QBManager

Limits, Continuity and Differentiability

80211 If $y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$ , then $d y / d x$ equals-

1 -1

2 1

3

x

4

\sqrt{x}

Explanation:

(A) : Given,
$y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$
$y = (1 + x^{1 / 4}) (1 - x^{1 / 4}) (1 + x^{1 / 2})$
We know that,
$(a + b) (a - b) = a^{2} - b^{2}$
Then,
$y = [1^{2} - {(x^{1 / 4})}^{2}] (1 + x^{1 / 2})$
$y = (1 - x^{1 / 2}) (1 + x^{1 / 2})$
$y = 1^{2} - {(x^{1 / 2})}^{2}$
$y = 1 - x$
On differentiating both side,
$∴ \frac{dy}{dx} = 0 - 1$
$\frac{dy}{dx} = - 1$

BITSAT-2006

Limits, Continuity and Differentiability

80212 If $f (t) = \frac{1 - t}{1 + t}$ , then $f^{'} (1 / t)$ is equal to

1

\frac{1}{(1 + t)^{2}}

2

\frac{1}{(1 - t)^{2}}

3

\frac{- 2 t^{2}}{(1 + t)^{2}}

4

\frac{2}{(1 - t)^{2}}

Explanation:

(C) : Given,
$f (t) = \frac{1 - t}{1 + t}$
Differentiating both sides with respect to $t$ ,
$f^{'} (t) = \frac{d}{dt} [\frac{1 - t}{1 + t}] = \frac{(1 + t) (- 1) - (1 - t) \times (1)}{(1 + t)^{2}}$
$f^{'} (t) = \frac{- 1 - t - 1 + t}{(1 + t)^{2}} = \frac{- 2}{(1 + t)^{2}}$
$∴ f^{'} (\frac{1}{t}) = \frac{- 2}{{(1 + \frac{1}{t})}^{2}}$
$f^{'} (\frac{1}{t}) = \frac{- 2 t^{2}}{(1 + t)^{2}}$

BITSAT-2009

Limits, Continuity and Differentiability

80213 Let $f (x + y) = f (x) . f (y)$ for all $x, y$ where $f (0) \neq 0$ . If $f (5) = 2$ and $f^{'} (0) = 3$ , then $f^{'} (5)$ is equal to-

1 6

2 0

3 1

4 None of these

Explanation:

(A) : Given,
$f (x + y) = f (x) \cdot f (y)$
Put, $y = 5$
Then,
$f (x + 5) = f (x) \cdot f (5)$
$f (x + 5) = 2 f (x) [∴ f (5) = 2]$
On differentiating both sides w.r.t. $x$ ,
Then,
$f^{'} (x + 5) = 2 f^{'} (x)$
Put,
$x = 0$
$f^{'} (0 + 5) = 2 f^{'} (0) [∵ f^{'} (0) = 3]$
$f^{'} (5) = 2 \times 3$
$f^{'} (5) = 6$

BITSAT-2012

Limits, Continuity and Differentiability

80214 Let $f (x) = (x^{5} - 1) (x^{3} + 1)$ , $g (x) = (x^{2} - 1) (x^{2} - x + 1)$ and let $h (x)$ be such that $f (x) = g (x) h (x)$ . Then $lim_{x \to 1} h (x)$ is

1 0

2 1

3 3

4 5

Explanation:

(D) : Given,
$f (x) = (x^{5} - 1) (x^{3} + 1)$
$g (x) = (x^{2} - 1) (x^{2} - x + 1)$
$f (x) = g (x) h (x)$
$h (x) = \frac{f (x)}{g (x)}$
$lim_{x \to 1} h (x) = lim_{x \to 1} \frac{f (x)}{g (x)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x^{3} + 1)}{(x^{2} - 1) (x^{2} - x + 1)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x + 1) (x^{2} + 1 - x)}{(x - 1) (x + 1) (x^{2} + 1 - x)}$
$lim_{x \to 1} \frac{x^{5} - 1^{5}}{x - 1} = 5 [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n]$

BITSAT-2013

Limits, Continuity and Differentiability

80215 The number of points, where the function $f : R$ $\to R, f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3)$
$| x^{2} - 5 x + 4 |$ , is NOT differentiable, is :

1 1

2 2

3 3

4 4

Explanation:

(C) : $f : R \to R$
$f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |$
$= | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 1 | | x - 4 |$
$= | x - 1 | [\cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 4 |]$
So, there are only three point $x = 1, 2$ and $x = 4$ where $f (x)$ is not differentiable.

JEE Main-2022-29.07.2022

Limits, Continuity and Differentiability

80211 If $y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$ , then $d y / d x$ equals-

1 -1

2 1

3

x

4

\sqrt{x}

Explanation:

(A) : Given,
$y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$
$y = (1 + x^{1 / 4}) (1 - x^{1 / 4}) (1 + x^{1 / 2})$
We know that,
$(a + b) (a - b) = a^{2} - b^{2}$
Then,
$y = [1^{2} - {(x^{1 / 4})}^{2}] (1 + x^{1 / 2})$
$y = (1 - x^{1 / 2}) (1 + x^{1 / 2})$
$y = 1^{2} - {(x^{1 / 2})}^{2}$
$y = 1 - x$
On differentiating both side,
$∴ \frac{dy}{dx} = 0 - 1$
$\frac{dy}{dx} = - 1$

BITSAT-2006

Limits, Continuity and Differentiability

80212 If $f (t) = \frac{1 - t}{1 + t}$ , then $f^{'} (1 / t)$ is equal to

1

\frac{1}{(1 + t)^{2}}

2

\frac{1}{(1 - t)^{2}}

3

\frac{- 2 t^{2}}{(1 + t)^{2}}

4

\frac{2}{(1 - t)^{2}}

Explanation:

(C) : Given,
$f (t) = \frac{1 - t}{1 + t}$
Differentiating both sides with respect to $t$ ,
$f^{'} (t) = \frac{d}{dt} [\frac{1 - t}{1 + t}] = \frac{(1 + t) (- 1) - (1 - t) \times (1)}{(1 + t)^{2}}$
$f^{'} (t) = \frac{- 1 - t - 1 + t}{(1 + t)^{2}} = \frac{- 2}{(1 + t)^{2}}$
$∴ f^{'} (\frac{1}{t}) = \frac{- 2}{{(1 + \frac{1}{t})}^{2}}$
$f^{'} (\frac{1}{t}) = \frac{- 2 t^{2}}{(1 + t)^{2}}$

BITSAT-2009

Limits, Continuity and Differentiability

80213 Let $f (x + y) = f (x) . f (y)$ for all $x, y$ where $f (0) \neq 0$ . If $f (5) = 2$ and $f^{'} (0) = 3$ , then $f^{'} (5)$ is equal to-

1 6

2 0

3 1

4 None of these

Explanation:

(A) : Given,
$f (x + y) = f (x) \cdot f (y)$
Put, $y = 5$
Then,
$f (x + 5) = f (x) \cdot f (5)$
$f (x + 5) = 2 f (x) [∴ f (5) = 2]$
On differentiating both sides w.r.t. $x$ ,
Then,
$f^{'} (x + 5) = 2 f^{'} (x)$
Put,
$x = 0$
$f^{'} (0 + 5) = 2 f^{'} (0) [∵ f^{'} (0) = 3]$
$f^{'} (5) = 2 \times 3$
$f^{'} (5) = 6$

BITSAT-2012

Limits, Continuity and Differentiability

80214 Let $f (x) = (x^{5} - 1) (x^{3} + 1)$ , $g (x) = (x^{2} - 1) (x^{2} - x + 1)$ and let $h (x)$ be such that $f (x) = g (x) h (x)$ . Then $lim_{x \to 1} h (x)$ is

1 0

2 1

3 3

4 5

Explanation:

(D) : Given,
$f (x) = (x^{5} - 1) (x^{3} + 1)$
$g (x) = (x^{2} - 1) (x^{2} - x + 1)$
$f (x) = g (x) h (x)$
$h (x) = \frac{f (x)}{g (x)}$
$lim_{x \to 1} h (x) = lim_{x \to 1} \frac{f (x)}{g (x)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x^{3} + 1)}{(x^{2} - 1) (x^{2} - x + 1)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x + 1) (x^{2} + 1 - x)}{(x - 1) (x + 1) (x^{2} + 1 - x)}$
$lim_{x \to 1} \frac{x^{5} - 1^{5}}{x - 1} = 5 [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n]$

BITSAT-2013

Limits, Continuity and Differentiability

80215 The number of points, where the function $f : R$ $\to R, f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3)$
$| x^{2} - 5 x + 4 |$ , is NOT differentiable, is :

1 1

2 2

3 3

4 4

Explanation:

(C) : $f : R \to R$
$f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |$
$= | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 1 | | x - 4 |$
$= | x - 1 | [\cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 4 |]$
So, there are only three point $x = 1, 2$ and $x = 4$ where $f (x)$ is not differentiable.

JEE Main-2022-29.07.2022

Limits, Continuity and Differentiability

80211 If $y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$ , then $d y / d x$ equals-

1 -1

2 1

3

x

4

\sqrt{x}

Explanation:

(A) : Given,
$y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$
$y = (1 + x^{1 / 4}) (1 - x^{1 / 4}) (1 + x^{1 / 2})$
We know that,
$(a + b) (a - b) = a^{2} - b^{2}$
Then,
$y = [1^{2} - {(x^{1 / 4})}^{2}] (1 + x^{1 / 2})$
$y = (1 - x^{1 / 2}) (1 + x^{1 / 2})$
$y = 1^{2} - {(x^{1 / 2})}^{2}$
$y = 1 - x$
On differentiating both side,
$∴ \frac{dy}{dx} = 0 - 1$
$\frac{dy}{dx} = - 1$

BITSAT-2006

Limits, Continuity and Differentiability

80212 If $f (t) = \frac{1 - t}{1 + t}$ , then $f^{'} (1 / t)$ is equal to

1

\frac{1}{(1 + t)^{2}}

2

\frac{1}{(1 - t)^{2}}

3

\frac{- 2 t^{2}}{(1 + t)^{2}}

4

\frac{2}{(1 - t)^{2}}

Explanation:

(C) : Given,
$f (t) = \frac{1 - t}{1 + t}$
Differentiating both sides with respect to $t$ ,
$f^{'} (t) = \frac{d}{dt} [\frac{1 - t}{1 + t}] = \frac{(1 + t) (- 1) - (1 - t) \times (1)}{(1 + t)^{2}}$
$f^{'} (t) = \frac{- 1 - t - 1 + t}{(1 + t)^{2}} = \frac{- 2}{(1 + t)^{2}}$
$∴ f^{'} (\frac{1}{t}) = \frac{- 2}{{(1 + \frac{1}{t})}^{2}}$
$f^{'} (\frac{1}{t}) = \frac{- 2 t^{2}}{(1 + t)^{2}}$

BITSAT-2009

Limits, Continuity and Differentiability

80213 Let $f (x + y) = f (x) . f (y)$ for all $x, y$ where $f (0) \neq 0$ . If $f (5) = 2$ and $f^{'} (0) = 3$ , then $f^{'} (5)$ is equal to-

1 6

2 0

3 1

4 None of these

Explanation:

(A) : Given,
$f (x + y) = f (x) \cdot f (y)$
Put, $y = 5$
Then,
$f (x + 5) = f (x) \cdot f (5)$
$f (x + 5) = 2 f (x) [∴ f (5) = 2]$
On differentiating both sides w.r.t. $x$ ,
Then,
$f^{'} (x + 5) = 2 f^{'} (x)$
Put,
$x = 0$
$f^{'} (0 + 5) = 2 f^{'} (0) [∵ f^{'} (0) = 3]$
$f^{'} (5) = 2 \times 3$
$f^{'} (5) = 6$

BITSAT-2012

Limits, Continuity and Differentiability

80214 Let $f (x) = (x^{5} - 1) (x^{3} + 1)$ , $g (x) = (x^{2} - 1) (x^{2} - x + 1)$ and let $h (x)$ be such that $f (x) = g (x) h (x)$ . Then $lim_{x \to 1} h (x)$ is

1 0

2 1

3 3

4 5

Explanation:

(D) : Given,
$f (x) = (x^{5} - 1) (x^{3} + 1)$
$g (x) = (x^{2} - 1) (x^{2} - x + 1)$
$f (x) = g (x) h (x)$
$h (x) = \frac{f (x)}{g (x)}$
$lim_{x \to 1} h (x) = lim_{x \to 1} \frac{f (x)}{g (x)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x^{3} + 1)}{(x^{2} - 1) (x^{2} - x + 1)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x + 1) (x^{2} + 1 - x)}{(x - 1) (x + 1) (x^{2} + 1 - x)}$
$lim_{x \to 1} \frac{x^{5} - 1^{5}}{x - 1} = 5 [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n]$

BITSAT-2013

Limits, Continuity and Differentiability

80215 The number of points, where the function $f : R$ $\to R, f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3)$
$| x^{2} - 5 x + 4 |$ , is NOT differentiable, is :

1 1

2 2

3 3

4 4

Explanation:

(C) : $f : R \to R$
$f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |$
$= | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 1 | | x - 4 |$
$= | x - 1 | [\cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 4 |]$
So, there are only three point $x = 1, 2$ and $x = 4$ where $f (x)$ is not differentiable.

JEE Main-2022-29.07.2022

Limits, Continuity and Differentiability

80211 If $y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$ , then $d y / d x$ equals-

1 -1

2 1

3

x

4

\sqrt{x}

Explanation:

(A) : Given,
$y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$
$y = (1 + x^{1 / 4}) (1 - x^{1 / 4}) (1 + x^{1 / 2})$
We know that,
$(a + b) (a - b) = a^{2} - b^{2}$
Then,
$y = [1^{2} - {(x^{1 / 4})}^{2}] (1 + x^{1 / 2})$
$y = (1 - x^{1 / 2}) (1 + x^{1 / 2})$
$y = 1^{2} - {(x^{1 / 2})}^{2}$
$y = 1 - x$
On differentiating both side,
$∴ \frac{dy}{dx} = 0 - 1$
$\frac{dy}{dx} = - 1$

BITSAT-2006

Limits, Continuity and Differentiability

80212 If $f (t) = \frac{1 - t}{1 + t}$ , then $f^{'} (1 / t)$ is equal to

1

\frac{1}{(1 + t)^{2}}

2

\frac{1}{(1 - t)^{2}}

3

\frac{- 2 t^{2}}{(1 + t)^{2}}

4

\frac{2}{(1 - t)^{2}}

Explanation:

(C) : Given,
$f (t) = \frac{1 - t}{1 + t}$
Differentiating both sides with respect to $t$ ,
$f^{'} (t) = \frac{d}{dt} [\frac{1 - t}{1 + t}] = \frac{(1 + t) (- 1) - (1 - t) \times (1)}{(1 + t)^{2}}$
$f^{'} (t) = \frac{- 1 - t - 1 + t}{(1 + t)^{2}} = \frac{- 2}{(1 + t)^{2}}$
$∴ f^{'} (\frac{1}{t}) = \frac{- 2}{{(1 + \frac{1}{t})}^{2}}$
$f^{'} (\frac{1}{t}) = \frac{- 2 t^{2}}{(1 + t)^{2}}$

BITSAT-2009

Limits, Continuity and Differentiability

80213 Let $f (x + y) = f (x) . f (y)$ for all $x, y$ where $f (0) \neq 0$ . If $f (5) = 2$ and $f^{'} (0) = 3$ , then $f^{'} (5)$ is equal to-

1 6

2 0

3 1

4 None of these

Explanation:

(A) : Given,
$f (x + y) = f (x) \cdot f (y)$
Put, $y = 5$
Then,
$f (x + 5) = f (x) \cdot f (5)$
$f (x + 5) = 2 f (x) [∴ f (5) = 2]$
On differentiating both sides w.r.t. $x$ ,
Then,
$f^{'} (x + 5) = 2 f^{'} (x)$
Put,
$x = 0$
$f^{'} (0 + 5) = 2 f^{'} (0) [∵ f^{'} (0) = 3]$
$f^{'} (5) = 2 \times 3$
$f^{'} (5) = 6$

BITSAT-2012

Limits, Continuity and Differentiability

80214 Let $f (x) = (x^{5} - 1) (x^{3} + 1)$ , $g (x) = (x^{2} - 1) (x^{2} - x + 1)$ and let $h (x)$ be such that $f (x) = g (x) h (x)$ . Then $lim_{x \to 1} h (x)$ is

1 0

2 1

3 3

4 5

Explanation:

(D) : Given,
$f (x) = (x^{5} - 1) (x^{3} + 1)$
$g (x) = (x^{2} - 1) (x^{2} - x + 1)$
$f (x) = g (x) h (x)$
$h (x) = \frac{f (x)}{g (x)}$
$lim_{x \to 1} h (x) = lim_{x \to 1} \frac{f (x)}{g (x)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x^{3} + 1)}{(x^{2} - 1) (x^{2} - x + 1)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x + 1) (x^{2} + 1 - x)}{(x - 1) (x + 1) (x^{2} + 1 - x)}$
$lim_{x \to 1} \frac{x^{5} - 1^{5}}{x - 1} = 5 [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n]$

BITSAT-2013

Limits, Continuity and Differentiability

80215 The number of points, where the function $f : R$ $\to R, f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3)$
$| x^{2} - 5 x + 4 |$ , is NOT differentiable, is :

1 1

2 2

3 3

4 4

Explanation:

(C) : $f : R \to R$
$f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |$
$= | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 1 | | x - 4 |$
$= | x - 1 | [\cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 4 |]$
So, there are only three point $x = 1, 2$ and $x = 4$ where $f (x)$ is not differentiable.

JEE Main-2022-29.07.2022

Limits, Continuity and Differentiability

80211 If $y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$ , then $d y / d x$ equals-

1 -1

2 1

3

x

4

\sqrt{x}

Explanation:

(A) : Given,
$y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4})$
$y = (1 + x^{1 / 4}) (1 - x^{1 / 4}) (1 + x^{1 / 2})$
We know that,
$(a + b) (a - b) = a^{2} - b^{2}$
Then,
$y = [1^{2} - {(x^{1 / 4})}^{2}] (1 + x^{1 / 2})$
$y = (1 - x^{1 / 2}) (1 + x^{1 / 2})$
$y = 1^{2} - {(x^{1 / 2})}^{2}$
$y = 1 - x$
On differentiating both side,
$∴ \frac{dy}{dx} = 0 - 1$
$\frac{dy}{dx} = - 1$

BITSAT-2006

Limits, Continuity and Differentiability

80212 If $f (t) = \frac{1 - t}{1 + t}$ , then $f^{'} (1 / t)$ is equal to

1

\frac{1}{(1 + t)^{2}}

2

\frac{1}{(1 - t)^{2}}

3

\frac{- 2 t^{2}}{(1 + t)^{2}}

4

\frac{2}{(1 - t)^{2}}

Explanation:

(C) : Given,
$f (t) = \frac{1 - t}{1 + t}$
Differentiating both sides with respect to $t$ ,
$f^{'} (t) = \frac{d}{dt} [\frac{1 - t}{1 + t}] = \frac{(1 + t) (- 1) - (1 - t) \times (1)}{(1 + t)^{2}}$
$f^{'} (t) = \frac{- 1 - t - 1 + t}{(1 + t)^{2}} = \frac{- 2}{(1 + t)^{2}}$
$∴ f^{'} (\frac{1}{t}) = \frac{- 2}{{(1 + \frac{1}{t})}^{2}}$
$f^{'} (\frac{1}{t}) = \frac{- 2 t^{2}}{(1 + t)^{2}}$

BITSAT-2009

Limits, Continuity and Differentiability

80213 Let $f (x + y) = f (x) . f (y)$ for all $x, y$ where $f (0) \neq 0$ . If $f (5) = 2$ and $f^{'} (0) = 3$ , then $f^{'} (5)$ is equal to-

1 6

2 0

3 1

4 None of these

Explanation:

(A) : Given,
$f (x + y) = f (x) \cdot f (y)$
Put, $y = 5$
Then,
$f (x + 5) = f (x) \cdot f (5)$
$f (x + 5) = 2 f (x) [∴ f (5) = 2]$
On differentiating both sides w.r.t. $x$ ,
Then,
$f^{'} (x + 5) = 2 f^{'} (x)$
Put,
$x = 0$
$f^{'} (0 + 5) = 2 f^{'} (0) [∵ f^{'} (0) = 3]$
$f^{'} (5) = 2 \times 3$
$f^{'} (5) = 6$

BITSAT-2012

Limits, Continuity and Differentiability

80214 Let $f (x) = (x^{5} - 1) (x^{3} + 1)$ , $g (x) = (x^{2} - 1) (x^{2} - x + 1)$ and let $h (x)$ be such that $f (x) = g (x) h (x)$ . Then $lim_{x \to 1} h (x)$ is

1 0

2 1

3 3

4 5

Explanation:

(D) : Given,
$f (x) = (x^{5} - 1) (x^{3} + 1)$
$g (x) = (x^{2} - 1) (x^{2} - x + 1)$
$f (x) = g (x) h (x)$
$h (x) = \frac{f (x)}{g (x)}$
$lim_{x \to 1} h (x) = lim_{x \to 1} \frac{f (x)}{g (x)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x^{3} + 1)}{(x^{2} - 1) (x^{2} - x + 1)}$
$lim_{x \to 1} \frac{(x^{5} - 1) (x + 1) (x^{2} + 1 - x)}{(x - 1) (x + 1) (x^{2} + 1 - x)}$
$lim_{x \to 1} \frac{x^{5} - 1^{5}}{x - 1} = 5 [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n]$

BITSAT-2013

Limits, Continuity and Differentiability

80215 The number of points, where the function $f : R$ $\to R, f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3)$
$| x^{2} - 5 x + 4 |$ , is NOT differentiable, is :

1 1

2 2

3 3

4 4

Explanation:

(C) : $f : R \to R$
$f (x) = | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |$
$= | x - 1 | \cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 1 | | x - 4 |$
$= | x - 1 | [\cos | x - 2 | \sin | x - 1 | + (x - 3) | x - 4 |]$
So, there are only three point $x = 1, 2$ and $x = 4$ where $f (x)$ is not differentiable.

JEE Main-2022-29.07.2022