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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

If f (x) = min \ 9 - x^ 2 , 1 + x^ 2 \ , x ∈ [-3,3], then the...

If f(x)=min{9−x2,1+x2},∀x∈[−3,3],thenf \left(x\right) = min \left\{\sqrt{9 - x^{2}}, \sqrt{1 + x^{2}}\right\}, \forall x \in \left[-3,3\right], then the number of point(s) wheref \left(x\right) is non-differentiable is/areIf f (x) = min \ 9 - x^ 2 , 1 + x^ 2 \ , x ∈ [-3,3], then the number of point(s) wheref (x) is non-differentiable is/are

Options

  1. A44
  2. B33
  3. C22
  4. D00

Correct answer

A. 44

Step-by-step solution

Using graph of
y=9−x2⇒x2+y2=9y = \sqrt{9 - x^{2}} ⇒ x^{2} + y^{2} = 9 ,
y=1+x2⇒y2−x2=1y = \sqrt{1 + x^{2}} ⇒ y^{2} - x^{2} = 1
In the figure, we can see the graph of f(x)f \left(x\right)
If f (x) = min \ 9 - x^ 2 , 1 + x^ 2 \ , x ∈ [-3,3], then the number of point(s) wheref (x) is non-differentiable is/are
Clearly, ff\left(x\right)isnon−differentiableat<br/>is non-differentiable at<br />x = - 3, {x}_{1}, {x}_{2}, 3(4(4 points)
because there are sharp points at x1,x2{x}_{1}, {x}_{2} and at x=3,x=–3x = 3,x = – 3 there are vertical tangents.
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