Let fx-x- - x-x2 - x- - -1- 1 - Then the number of points at which fx is discontinuous is

The correct option is A 0 case 1: x x^2≥ 0 ⇒ xϵ [0,1] f(x)=x^2 case 2: x x^2≤ 0 ⇒ xϵ [ 1,0) f(x)= 2x x^2 At x=0 both the functions tend ...

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Standard XII

Mathematics

Algebra of Derivatives

Let fx=x- |...

Question

Let $f (x) = x - ∣ ∣ x - x^{2} ∣ ∣, x ϵ [- 1, 1]$ . Then the number of points at which $f (x)$ is discontinuous is

1

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Solution

The correct option is A $0$
case 1:
$x - x^{2} \geq 0 \Rightarrow x ϵ [0, 1]$
$f (x) = x^{2}$
case 2:
$x - x^{2} \leq 0 \Rightarrow x ϵ [- 1, 0)$
$f (x) = 2 x - x^{2}$
At x=0 both the functions tend to zero. So the function f(x) is continuous at x=0.
Both the functions are continuous in respective intervals.
Hence the number of discontinuous are zero.

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