + 6x

-

15, x

\in

R is increasing in

\left( { - \infty ,{3 \over 4}} \right)

and decreasing in

\left( {{3 \over 4},\infty } \right)

. Then the function g(x) = ax2

-

6x + 15, x

\in

R has a :

A local maximum at x =

-

{{3 \over 4}}

B local minimum at x =

-

{{3 \over 4}}

C local maximum at x =

{{3 \over 4}}

D local minimum at x =

{{3 \over 4}}

Correct Answer

Option A

Solution

{{ - B} \over {2A}} = {3 \over 4}

\Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}

\Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4

$\therefore$

g(x) = 4{x^2} - 6x + 15

Local max. at

x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}

= {{ - 3} \over 4}