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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Advanced 2016 · Paper 2 · Mathematics · Continuity and Differentiability

Let f: [- (1)/(2), 2] → ℝ and g: [- (1)/(2), 2] → ℝ be functions...

Let f:[−12,2]→Randf: \left[- \dfrac{1}{2}, 2\right] \to \mathbb{R} andg: \left[- \dfrac{1}{2}, 2\right] \to \mathbb{R}befunctionsdefinedbybe functions defined byf\left(x\right) = \left[{x}^{2} - 3\right]andg(x)=and g \left(x\right) =\left|x\right|f\left(x\right) + \left|4x - 7\right|f(x)f\left(x\right) , where \left[y\right]denotesthegreatestintegerlessthanorequaltodenotes the greatest integer less than or equal toyforfory \in \mathbb{R} . Then

Options

  1. Afisdiscontinuousexactlyatthreepointsin[−12,2]is discontinuous exactly at three points in \left[- \dfrac{1}{2}, 2\right]
  2. Bfisdiscontinuousexactlyatfourpointsin[−12,2]is discontinuous exactly at four points in \left[- \dfrac{1}{2}, 2\right]
  3. Cgg is NOT differentiable exactly at four points in (−12,2)\left(- \dfrac{1}{2}, 2\right)
  4. Dgg is NOT differentiable exactly at five points in (−12,2)\left(- \dfrac{1}{2}, 2\right)

Correct answer

B. fisdiscontinuousexactlyatfourpointsin[−12,2]is discontinuous exactly at four points in \left[- \dfrac{1}{2}, 2\right]

Step-by-step solution

f(x)=[x2]−3f\left(x\right) = \left[{x}^{2}\right] - 3
g(x)=(∣x∣+∣4x−7∣)([x2]−3)g\left(x\right) = \left(\left|x\right| + \left|4x - 7\right|\right)\left(\left[{x}^{2}\right] - 3\right)
∵∵ ff is discontinuous at (when x2 is an integer)
x=1,2,3,2in[−12,2]x = 1, \sqrt{2}, \sqrt{3}, 2in\left[- \dfrac{1}{2}, 2\right]
and \left|x\right| + \left|4x - 7\right| \ne 0atatx = 1, \sqrt{2}, \sqrt{3}, 2<br/><br />⇒ g\left(x\right)isdiscontinuousatis discontinuous atx = 1, \sqrt{2}, \sqrt{3}in(−12,2)in \left(- \dfrac{1}{2}, 2\right)
InIn \left(0 - \delta, 0 + \delta\right)((\deltaisverysmall)<br/>is very small)<br />⇒ g\left(x\right) = \left(\left|x\right| + \left|4x - 7\right|\right) . \left(- 3\right)<br/><br />⇒′g′isnonderivableat'g' is non derivable atx = 0.as.as| x |istherewhichisnotdiff.atis there which is not diff. atx = 0<br/><br />In(74−δ,74+δ)\left(\dfrac{7}{4} - \delta, \dfrac{7}{4} + \delta\right)
⇒g(x)=0⇒ g\left(x\right) = 0 as f(x)=0f\left(x\right) = 0
⇒⇒ Derivable at x=74x = \dfrac{7}{4}
∴∴ 'g' is non - derivable at 0,1,2,30, 1, \sqrt{2}, \sqrt{3}
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