Let the functions f: (- 1, 1) → ℝ and g: (- 1, 1) → (- 1, 1) be defined by f(x) = |2x - 1| + |2x + 1| and g(x) = x - [x] , where [x] denotes the greatest integer less than or equal to x . Let f ∘ g: (- 1, 1) → ℝ be the

Correct answer: 0 f (x) = |2x - 1| + |2x + 1| = \ cases - 4x & , & x (1)/(2) cases ) g(x) = \ x\ f(g(x)) = \ cases - 4g(x), & g(x) (1)/(2) cases ) Graph of g(x) is f(g(x)) = \ cases 2 & , x ∈ (- 1, - (1)/(2)) (0, (1)/(2)) 4(x + 1) & x ∈ (- (1)/(2), 0) 4x & x ∈ ((1)/(2), 1) cases ) c = 1 at x = 0 d = 3 at x = - (1)/(2), 0, (1)/(2) , ∴ c + d = 4 .

JEE Advanced 2020Paper 2

Exam: JEE Advanced
Year: 2020
Paper: Paper 2
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Advanced 2020 · Paper 2 · Mathematics · Continuity and Differentiability

Let the functions f: (- 1, 1) → ℝ and g: (- 1, 1) → (- 1, 1) be...

Let the functions $f: \left(- 1, 1\right) \to \mathbb{R}$ and $g: \left(- 1, 1\right) \to \left(- 1, 1\right)$ be defined by $f\left(x\right) = \left|2x - 1\right| + \left|2x + 1\right| \text{\text{ and } } g\left(x\right) = x - \left[x\right]$ , where \left[x\right] $denotes the greatest integer less than or equal to$ x $. Let$ f ∘ g: \left(- 1, 1\right) \to \mathbb{R} $be the composite function defined by \left(f ∘ g\right)\left(x\right) = f\left(g\left(x\right)\right)$ . Suppose $c$ is the number of points in the interval \left(- 1, 1\right) $at which$ f ∘ g $is not continuous, and suppose$ d $is the number of points in the interval \left(- 1, 1\right)$ at which $f ∘ g$ is not differentiable. Then the value of $c + d$ is _____________

Correct answer

Step-by-step solution

$f$ \left(x\right) = \left|2x - 1\right| + \left|2x + 1\right| = \left\{ $\begin{cases}- 4x & , & x \dfrac{1}{2}\end{cases}$ \right)

$g\left(x\right) = \left\{x\right\}$

$f\left(g\left(x\right)\right) = \left\{$ \begin{cases}- 4g\left(x\right), & g\left(x\right) \dfrac{1}{2}\end{cases} $\right)Graph of$ g\left(x\right) $is<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/21bbae00-f37c-428f-8f3a-344dcb0aed0b-image.png"/>$ f\left(g\left(x\right)\right) = \left\{ $\begin{cases}2 & , x \in \left(- 1, - \dfrac{1}{2}\right) \cup \left(0, \dfrac{1}{2}\right) \\ 4\left(x + 1\right) & x \in \left(- \dfrac{1}{2}, 0\right) \\ 4x & x \in \left(\dfrac{1}{2}, 1\right)\end{cases}$ \right)

Let the functions f: (- 1, 1) → ℝ and g: (- 1, 1) → (- 1, 1) be defined by f(x) = |2x - 1| + |2x + 1| and g(x) = x -

$c = 1$ at $x = 0$

$d = 3$ at $x = - \dfrac{1}{2}, 0, \dfrac{1}{2}$ ,

$∴ c + d = 4$ .

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