Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

If fx=x|x|, t...

Question

If $f (x) = x | x |$ , then the value of $f^{'} (x) - 2 | x |$ at $x = 2021$ is

0.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

We have,
$f (x) = x | x |$
$\Rightarrow f (x) = {\begin{matrix} - x^{2} x < 0 x^{2} x \geq 0 \end{matrix}$

$\Rightarrow f^{'} (x) = {\begin{matrix} - 2 x x < 0 2 x x \geq 0 \end{matrix}$

$\Rightarrow f^{'} (x) = 2 | x |$
$∴ f^{'} (x) - 2 | x | = 0$
$∴$ at $x = 2021, f^{'} (x) - 2 | x | = 0$

Suggest Corrections

Similar questions

Q. If

f (x) = \{\begin{cases} 2 x^{2} + k, & if x \geq 0 \\ - 2 x^{2} + k, & if x < 0 \end{cases}

, then what should be the value of
k so that f(x) is continuous at x = 0.

Q. If

f (x) = x | x |

, then the value of

f^{'} (x) - 2 | x |

x = 2021

Q. If

f (x) = a log | x | + b x^{2} + x

has is extremum value at

x = - 1

and

x = 2

, then

$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2^{x + 2} - 16 & i f x \neq 2 \begin{matrix} 4^{x} - 16 k \end{matrix} & i f x = 2 \end{matrix} a t x = 2$

If f(x) is continuous at x=2, then find the value of k.

Q. If

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{sin 3 x}{x^{2} + x}, & x \neq 0 λ + 2, & x = 0 \end{matrix}

is continuous at

x = 0,

then the value of

λ

is

[2 marks]

Explore more

Chain Rule of Differentiation

Standard XII Mathematics

If f(x)=x|x|, then the value of f′(x)−2|x| at x=2021 is

We have, f(x)=x|x| ⇒f(x)={−x2 x<0 x2 x≥0 ⇒f′(x)={−2x x<0 2x x≥0 ⇒f′(x)=2|x| ∴f′(x)−2|x|=0 ∴ at x=2021,f′(x)−2|x|=0

If $f (x) = x | x |$ , then the value of $f^{'} (x) - 2 | x |$ at $x = 2021$ is

We have,
$f (x) = x | x |$
$\Rightarrow f (x) = {\begin{matrix} - x^{2} x < 0 x^{2} x \geq 0 \end{matrix}$

$\Rightarrow f^{'} (x) = {\begin{matrix} - 2 x x < 0 2 x x \geq 0 \end{matrix}$

$\Rightarrow f^{'} (x) = 2 | x |$
$∴ f^{'} (x) - 2 | x | = 0$
$∴$ at $x = 2021, f^{'} (x) - 2 | x | = 0$