NDA 2013
Exam NDA
Year 2013
Subject Mathematics
Chapter Continuity and Differentiability NDA 2013 · Mathematics · Continuity and Differentiability
A function f: ℝ arrow ℝ is defined as f(x)=x^2 for x ≥ 0 and...
A function
f : R → R f: \mathbb{R} \rightarrow \mathbb{R} f : R → R is defined as
f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 for
x ≥ 0 a n d x \geq 0 and x ≥ 0 an d f(x)=-x
f o r for f or x < 0
. C o n s i d e r t h e f o l l o w i n g s t a t e m e n t s i n r e s p e c t o f t h e a b o v e f u n c t i o n : 1. T h e f u n c t i o n i s c o n t i n u o u s a t .
Consider the following statements in respect of the above function:
1. The function is continuous at . C o n s i d er t h e f o ll o w in g s t a t e m e n t s in res p ec t o f t h e ab o v e f u n c t i o n : 1. T h e f u n c t i o ni sco n t in u o u s a t x=0
. 2. T h e f u n c t i o n i s d i f f e r e n t i a b l e a t .
2. The function is differentiable at .2. T h e f u n c t i o ni s d i ff ere n t iab l e a t x=0
. W h i c h o f t h e a b o v e s t a t e m e n t s i s / a r e c o r r e c t ? .
Which of the above statements is/are correct? . Whi c h o f t h e ab o v es t a t e m e n t s i s / a recorrec t ? Options A 1 only B 2 only C Both 1 and 2 D Neither 1 nor 2
Step-by-step solution
Test of continuity at
x = 0 R f ( x ) = lim x → 0 + x 2 = 0 L f ( x ) = lim x → 0 − ( − x ) = 0 x=0\mathrm{R} f(x)=\lim _{x \rightarrow 0^{+}} x^2=0\mathrm{L} f(x)=\lim _{x \rightarrow 0^{-}}(-x)=0 x = 0 R f ( x ) = lim x → 0 + x 2 = 0 L f ( x ) = lim x → 0 − ( − x ) = 0
also,
f ( 0 ) = ( 0 ) 2 = 0 f(0)=(0)^2=0 f ( 0 ) = ( 0 ) 2 = 0
Hence,
f ( x ) f(x) f ( x ) is continuous at
x = 0 x=0 x = 0
Differentiability
f ′ ( x ) f^{\prime}(x) f ′ ( x ) for
( x > 0 ) = 2 x ⇒ f ′ ( 0 ) = 0 f ′ ( x ) (x > 0)=2 x \Rightarrow f^{\prime}(0)=0f^{\prime}(x) ( x > 0 ) = 2 x ⇒ f ′ ( 0 ) = 0 f ′ ( x ) for
( x < 0 ) = − 1 ⇒ f ′ ( 0 ) = − 1 (x < 0)=-1 \Rightarrow f^{\prime}(0)=-1 ( x < 0 ) = − 1 ⇒ f ′ ( 0 ) = − 1
Hence,
f ( x ) f(x) f ( x ) is not differentiable at
x = 0 x=0 x = 0 .
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