QBManager

Limits, Continuity and Differentiability

80241 Let $f (x) = 15 - | x - 10 |; x \in R$ . Then, the set of all values of $x$ , at which the function, $g (x) =$ $f (f (x))$ is not differentiable, is

1

{5, 10, 15, 20}

2

{5, 10, 15}

3

{10}

4

{10, 15}

Explanation:

(B)Given function is -
$f (x) = 15 - | x - 10 |, x \in R$
$and g (x) = f (f (x)) = f (15 - | x - 10 |)$
$= 15 - | 15 - | x - 10 | - 10 |$
$= 15 - | 5 - | x - 10 | |$
$\begin{aligned} = {\begin{matrix} 15 - | 5 - (x - 10) |, x \geq 10 \\ 15 - | 5 + (x - 10) |, x < 10 \end{matrix} \\ = {\begin{array}{c} 15 - | (15 - x) |, x \geq 10 \\ 15 - | (x - 5) |, x < 10 \end{array} \\ = {\begin{array}{c} 15 + (x - 10) = 10 + x, x < 5 \\ 15 - (x - 5) = 20 - x, 5 \leq x < 10 \\ 15 + (x - 15) = x, 10 \leq x < 15 \\ 15 - (x - 15) = 30 - x, x \geq 15 \end{array} \end{aligned}$
From the above definition it is clear that $g (x)$ is not differentiable at $x = 5, 10, 15$ .

JEE Main-2019-09.4.2019

Limits, Continuity and Differentiability

80242 If $f (x) = {\begin{array}{cl} \frac{1}{| x |}; | x | \geq 1 \\ a x^{2} + b; | x | < 1 \end{array}$ is differentiable at every point of the domain, then the values of a and $b$ are respectively

1

\frac{1}{2}, \frac{1}{2}

2

\frac{1}{2}, - \frac{3}{2}

3

\frac{5}{2}, - \frac{3}{2}

4

- \frac{1}{2}, \frac{3}{2}

Explanation:

(D) : Given,
$\begin{aligned} f (x) = {\begin{array}{cl} \frac{1}{| x |}, & | x | \geq 1 \\ {ax}^{2} + b, & | x | < 1 \end{array} \\ f (x) = {\begin{array}{cl} \frac{1}{| x |}, & x \leq - 1 or x \geq 1 \\ {ax}^{2} + b, & - 1 < x < 1 \end{array} \\ f (x) = {\begin{array}{cc} \frac{- 1}{x}, & x \leq - 1 \\ {ax}^{2} + b, & - 1 < x < 1 \\ \frac{1}{x}, & x \geq 1 \end{array} \end{aligned}$
Given $f (x)$ is differentiable at every point of domain
$∴ f^{'} (x) = {\begin{array}{cc} \frac{1}{x^{2}}, x < - 1 \\ 2 a x, - 1 < x < 1 \\ \frac{- 1}{x^{2}} x > 1 \end{array}$
$∵ f (x)$ is differentiable at $x = 1$
$∴ (LHD at x = 1) = (RHD at x = 1)$
$f^{'} (1^{-}) = f^{'} (1^{+})$
$2 a = - 1$
$a = - \frac{1}{2}$
As we know that, a function is differentiable at $x = a$ if it is continuous at $x = a$
Hence, $f (x)$ is also continuous at $x = 1$
i.e. $($ LHL at $x = 1) = ($ RHL at $x = 1) = f (1)$
$a + b = 1$
$(\frac{- 1}{2}) + b = 1 \Rightarrow b = \frac{3}{2}$
Hence, $a = - \frac{1}{3}, b = \frac{3}{2}$

JEE Main-2021-18.03.2021

Limits, Continuity and Differentiability

80243 The function that is not differentiable at $x = 1$ is

1

f_{1} (x) = | x |, - \infty < x < \infty

2

f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), - \infty < x \leq 1 \\ x, x \geq 1 \end{array}

3

f_{3} (x) = (\begin{matrix} x^{2} + 7 x - 7, - \infty < x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{matrix}

4

f_{4} (x) = {\begin{matrix} | x - 1 | + | x - 2 |, - \infty < x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{matrix}

Explanation:

(C) : We have,
$f_{1} (x) = | x | = {\begin{array}{cc} - x, x < 0 \\ x, x \geq 0 \end{array}$
$f_{1}^{'} (x) = {\begin{array}{cc} - 1, x < 0 \\ 1, x \geq 0 \end{array}$
$∴ f_{1}^{'} (1) = 1$
$f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), x \leq 1 \\ x, x \geq 1 \end{array}$
$f_{2}^{'} (x) = {\begin{array}{cc} \cos (x - 1), x \leq 1 \\ 1, x \geq 1 \end{array}$
$∴ f_{2}^{'} (1) = 1$
$f_{3} (x) = {\begin{array}{cc} x^{2} + 7 x - 7, x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{array}$
$f_{3}^{'} (x) = {\begin{array}{cc} 2 x + 7, x \leq 1 \\ \frac{3}{2}, x \geq 1 \end{array}$
Since, $LHD ($ at $x = 1) = 2 \times 1 + 7 = 9$ and $RHD ($ at $x = 1) = \frac{3}{2}$
$∴$ LHD $\neq$ RHD
Hence, $f_{3} (x)$ is not differentiable at $x = 1$
$f_{4} (x) = {\begin{array}{cc} | x - 1 | + | x - 2 |, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} - (x - 1) - (x - 2), x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} 3 - 2 x, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$f_{4}^{'} (x) = {\begin{array}{cc} - 2, x \leq 1 \\ 1 - 3 x^{2}, x \geq 1 \end{array}$
$∴ f_{4}^{'} (1) = - 2$

TS EAMCET-2018-07.05.2018

Limits, Continuity and Differentiability

80244 The function $f (x) = | x^{2} - 2 x - 3 | \cdot e^{| 9 x^{2} - 12 x + 4 |}$ is not differentiable at exactly

1 four points

2 three points

3 two points

4 one point

Explanation:

(C) : Given,
$f (x) = | x^{2} - 2 x - 3 | e^{| 9 x^{2} - 12 x + 4 |}$
$f (x) = {\begin{array}{cc} (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x > 3 \\ - (x - 3) (x + 1) e^{(3 x - 2)^{2}} & - 1 \leq x \leq 3 \\ (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x < - 1 \end{array}$
At, $x = - 1$ let LHD be $α$ , then its clear that RHD be $= -$ Similarly, at $x = 3$ , if LHD is $β$ , then RHD at $x = 3$ will be $- β$
So, $f (x)$ is not differentiable at $x = - 1, x = 3$
At all other points $f (x)$ will be differentiable.
Hence, option (c) is correct.

JEE Main-2021-31.08.2021

Limits, Continuity and Differentiability

80245 $f : R \to R$ is a function such that
$| f (x) - f (y) | \leq \frac{1}{2} | x - y | \forall x, y \in R$ and $f^{'} (x) \geq \frac{1}{2}$ $\forall x \in R, f (1) = \frac{1}{2}$ Then the number of points of intersection of curve $y = f (x)$ and the curve $y =$ $x^{2} - 2 x - 5$ is

1 1

2 0

3 2

4 infinite

Explanation:

(C) : We have,
$= lim_{h \to 0} | f (x + h) | - f (x) | \leq \frac{1}{2} | | x + h - x |$
$= lim_{h \to 0} \frac{| f (x) + h - f (x) |}{h} \leq \frac{1}{2}$
$f^{'} (x) \leq \frac{1}{2}$
But $f^{'} (x) \geq \frac{1}{2}$ (given)
$f^{'} (x) = \frac{1}{2} \Rightarrow f (x) = \frac{1}{2} x + c$
$f (1) = \frac{1}{2} \times 1 + c$
$\frac{1}{2} = \frac{1}{2} + c [∵ f (1) = \frac{1}{2}]$
$c = 0$
$f (x) = \frac{1}{2} x$ or $y = \frac{1}{2} x$
Substitute $y = \frac{1}{2} x$ in equation $y = x^{2} - 2 x - 5$ to determine the number of points of intersection
$\frac{1}{2} x = x^{2} - 2 x - 5$
$2 x^{2} - 5 x - 10 = 0$
$x = \frac{5 \pm \sqrt{25 + 80}}{4} = \frac{5 \pm \sqrt{105}}{4}$
There are two point of intersection

AP EAMCET-2022-05.07.2022

Limits, Continuity and Differentiability

80241 Let $f (x) = 15 - | x - 10 |; x \in R$ . Then, the set of all values of $x$ , at which the function, $g (x) =$ $f (f (x))$ is not differentiable, is

1

{5, 10, 15, 20}

2

{5, 10, 15}

3

{10}

4

{10, 15}

Explanation:

(B)Given function is -
$f (x) = 15 - | x - 10 |, x \in R$
$and g (x) = f (f (x)) = f (15 - | x - 10 |)$
$= 15 - | 15 - | x - 10 | - 10 |$
$= 15 - | 5 - | x - 10 | |$
$\begin{aligned} = {\begin{matrix} 15 - | 5 - (x - 10) |, x \geq 10 \\ 15 - | 5 + (x - 10) |, x < 10 \end{matrix} \\ = {\begin{array}{c} 15 - | (15 - x) |, x \geq 10 \\ 15 - | (x - 5) |, x < 10 \end{array} \\ = {\begin{array}{c} 15 + (x - 10) = 10 + x, x < 5 \\ 15 - (x - 5) = 20 - x, 5 \leq x < 10 \\ 15 + (x - 15) = x, 10 \leq x < 15 \\ 15 - (x - 15) = 30 - x, x \geq 15 \end{array} \end{aligned}$
From the above definition it is clear that $g (x)$ is not differentiable at $x = 5, 10, 15$ .

JEE Main-2019-09.4.2019

Limits, Continuity and Differentiability

80242 If $f (x) = {\begin{array}{cl} \frac{1}{| x |}; | x | \geq 1 \\ a x^{2} + b; | x | < 1 \end{array}$ is differentiable at every point of the domain, then the values of a and $b$ are respectively

1

\frac{1}{2}, \frac{1}{2}

2

\frac{1}{2}, - \frac{3}{2}

3

\frac{5}{2}, - \frac{3}{2}

4

- \frac{1}{2}, \frac{3}{2}

Explanation:

(D) : Given,
$\begin{aligned} f (x) = {\begin{array}{cl} \frac{1}{| x |}, & | x | \geq 1 \\ {ax}^{2} + b, & | x | < 1 \end{array} \\ f (x) = {\begin{array}{cl} \frac{1}{| x |}, & x \leq - 1 or x \geq 1 \\ {ax}^{2} + b, & - 1 < x < 1 \end{array} \\ f (x) = {\begin{array}{cc} \frac{- 1}{x}, & x \leq - 1 \\ {ax}^{2} + b, & - 1 < x < 1 \\ \frac{1}{x}, & x \geq 1 \end{array} \end{aligned}$
Given $f (x)$ is differentiable at every point of domain
$∴ f^{'} (x) = {\begin{array}{cc} \frac{1}{x^{2}}, x < - 1 \\ 2 a x, - 1 < x < 1 \\ \frac{- 1}{x^{2}} x > 1 \end{array}$
$∵ f (x)$ is differentiable at $x = 1$
$∴ (LHD at x = 1) = (RHD at x = 1)$
$f^{'} (1^{-}) = f^{'} (1^{+})$
$2 a = - 1$
$a = - \frac{1}{2}$
As we know that, a function is differentiable at $x = a$ if it is continuous at $x = a$
Hence, $f (x)$ is also continuous at $x = 1$
i.e. $($ LHL at $x = 1) = ($ RHL at $x = 1) = f (1)$
$a + b = 1$
$(\frac{- 1}{2}) + b = 1 \Rightarrow b = \frac{3}{2}$
Hence, $a = - \frac{1}{3}, b = \frac{3}{2}$

JEE Main-2021-18.03.2021

Limits, Continuity and Differentiability

80243 The function that is not differentiable at $x = 1$ is

1

f_{1} (x) = | x |, - \infty < x < \infty

2

f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), - \infty < x \leq 1 \\ x, x \geq 1 \end{array}

3

f_{3} (x) = (\begin{matrix} x^{2} + 7 x - 7, - \infty < x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{matrix}

4

f_{4} (x) = {\begin{matrix} | x - 1 | + | x - 2 |, - \infty < x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{matrix}

Explanation:

(C) : We have,
$f_{1} (x) = | x | = {\begin{array}{cc} - x, x < 0 \\ x, x \geq 0 \end{array}$
$f_{1}^{'} (x) = {\begin{array}{cc} - 1, x < 0 \\ 1, x \geq 0 \end{array}$
$∴ f_{1}^{'} (1) = 1$
$f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), x \leq 1 \\ x, x \geq 1 \end{array}$
$f_{2}^{'} (x) = {\begin{array}{cc} \cos (x - 1), x \leq 1 \\ 1, x \geq 1 \end{array}$
$∴ f_{2}^{'} (1) = 1$
$f_{3} (x) = {\begin{array}{cc} x^{2} + 7 x - 7, x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{array}$
$f_{3}^{'} (x) = {\begin{array}{cc} 2 x + 7, x \leq 1 \\ \frac{3}{2}, x \geq 1 \end{array}$
Since, $LHD ($ at $x = 1) = 2 \times 1 + 7 = 9$ and $RHD ($ at $x = 1) = \frac{3}{2}$
$∴$ LHD $\neq$ RHD
Hence, $f_{3} (x)$ is not differentiable at $x = 1$
$f_{4} (x) = {\begin{array}{cc} | x - 1 | + | x - 2 |, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} - (x - 1) - (x - 2), x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} 3 - 2 x, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$f_{4}^{'} (x) = {\begin{array}{cc} - 2, x \leq 1 \\ 1 - 3 x^{2}, x \geq 1 \end{array}$
$∴ f_{4}^{'} (1) = - 2$

TS EAMCET-2018-07.05.2018

Limits, Continuity and Differentiability

80244 The function $f (x) = | x^{2} - 2 x - 3 | \cdot e^{| 9 x^{2} - 12 x + 4 |}$ is not differentiable at exactly

1 four points

2 three points

3 two points

4 one point

Explanation:

(C) : Given,
$f (x) = | x^{2} - 2 x - 3 | e^{| 9 x^{2} - 12 x + 4 |}$
$f (x) = {\begin{array}{cc} (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x > 3 \\ - (x - 3) (x + 1) e^{(3 x - 2)^{2}} & - 1 \leq x \leq 3 \\ (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x < - 1 \end{array}$
At, $x = - 1$ let LHD be $α$ , then its clear that RHD be $= -$ Similarly, at $x = 3$ , if LHD is $β$ , then RHD at $x = 3$ will be $- β$
So, $f (x)$ is not differentiable at $x = - 1, x = 3$
At all other points $f (x)$ will be differentiable.
Hence, option (c) is correct.

JEE Main-2021-31.08.2021

Limits, Continuity and Differentiability

80245 $f : R \to R$ is a function such that
$| f (x) - f (y) | \leq \frac{1}{2} | x - y | \forall x, y \in R$ and $f^{'} (x) \geq \frac{1}{2}$ $\forall x \in R, f (1) = \frac{1}{2}$ Then the number of points of intersection of curve $y = f (x)$ and the curve $y =$ $x^{2} - 2 x - 5$ is

1 1

2 0

3 2

4 infinite

Explanation:

(C) : We have,
$= lim_{h \to 0} | f (x + h) | - f (x) | \leq \frac{1}{2} | | x + h - x |$
$= lim_{h \to 0} \frac{| f (x) + h - f (x) |}{h} \leq \frac{1}{2}$
$f^{'} (x) \leq \frac{1}{2}$
But $f^{'} (x) \geq \frac{1}{2}$ (given)
$f^{'} (x) = \frac{1}{2} \Rightarrow f (x) = \frac{1}{2} x + c$
$f (1) = \frac{1}{2} \times 1 + c$
$\frac{1}{2} = \frac{1}{2} + c [∵ f (1) = \frac{1}{2}]$
$c = 0$
$f (x) = \frac{1}{2} x$ or $y = \frac{1}{2} x$
Substitute $y = \frac{1}{2} x$ in equation $y = x^{2} - 2 x - 5$ to determine the number of points of intersection
$\frac{1}{2} x = x^{2} - 2 x - 5$
$2 x^{2} - 5 x - 10 = 0$
$x = \frac{5 \pm \sqrt{25 + 80}}{4} = \frac{5 \pm \sqrt{105}}{4}$
There are two point of intersection

AP EAMCET-2022-05.07.2022

Limits, Continuity and Differentiability

80241 Let $f (x) = 15 - | x - 10 |; x \in R$ . Then, the set of all values of $x$ , at which the function, $g (x) =$ $f (f (x))$ is not differentiable, is

1

{5, 10, 15, 20}

2

{5, 10, 15}

3

{10}

4

{10, 15}

Explanation:

(B)Given function is -
$f (x) = 15 - | x - 10 |, x \in R$
$and g (x) = f (f (x)) = f (15 - | x - 10 |)$
$= 15 - | 15 - | x - 10 | - 10 |$
$= 15 - | 5 - | x - 10 | |$
$\begin{aligned} = {\begin{matrix} 15 - | 5 - (x - 10) |, x \geq 10 \\ 15 - | 5 + (x - 10) |, x < 10 \end{matrix} \\ = {\begin{array}{c} 15 - | (15 - x) |, x \geq 10 \\ 15 - | (x - 5) |, x < 10 \end{array} \\ = {\begin{array}{c} 15 + (x - 10) = 10 + x, x < 5 \\ 15 - (x - 5) = 20 - x, 5 \leq x < 10 \\ 15 + (x - 15) = x, 10 \leq x < 15 \\ 15 - (x - 15) = 30 - x, x \geq 15 \end{array} \end{aligned}$
From the above definition it is clear that $g (x)$ is not differentiable at $x = 5, 10, 15$ .

JEE Main-2019-09.4.2019

Limits, Continuity and Differentiability

80242 If $f (x) = {\begin{array}{cl} \frac{1}{| x |}; | x | \geq 1 \\ a x^{2} + b; | x | < 1 \end{array}$ is differentiable at every point of the domain, then the values of a and $b$ are respectively

1

\frac{1}{2}, \frac{1}{2}

2

\frac{1}{2}, - \frac{3}{2}

3

\frac{5}{2}, - \frac{3}{2}

4

- \frac{1}{2}, \frac{3}{2}

Explanation:

(D) : Given,
$\begin{aligned} f (x) = {\begin{array}{cl} \frac{1}{| x |}, & | x | \geq 1 \\ {ax}^{2} + b, & | x | < 1 \end{array} \\ f (x) = {\begin{array}{cl} \frac{1}{| x |}, & x \leq - 1 or x \geq 1 \\ {ax}^{2} + b, & - 1 < x < 1 \end{array} \\ f (x) = {\begin{array}{cc} \frac{- 1}{x}, & x \leq - 1 \\ {ax}^{2} + b, & - 1 < x < 1 \\ \frac{1}{x}, & x \geq 1 \end{array} \end{aligned}$
Given $f (x)$ is differentiable at every point of domain
$∴ f^{'} (x) = {\begin{array}{cc} \frac{1}{x^{2}}, x < - 1 \\ 2 a x, - 1 < x < 1 \\ \frac{- 1}{x^{2}} x > 1 \end{array}$
$∵ f (x)$ is differentiable at $x = 1$
$∴ (LHD at x = 1) = (RHD at x = 1)$
$f^{'} (1^{-}) = f^{'} (1^{+})$
$2 a = - 1$
$a = - \frac{1}{2}$
As we know that, a function is differentiable at $x = a$ if it is continuous at $x = a$
Hence, $f (x)$ is also continuous at $x = 1$
i.e. $($ LHL at $x = 1) = ($ RHL at $x = 1) = f (1)$
$a + b = 1$
$(\frac{- 1}{2}) + b = 1 \Rightarrow b = \frac{3}{2}$
Hence, $a = - \frac{1}{3}, b = \frac{3}{2}$

JEE Main-2021-18.03.2021

Limits, Continuity and Differentiability

80243 The function that is not differentiable at $x = 1$ is

1

f_{1} (x) = | x |, - \infty < x < \infty

2

f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), - \infty < x \leq 1 \\ x, x \geq 1 \end{array}

3

f_{3} (x) = (\begin{matrix} x^{2} + 7 x - 7, - \infty < x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{matrix}

4

f_{4} (x) = {\begin{matrix} | x - 1 | + | x - 2 |, - \infty < x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{matrix}

Explanation:

(C) : We have,
$f_{1} (x) = | x | = {\begin{array}{cc} - x, x < 0 \\ x, x \geq 0 \end{array}$
$f_{1}^{'} (x) = {\begin{array}{cc} - 1, x < 0 \\ 1, x \geq 0 \end{array}$
$∴ f_{1}^{'} (1) = 1$
$f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), x \leq 1 \\ x, x \geq 1 \end{array}$
$f_{2}^{'} (x) = {\begin{array}{cc} \cos (x - 1), x \leq 1 \\ 1, x \geq 1 \end{array}$
$∴ f_{2}^{'} (1) = 1$
$f_{3} (x) = {\begin{array}{cc} x^{2} + 7 x - 7, x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{array}$
$f_{3}^{'} (x) = {\begin{array}{cc} 2 x + 7, x \leq 1 \\ \frac{3}{2}, x \geq 1 \end{array}$
Since, $LHD ($ at $x = 1) = 2 \times 1 + 7 = 9$ and $RHD ($ at $x = 1) = \frac{3}{2}$
$∴$ LHD $\neq$ RHD
Hence, $f_{3} (x)$ is not differentiable at $x = 1$
$f_{4} (x) = {\begin{array}{cc} | x - 1 | + | x - 2 |, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} - (x - 1) - (x - 2), x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} 3 - 2 x, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$f_{4}^{'} (x) = {\begin{array}{cc} - 2, x \leq 1 \\ 1 - 3 x^{2}, x \geq 1 \end{array}$
$∴ f_{4}^{'} (1) = - 2$

TS EAMCET-2018-07.05.2018

Limits, Continuity and Differentiability

80244 The function $f (x) = | x^{2} - 2 x - 3 | \cdot e^{| 9 x^{2} - 12 x + 4 |}$ is not differentiable at exactly

1 four points

2 three points

3 two points

4 one point

Explanation:

(C) : Given,
$f (x) = | x^{2} - 2 x - 3 | e^{| 9 x^{2} - 12 x + 4 |}$
$f (x) = {\begin{array}{cc} (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x > 3 \\ - (x - 3) (x + 1) e^{(3 x - 2)^{2}} & - 1 \leq x \leq 3 \\ (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x < - 1 \end{array}$
At, $x = - 1$ let LHD be $α$ , then its clear that RHD be $= -$ Similarly, at $x = 3$ , if LHD is $β$ , then RHD at $x = 3$ will be $- β$
So, $f (x)$ is not differentiable at $x = - 1, x = 3$
At all other points $f (x)$ will be differentiable.
Hence, option (c) is correct.

JEE Main-2021-31.08.2021

Limits, Continuity and Differentiability

80245 $f : R \to R$ is a function such that
$| f (x) - f (y) | \leq \frac{1}{2} | x - y | \forall x, y \in R$ and $f^{'} (x) \geq \frac{1}{2}$ $\forall x \in R, f (1) = \frac{1}{2}$ Then the number of points of intersection of curve $y = f (x)$ and the curve $y =$ $x^{2} - 2 x - 5$ is

1 1

2 0

3 2

4 infinite

Explanation:

(C) : We have,
$= lim_{h \to 0} | f (x + h) | - f (x) | \leq \frac{1}{2} | | x + h - x |$
$= lim_{h \to 0} \frac{| f (x) + h - f (x) |}{h} \leq \frac{1}{2}$
$f^{'} (x) \leq \frac{1}{2}$
But $f^{'} (x) \geq \frac{1}{2}$ (given)
$f^{'} (x) = \frac{1}{2} \Rightarrow f (x) = \frac{1}{2} x + c$
$f (1) = \frac{1}{2} \times 1 + c$
$\frac{1}{2} = \frac{1}{2} + c [∵ f (1) = \frac{1}{2}]$
$c = 0$
$f (x) = \frac{1}{2} x$ or $y = \frac{1}{2} x$
Substitute $y = \frac{1}{2} x$ in equation $y = x^{2} - 2 x - 5$ to determine the number of points of intersection
$\frac{1}{2} x = x^{2} - 2 x - 5$
$2 x^{2} - 5 x - 10 = 0$
$x = \frac{5 \pm \sqrt{25 + 80}}{4} = \frac{5 \pm \sqrt{105}}{4}$
There are two point of intersection

AP EAMCET-2022-05.07.2022

Limits, Continuity and Differentiability

80241 Let $f (x) = 15 - | x - 10 |; x \in R$ . Then, the set of all values of $x$ , at which the function, $g (x) =$ $f (f (x))$ is not differentiable, is

1

{5, 10, 15, 20}

2

{5, 10, 15}

3

{10}

4

{10, 15}

Explanation:

(B)Given function is -
$f (x) = 15 - | x - 10 |, x \in R$
$and g (x) = f (f (x)) = f (15 - | x - 10 |)$
$= 15 - | 15 - | x - 10 | - 10 |$
$= 15 - | 5 - | x - 10 | |$
$\begin{aligned} = {\begin{matrix} 15 - | 5 - (x - 10) |, x \geq 10 \\ 15 - | 5 + (x - 10) |, x < 10 \end{matrix} \\ = {\begin{array}{c} 15 - | (15 - x) |, x \geq 10 \\ 15 - | (x - 5) |, x < 10 \end{array} \\ = {\begin{array}{c} 15 + (x - 10) = 10 + x, x < 5 \\ 15 - (x - 5) = 20 - x, 5 \leq x < 10 \\ 15 + (x - 15) = x, 10 \leq x < 15 \\ 15 - (x - 15) = 30 - x, x \geq 15 \end{array} \end{aligned}$
From the above definition it is clear that $g (x)$ is not differentiable at $x = 5, 10, 15$ .

JEE Main-2019-09.4.2019

Limits, Continuity and Differentiability

80242 If $f (x) = {\begin{array}{cl} \frac{1}{| x |}; | x | \geq 1 \\ a x^{2} + b; | x | < 1 \end{array}$ is differentiable at every point of the domain, then the values of a and $b$ are respectively

1

\frac{1}{2}, \frac{1}{2}

2

\frac{1}{2}, - \frac{3}{2}

3

\frac{5}{2}, - \frac{3}{2}

4

- \frac{1}{2}, \frac{3}{2}

Explanation:

(D) : Given,
$\begin{aligned} f (x) = {\begin{array}{cl} \frac{1}{| x |}, & | x | \geq 1 \\ {ax}^{2} + b, & | x | < 1 \end{array} \\ f (x) = {\begin{array}{cl} \frac{1}{| x |}, & x \leq - 1 or x \geq 1 \\ {ax}^{2} + b, & - 1 < x < 1 \end{array} \\ f (x) = {\begin{array}{cc} \frac{- 1}{x}, & x \leq - 1 \\ {ax}^{2} + b, & - 1 < x < 1 \\ \frac{1}{x}, & x \geq 1 \end{array} \end{aligned}$
Given $f (x)$ is differentiable at every point of domain
$∴ f^{'} (x) = {\begin{array}{cc} \frac{1}{x^{2}}, x < - 1 \\ 2 a x, - 1 < x < 1 \\ \frac{- 1}{x^{2}} x > 1 \end{array}$
$∵ f (x)$ is differentiable at $x = 1$
$∴ (LHD at x = 1) = (RHD at x = 1)$
$f^{'} (1^{-}) = f^{'} (1^{+})$
$2 a = - 1$
$a = - \frac{1}{2}$
As we know that, a function is differentiable at $x = a$ if it is continuous at $x = a$
Hence, $f (x)$ is also continuous at $x = 1$
i.e. $($ LHL at $x = 1) = ($ RHL at $x = 1) = f (1)$
$a + b = 1$
$(\frac{- 1}{2}) + b = 1 \Rightarrow b = \frac{3}{2}$
Hence, $a = - \frac{1}{3}, b = \frac{3}{2}$

JEE Main-2021-18.03.2021

Limits, Continuity and Differentiability

80243 The function that is not differentiable at $x = 1$ is

1

f_{1} (x) = | x |, - \infty < x < \infty

2

f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), - \infty < x \leq 1 \\ x, x \geq 1 \end{array}

3

f_{3} (x) = (\begin{matrix} x^{2} + 7 x - 7, - \infty < x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{matrix}

4

f_{4} (x) = {\begin{matrix} | x - 1 | + | x - 2 |, - \infty < x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{matrix}

Explanation:

(C) : We have,
$f_{1} (x) = | x | = {\begin{array}{cc} - x, x < 0 \\ x, x \geq 0 \end{array}$
$f_{1}^{'} (x) = {\begin{array}{cc} - 1, x < 0 \\ 1, x \geq 0 \end{array}$
$∴ f_{1}^{'} (1) = 1$
$f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), x \leq 1 \\ x, x \geq 1 \end{array}$
$f_{2}^{'} (x) = {\begin{array}{cc} \cos (x - 1), x \leq 1 \\ 1, x \geq 1 \end{array}$
$∴ f_{2}^{'} (1) = 1$
$f_{3} (x) = {\begin{array}{cc} x^{2} + 7 x - 7, x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{array}$
$f_{3}^{'} (x) = {\begin{array}{cc} 2 x + 7, x \leq 1 \\ \frac{3}{2}, x \geq 1 \end{array}$
Since, $LHD ($ at $x = 1) = 2 \times 1 + 7 = 9$ and $RHD ($ at $x = 1) = \frac{3}{2}$
$∴$ LHD $\neq$ RHD
Hence, $f_{3} (x)$ is not differentiable at $x = 1$
$f_{4} (x) = {\begin{array}{cc} | x - 1 | + | x - 2 |, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} - (x - 1) - (x - 2), x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} 3 - 2 x, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$f_{4}^{'} (x) = {\begin{array}{cc} - 2, x \leq 1 \\ 1 - 3 x^{2}, x \geq 1 \end{array}$
$∴ f_{4}^{'} (1) = - 2$

TS EAMCET-2018-07.05.2018

Limits, Continuity and Differentiability

80244 The function $f (x) = | x^{2} - 2 x - 3 | \cdot e^{| 9 x^{2} - 12 x + 4 |}$ is not differentiable at exactly

1 four points

2 three points

3 two points

4 one point

Explanation:

(C) : Given,
$f (x) = | x^{2} - 2 x - 3 | e^{| 9 x^{2} - 12 x + 4 |}$
$f (x) = {\begin{array}{cc} (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x > 3 \\ - (x - 3) (x + 1) e^{(3 x - 2)^{2}} & - 1 \leq x \leq 3 \\ (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x < - 1 \end{array}$
At, $x = - 1$ let LHD be $α$ , then its clear that RHD be $= -$ Similarly, at $x = 3$ , if LHD is $β$ , then RHD at $x = 3$ will be $- β$
So, $f (x)$ is not differentiable at $x = - 1, x = 3$
At all other points $f (x)$ will be differentiable.
Hence, option (c) is correct.

JEE Main-2021-31.08.2021

Limits, Continuity and Differentiability

80245 $f : R \to R$ is a function such that
$| f (x) - f (y) | \leq \frac{1}{2} | x - y | \forall x, y \in R$ and $f^{'} (x) \geq \frac{1}{2}$ $\forall x \in R, f (1) = \frac{1}{2}$ Then the number of points of intersection of curve $y = f (x)$ and the curve $y =$ $x^{2} - 2 x - 5$ is

1 1

2 0

3 2

4 infinite

Explanation:

(C) : We have,
$= lim_{h \to 0} | f (x + h) | - f (x) | \leq \frac{1}{2} | | x + h - x |$
$= lim_{h \to 0} \frac{| f (x) + h - f (x) |}{h} \leq \frac{1}{2}$
$f^{'} (x) \leq \frac{1}{2}$
But $f^{'} (x) \geq \frac{1}{2}$ (given)
$f^{'} (x) = \frac{1}{2} \Rightarrow f (x) = \frac{1}{2} x + c$
$f (1) = \frac{1}{2} \times 1 + c$
$\frac{1}{2} = \frac{1}{2} + c [∵ f (1) = \frac{1}{2}]$
$c = 0$
$f (x) = \frac{1}{2} x$ or $y = \frac{1}{2} x$
Substitute $y = \frac{1}{2} x$ in equation $y = x^{2} - 2 x - 5$ to determine the number of points of intersection
$\frac{1}{2} x = x^{2} - 2 x - 5$
$2 x^{2} - 5 x - 10 = 0$
$x = \frac{5 \pm \sqrt{25 + 80}}{4} = \frac{5 \pm \sqrt{105}}{4}$
There are two point of intersection

AP EAMCET-2022-05.07.2022

Limits, Continuity and Differentiability

80241 Let $f (x) = 15 - | x - 10 |; x \in R$ . Then, the set of all values of $x$ , at which the function, $g (x) =$ $f (f (x))$ is not differentiable, is

1

{5, 10, 15, 20}

2

{5, 10, 15}

3

{10}

4

{10, 15}

Explanation:

(B)Given function is -
$f (x) = 15 - | x - 10 |, x \in R$
$and g (x) = f (f (x)) = f (15 - | x - 10 |)$
$= 15 - | 15 - | x - 10 | - 10 |$
$= 15 - | 5 - | x - 10 | |$
$\begin{aligned} = {\begin{matrix} 15 - | 5 - (x - 10) |, x \geq 10 \\ 15 - | 5 + (x - 10) |, x < 10 \end{matrix} \\ = {\begin{array}{c} 15 - | (15 - x) |, x \geq 10 \\ 15 - | (x - 5) |, x < 10 \end{array} \\ = {\begin{array}{c} 15 + (x - 10) = 10 + x, x < 5 \\ 15 - (x - 5) = 20 - x, 5 \leq x < 10 \\ 15 + (x - 15) = x, 10 \leq x < 15 \\ 15 - (x - 15) = 30 - x, x \geq 15 \end{array} \end{aligned}$
From the above definition it is clear that $g (x)$ is not differentiable at $x = 5, 10, 15$ .

JEE Main-2019-09.4.2019

Limits, Continuity and Differentiability

80242 If $f (x) = {\begin{array}{cl} \frac{1}{| x |}; | x | \geq 1 \\ a x^{2} + b; | x | < 1 \end{array}$ is differentiable at every point of the domain, then the values of a and $b$ are respectively

1

\frac{1}{2}, \frac{1}{2}

2

\frac{1}{2}, - \frac{3}{2}

3

\frac{5}{2}, - \frac{3}{2}

4

- \frac{1}{2}, \frac{3}{2}

Explanation:

(D) : Given,
$\begin{aligned} f (x) = {\begin{array}{cl} \frac{1}{| x |}, & | x | \geq 1 \\ {ax}^{2} + b, & | x | < 1 \end{array} \\ f (x) = {\begin{array}{cl} \frac{1}{| x |}, & x \leq - 1 or x \geq 1 \\ {ax}^{2} + b, & - 1 < x < 1 \end{array} \\ f (x) = {\begin{array}{cc} \frac{- 1}{x}, & x \leq - 1 \\ {ax}^{2} + b, & - 1 < x < 1 \\ \frac{1}{x}, & x \geq 1 \end{array} \end{aligned}$
Given $f (x)$ is differentiable at every point of domain
$∴ f^{'} (x) = {\begin{array}{cc} \frac{1}{x^{2}}, x < - 1 \\ 2 a x, - 1 < x < 1 \\ \frac{- 1}{x^{2}} x > 1 \end{array}$
$∵ f (x)$ is differentiable at $x = 1$
$∴ (LHD at x = 1) = (RHD at x = 1)$
$f^{'} (1^{-}) = f^{'} (1^{+})$
$2 a = - 1$
$a = - \frac{1}{2}$
As we know that, a function is differentiable at $x = a$ if it is continuous at $x = a$
Hence, $f (x)$ is also continuous at $x = 1$
i.e. $($ LHL at $x = 1) = ($ RHL at $x = 1) = f (1)$
$a + b = 1$
$(\frac{- 1}{2}) + b = 1 \Rightarrow b = \frac{3}{2}$
Hence, $a = - \frac{1}{3}, b = \frac{3}{2}$

JEE Main-2021-18.03.2021

Limits, Continuity and Differentiability

80243 The function that is not differentiable at $x = 1$ is

1

f_{1} (x) = | x |, - \infty < x < \infty

2

f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), - \infty < x \leq 1 \\ x, x \geq 1 \end{array}

3

f_{3} (x) = (\begin{matrix} x^{2} + 7 x - 7, - \infty < x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{matrix}

4

f_{4} (x) = {\begin{matrix} | x - 1 | + | x - 2 |, - \infty < x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{matrix}

Explanation:

(C) : We have,
$f_{1} (x) = | x | = {\begin{array}{cc} - x, x < 0 \\ x, x \geq 0 \end{array}$
$f_{1}^{'} (x) = {\begin{array}{cc} - 1, x < 0 \\ 1, x \geq 0 \end{array}$
$∴ f_{1}^{'} (1) = 1$
$f_{2} (x) = {\begin{array}{cc} 1 + \sin (x - 1), x \leq 1 \\ x, x \geq 1 \end{array}$
$f_{2}^{'} (x) = {\begin{array}{cc} \cos (x - 1), x \leq 1 \\ 1, x \geq 1 \end{array}$
$∴ f_{2}^{'} (1) = 1$
$f_{3} (x) = {\begin{array}{cc} x^{2} + 7 x - 7, x \leq 1 \\ \frac{3 x - 1}{2}, x \geq 1 \end{array}$
$f_{3}^{'} (x) = {\begin{array}{cc} 2 x + 7, x \leq 1 \\ \frac{3}{2}, x \geq 1 \end{array}$
Since, $LHD ($ at $x = 1) = 2 \times 1 + 7 = 9$ and $RHD ($ at $x = 1) = \frac{3}{2}$
$∴$ LHD $\neq$ RHD
Hence, $f_{3} (x)$ is not differentiable at $x = 1$
$f_{4} (x) = {\begin{array}{cc} | x - 1 | + | x - 2 |, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} - (x - 1) - (x - 2), x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$= {\begin{array}{cc} 3 - 2 x, x \leq 1 \\ 1 + x - x^{3}, x \geq 1 \end{array}$
$f_{4}^{'} (x) = {\begin{array}{cc} - 2, x \leq 1 \\ 1 - 3 x^{2}, x \geq 1 \end{array}$
$∴ f_{4}^{'} (1) = - 2$

TS EAMCET-2018-07.05.2018

Limits, Continuity and Differentiability

80244 The function $f (x) = | x^{2} - 2 x - 3 | \cdot e^{| 9 x^{2} - 12 x + 4 |}$ is not differentiable at exactly

1 four points

2 three points

3 two points

4 one point

Explanation:

(C) : Given,
$f (x) = | x^{2} - 2 x - 3 | e^{| 9 x^{2} - 12 x + 4 |}$
$f (x) = {\begin{array}{cc} (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x > 3 \\ - (x - 3) (x + 1) e^{(3 x - 2)^{2}} & - 1 \leq x \leq 3 \\ (x - 3) (x + 1) e^{(3 x - 2)^{2}} & x < - 1 \end{array}$
At, $x = - 1$ let LHD be $α$ , then its clear that RHD be $= -$ Similarly, at $x = 3$ , if LHD is $β$ , then RHD at $x = 3$ will be $- β$
So, $f (x)$ is not differentiable at $x = - 1, x = 3$
At all other points $f (x)$ will be differentiable.
Hence, option (c) is correct.

JEE Main-2021-31.08.2021

Limits, Continuity and Differentiability

80245 $f : R \to R$ is a function such that
$| f (x) - f (y) | \leq \frac{1}{2} | x - y | \forall x, y \in R$ and $f^{'} (x) \geq \frac{1}{2}$ $\forall x \in R, f (1) = \frac{1}{2}$ Then the number of points of intersection of curve $y = f (x)$ and the curve $y =$ $x^{2} - 2 x - 5$ is

1 1

2 0

3 2

4 infinite

Explanation:

(C) : We have,
$= lim_{h \to 0} | f (x + h) | - f (x) | \leq \frac{1}{2} | | x + h - x |$
$= lim_{h \to 0} \frac{| f (x) + h - f (x) |}{h} \leq \frac{1}{2}$
$f^{'} (x) \leq \frac{1}{2}$
But $f^{'} (x) \geq \frac{1}{2}$ (given)
$f^{'} (x) = \frac{1}{2} \Rightarrow f (x) = \frac{1}{2} x + c$
$f (1) = \frac{1}{2} \times 1 + c$
$\frac{1}{2} = \frac{1}{2} + c [∵ f (1) = \frac{1}{2}]$
$c = 0$
$f (x) = \frac{1}{2} x$ or $y = \frac{1}{2} x$
Substitute $y = \frac{1}{2} x$ in equation $y = x^{2} - 2 x - 5$ to determine the number of points of intersection
$\frac{1}{2} x = x^{2} - 2 x - 5$
$2 x^{2} - 5 x - 10 = 0$
$x = \frac{5 \pm \sqrt{25 + 80}}{4} = \frac{5 \pm \sqrt{105}}{4}$
There are two point of intersection

AP EAMCET-2022-05.07.2022