Let g (x) = | x - 2| and h(x) = g (g (x)) be two functions, then the value of h^ ' (-1) + h^ ' (1)+ h^ ' (3) + h^ ' (5) is equal to (where,h^ ' denotes the derivative of h )

Correct answer: C. 0 g(x) = \ cases x - 2, & x > 2 2 - x, & x h(x) = \ cases g (x) - 2, & g (x) > 2 2 - g(x), g(x) h(x) = \ array ll x - 4, & x ≥ 4 - x, & x ≤ 0 4 - x, x, array array ll x ∈ [2,4) x ∈ (0,2) array Then, h^ ' (-1) + h^ ' (1) + h^ ' (3) + h^ ' (5) =0

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

Let g (x) = | x - 2| and h(x) = g (g (x)) be two functions, then...

Let

g \left(x\right) = \left|

x - 2\right| and

h\left(x\right) =

g\lleft(g \left(x\right))

be two functions, then the value of

h^{'} \left(-1\right)

+ h^{'} \left(1\right)+ h^{'}

\left(3\right)

+ h^{'}

\left(5\right) is equal

to (where,h^{'}

denotes the derivative of

h

)

Options

A $2$
B $- 1$
C $0$
D $1$

Correct answer

C. $0$

Step-by-step solution

$g\left(x\right) = \left\{$ \begin{cases}x - 2, & x > 2 \\ 2 - x, & x < 2\end{cases} $\right)<br /><br />$ h\left(x\right) = \left\{ $\begin{cases}g \left(x\right) - 2, & g \left(x\right) > 2 \\ 2 - g\left(x\right), g\left(x\right) < 2\end{cases}$ \right)

$h\left(x\right) = \left\{$ $\begin{array}{ll}x - 4, & x \ge 4 \\ - x, & x \le 0 \\ 4 - x, \\ x,\end{array}$ \begin{array}{ll}x \in \left[2,4\right) \\ x \in \left(0,2\right)\end{array}$ $\right</p><p>Then,$ h^{'} $\left(-1\right) +$ h^{'} $\left(1\right)$ + h^{'} $\left(3\right) + h^{'$ }\left(5\right) $=0$

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