Let $$f\left( x \right) = \left\{ {\matrix{ {\max \left\{ {\left| x \right|,{x^2}} \right\}} & {\left| x \right| \le 2} \cr {8 - 2\left| x \right|} & {2 < \left| x \right| \le 4} \cr } } \

$$f\left( x \right) = \left\{ {\matrix{ {8 + 2x,} & { - 4 \le x \le - 2} \cr {{x^2},} & { - 2 \le x \le - 1} \cr {\left| x \right|,} & { - 1 < x < 1} \cr {{x^2},} & {1 \le x \le 2} \cr {8 - 2x,} & {2 < x \le 4} \cr } } \right.$$ f(x) is not differentiable at x = $$\left\{ { - 2, - 1,0,1,2} \right\}$$ $$ \Rightarrow $$ S = {$$-$$2, $$-$$ 1, 0, 1, 2}

Let ƒ : [–1,3] $$ \to $$ R be defined as $$f(x) = \left\{ {\matrix{ {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr {x + \left| x \right|} & , & {1 \le x < 2} \c

$$f(x) = \left\{ {\matrix{ {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr {x + \left| x \right|} & , & {1 \le x < 2} \cr {x + \left[ x \right]} & , & {2 \le x \le 3} \cr } } \right.$$ = $$ = \left\{ {\matrix{ { - x - 1,} & { - 1 \le x < 0} \cr {x,} & {0 \le x < 1} \cr {2x,} & {1 \le x < 2} \cr {x + 2,} & {2 \le x < 3} \cr {6,} & {x = 3} \cr } } \right.$$ f(-1) = $$\mathop {\lim }\limits_{x \to - 1} \left( { - x - 1}

Limits, Continuity and Differentiability — Page 19 | JEE Mathematics

Q181

Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x –

\pi

) cos |x| is not differentiable. Then the set K is equal to :

A {0,

\pi

}

B

\phi

(an empty set)

C { r }

D {0}

Correct Answer

Option B

Solution

f(x) = sin

\left| x \right| - \left| x \right|

+ 2(x $-$ $\pi$ ) cosx $\because$ sin

\left| x \right|

$-$

\left| x \right|

is differentiable function at c = 0 $\therefore$ k = $\phi$

Q182

Let

f\left( x \right) = \left\{ \begin{array}{ll}{ - 1} & { - 2 \le x < 0} \\ {{x^2} - 1,} & {0 \le x \le 2} \end{array} \right.

and

g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).

Then, in the interval (–2, 2), g is :

A non continuous

B differentiable at all points

C not differentiable at two points

D not differentiable at one point

Correct Answer

Option D

Solution

\left| {f\left( x \right)} \right| = \left\{ \begin{array}{lll}1 & , & { - 2 \le x < 0} \\ {1 - {x^2}} & , & {0 \le x < 1} \\ {{x^2} - 1} & , & {1 \le x \le 2} \end{array} \right.

and

f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]

Hence

g(x) = \left\{ \begin{array}{lll}{{x^2}} & , & {x \in \left[ { - 2,0} \right]} \\ 0 & , & {x \in \left[ {0,1} \right)} \\ {2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \end{array} \right.

It is not differentiable at x = 1

Q183

If

\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3},

then 'a' is equal to :

A 2

B

{3 \over 2}

C

{2 \over 3}

D

{1 \over 2}

Correct Answer

Option B

Solution

Given,

\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3}

So,

\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}}\,\left[ {{1^ \propto }\,\,\,form} \right]

= {e^{\mathop {\lim }\limits_{x \to \propto } \left[ {\left( {1 + {a \over x} - {4 \over {{x^2}}} - 1} \right)2x} \right]}}

= {e^{\mathop {\lim }\limits_{x \to \propto } \left( {2a - {8 \over x}} \right)}}

= {e^{2a}}

$\therefore$ e2a = e3 $\therefore$ 2a = 3 $\Rightarrow$ a $=$

{3 \over 2}

Q184

Let f(x) be a polynomial of degree

4

having extreme values at

x = 1

and

x = 2.

If

\mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3

then f(

-

1) is equal to :

A

{9 \over 2}

B

{5 \over 2}

C

{3 \over 2}

D

{1 \over 2}

Correct Answer

Option A

Solution

$\because$ f(x) has extremum values at x = 1, and x = 2 $\because$ f'(1) = 0 and f'(2) = 0 As, f(x) is a polynomial of degree 4.

Suppose f(x) = Ax4 + Bx3 + cx2 + Dx + E $\because$

\mathop {\lim }\limits_{x \to 0}

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

= 3 $\Rightarrow$

\,\,\,

\mathop {\lim }\limits_{x \to 0} \left( {{{A{x^4} + B{x^3} + C{x^2} + Dx + E} \over {{x^2}}} + 1} \right)

= 3 $\Rightarrow$

\mathop {\lim }\limits_{x \to 0}

\left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right)

= 3 As limit has finite value, so D = 0 and E = 0 Now A(0)2 + B(0) + C + 0 + 0 + 1 = 3 $\Rightarrow$ c + 1 = 3 $\Rightarrow$ c = 2 f'(x) = 4Ax3 + 3Bx2 + 2Cx + D f'(1) = 0 $\Rightarrow$ 4A(1) + 3B(1) + 2C(1) + D = 0 $\Rightarrow$

\,\,\,

4A + 3B = $-$ 4 . . . .(1) f'(2) = 0 $\Rightarrow$ 4A(8) + 3B(4) + 2C(2) + D = 0 $\Rightarrow$ 8A + 3B = $-$ 2 . . . . .(

2) From equations (1) and (2), we get A =

{1 \over 2}

and B = $-$ 2 So, f(x) =

{{{x_4}} \over 2} - 2{x^3} + 2x{}^2

Therefore, f( $-$ 1) =

{{{{( - 1)}^4}} \over 2} - 2{( - 1)^3} + 2{( - 1)^2}

=

{1 \over 2} + 2 + 2 = {9 \over 2}

Hence f( $-$ 1) =

{9 \over 2}

Q185

Let

f\left( x \right) = \left\{ \begin{array}{ll}{\max \left\{ {\left| x \right|,{x^2}} \right\}} & {\left| x \right| \le 2} \\ {8 - 2\left| x \right|} & {2 < \left| x \right| \le 4} \end{array} \right.

Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S

A equals

\left\{ { - 2, - 1,1,2} \right\}

B equals

\left\{ { - 2, - 1,0,1,2} \right\}

C equals

\left\{ { - 2,2} \right\}

D is an empty set

Correct Answer

Option B

Solution

f\left( x \right) = \left\{ \begin{array}{ll}{8 + 2x,} & { - 4 \le x \le - 2} \\ {{x^2},} & { - 2 \le x \le - 1} \\ {\left| x \right|,} & { - 1 < x < 1} \\ {{x^2},} & {1 \le x \le 2} \\ {8 - 2x,} & {2 < x \le 4} \end{array} \right.

f(x) is not differentiable at x =

\left\{ { - 2, - 1,0,1,2} \right\}

$\Rightarrow$ S = { $-$ 2, $-$ 1, 0, 1, 2}

Q186

Let ƒ : [–1,3]

\to

R be defined as

f(x) = \left\{ \begin{array}{lll}{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \\ {x + \left| x \right|} & , & {1 \le x < 2} \\ {x + \left[ x \right]} & , & {2 \le x \le 3} \end{array} \right.

where [t] denotes the greatest integer less than or equal to t. Then, ƒ is discontinuous at:

A only three points

B four or more points

C only two points

D only one point

Correct Answer

Option A

Solution

f(x) = \left\{ \begin{array}{lll}{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \\ {x + \left| x \right|} & , & {1 \le x < 2} \\ {x + \left[ x \right]} & , & {2 \le x \le 3} \end{array} \right.

=

= \left\{ \begin{array}{ll}{ - x - 1,} & { - 1 \le x < 0} \\ {x,} & {0 \le x < 1} \\ {2x,} & {1 \le x < 2} \\ {x + 2,} & {2 \le x < 3} \\ {6,} & {x = 3} \end{array} \right.

f(-1) =

\mathop {\lim }\limits_{x \to - 1} \left( { - x - 1} \right)

= -( -1) - 1 = 0 f(-1+) =

\mathop {\lim }\limits_{x \to - {1^ + }} \left( { - x - 1} \right)

= 0 $\therefore$ f(x) is continuous at x = - 1 f(0-) =

\mathop {\lim }\limits_{x \to {0^ - }} \left( { - x - 1} \right)

= -(0) - 1 = - 1 f(0) =

\mathop {\lim }\limits_{x \to 0} \left( x \right)

= 0 f(0+) =

\mathop {\lim }\limits_{x \to {0^ + }} \left( x \right)

= 0 $\therefore$ f(x) is discontinuous at x = 0 f(1-) =

\mathop {\lim }\limits_{x \to {1^ - }} \left( x \right)

= 1 f(1) =

\mathop {\lim }\limits_{x \to 1} \left( {2x} \right)

= 2 f(1+) =

\mathop {\lim }\limits_{x \to {1^ + }} \left( {2x} \right)

= 2 $\therefore$ f(x) is discontinuous at x = 1 f(3-) =

\mathop {\lim }\limits_{x \to {3^ - }} \left( {x + 2} \right)

= 3 + 2 = 5 f(3) = 6 $\therefore$ f(x) is discontinuous at x = 3 So, f(x) is discontinuous at x = {0, 1, 3}.

Q187

Let ƒ(x) = 15 – |x – 10|; x

\in

R. Then the set of all values of x, at which the function, g(x) = ƒ(ƒ(x)) is not differentiable, is :

A {10,15}

B {5,10,15,20}

C {10}

D {5,10,15}

Correct Answer

Option D

Solution

ƒ(x) = 15 – |x – 10| g(x) = ƒ(ƒ(x)) = 15 – |ƒ(x) – 10| = 15 – |15 – |x – 10| – 10| = 15 – |5 – |x – 10|| As this is a linear expression so it is non differentiable when value inside the modulus is zero.

So non differentiable when x – 10 = 0 $\Rightarrow$ x = 10 and 5 – |x – 10| = 0 $\Rightarrow$ |x – 10| = 5 $\Rightarrow$ x - 10 = $\pm$ 5 $\Rightarrow$ x = 5, 15 $\therefore$ g(x) is not differentiable at x = 5, 10, 15.

Q188

If the function ƒ defined on ,

\left( {{\pi \over 6},{\pi \over 3}} \right)

by

f(x) = \left\{ \begin{array}{ll}{{{\sqrt 2 {\mathop{\rm cosx}\nolimits} - 1} \over {\cot x - 1}},} & {x \ne {\pi \over 4}} \\ {k,} & {x = {\pi \over 4}} \end{array} \right.

$ is continuous, then k is equal to

A 1

B 1 /

\sqrt 2

C

{1 \over 2}

D 2

Correct Answer

Option C

Solution

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}

= f(

{{\pi \over 4}}

) = k

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}

(

{0 \over 0}

form) = k $\Rightarrow$

\mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - \sqrt 2 \sin x} \over {-\cos e{c^2}x}}

(Using L Hospital Rule) $\Rightarrow$

\mathop {\lim }\limits_{x \to {\pi \over 4}} \sqrt 2 {\sin ^3}x

= k $\Rightarrow$ k =

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

=

{1 \over 2}

Q189

The value of

p

and

q

for which the function

f\left( x \right) = \left\{ \begin{array}{ll}{{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \\ q & {,x = 0} \\ {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{3/2}}}}} & {,x > 0} \end{array} \right.

is continuous for all

x

in R, are

A

p =

{5 \over 2}

,

q =

{1 \over 2}

B

p =

-{3 \over 2}

,

q =

{1 \over 2}

C

p =

{1 \over 2}

,

q =

{3 \over 2}

D

p =

{1 \over 2}

,

q =

-{3 \over 2}

Correct Answer

Option B

Solution

L.H.L. = \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sin \left\{ {\left( {p + 1} \right)\left( { - h} \right)} \right\} - \sin \left( { - h} \right)} \over { - h}}

= \mathop {\lim }\limits_{h \to 0} {{ - \sin \left( {p + 1} \right)h} \over { - h}} + {{\sin \left( { - h} \right)} \over { - h}} = p + 1 + 1 = p + 2

R.H.L.

= \mathop {\lim }\limits_{x \to {\sigma ^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {1 + h} - 1} \over h}

= \mathop {\lim }\limits_{h \to 0} {1 \over {\left( {\sqrt {1 + h} + 1} \right)}} = {1 \over 2}

and

f\left( 0 \right) = q \Rightarrow p = - {3 \over 2},q = {1 \over 2}

Q190

If

\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to k} {{{x^3} - {k^3}} \over {{x^2} - {k^2}}}

, then k is :

A

{3 \over 2}

B

{8 \over 3}

C

{4 \over 3}

D

{3 \over 8}

Correct Answer

Option B

Solution

If

\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)

L·H·S·

\mathop {Lt}\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \left( {{0 \over 0}form} \right)

\Rightarrow \mathop {Lt}\limits_{x \to 1} {{4{x^3}} \over 1} = 4

Now,

\mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)

= 4

\Rightarrow \mathop {\lim }\limits_{x \to K} {{3{x^2}} \over {2x}} = 4

\Rightarrow {3 \over 2}k = 4 \Rightarrow k = {8 \over 3}