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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) = \ cases x^ p + 1 (1)/(x) & :x ≠ 0 0 & :x = 0 cases ,)...

If \text{f}\left(\text{x}\right) = \left\{\begin{cases}x^{\text{p} + 1}\text\cos\dfrac{1}{x} & \text{:x} \ne 0 \\ 0 & \text{:x} = 0\end{cases},\right) then at\text{x} = 0thefunctionthe function\text{f} \left(\text{x}\right) is

Options

  1. AContinuous if p>−1p > - 1 and differentiable if p>0p > 0
  2. BContinuous if p>0p > 0 and differentiable if p>−1p > - 1
  3. CContinuous and differentiable if p>−1p > - 1
  4. DNone of these

Correct answer

A. Continuous if p>−1p > - 1 and differentiable if p>0p > 0

Step-by-step solution

Continuity at x=0,\text{x} = 0,
L.H.L.=limh→0+f(0−h)=limh→0+(−h)p+1cos−11h\text{L} . \text{H} . \text{L} . = \mathop{\text{l}\text{i}\text{m}}_{\text{h} \to {0}^{+}}\text{f}\left(0 - \text{h}\right) = \mathop{\text{l}\text{i}\text{m}}_{\text{h} \to {0}^{+}}{\left(- \text{h}\right)}^{\text{p} + 1}{\text{cos}}^{- 1} \dfrac{1}{\text{h}}
=0if p>−1= 0\text{i}\text{f p} > - 1
R.H.L.=limh→0+f(0+h)=limh→0+(h)p+1cos−11h\text{R} . \text{H} . \text{L} . = \mathop{\text{l}\text{i}\text{m}}_{\text{h} \to {0}^{+}}\text{f}\left(0 + \text{h}\right) = \mathop{\text{l}\text{i}\text{m}}_{\text{h} \to {0}^{+}}{\left(\text{h}\right)}^{\text{p} + 1}{\text{cos}}^{- 1} \dfrac{1}{\text{h}}
=0if p>−1= 0\text{i}\text{f p} > - 1
and \text{f} \left(0\right) = 0 .<br/><br />∴ f(x)iscontinuousatx=\text{f}\left(\text{x}\right) is continuous at \text{x} =0ifp\text{i}\text{f}\text{p}> - 1<br/>Differentiabilityatx=<br />Differentiability at \text{x} =0:<br/>L.H.D.=lim⁡h→0+f(0−h)−f(0)−h<br/>:<br />\text{L.H.D.} = \lim_{\text{h} \to {0}^{+}}\dfrac{\text{f}\left(0 - \text{h}\right) - \text{f}\left(0\right)}{- \text{h}}<br />= \mathop{lim\text{l}\text{i}\text{m}}_{h→0+\text{h} \to {0}^{+}}(−h)p+1cos−11h−0−h<br/>\dfrac{{\left(- \text{h}\right)}^{\text{p} + 1}{\text{cos}}^{- 1} \dfrac{1}{\text{h}} - 0}{- \text{h}}<br />= \mathop{lim\text{l}\text{i}\text{m}}_{h→0+\text{h} \to {0}^{+}}{\left(- \text{h}\right)}^{\text{p}}{\text{cos}}^{- 1}\dfrac{1}{\text{h}} =00 if \text{p}>0,> 0,
\text{R.H.D.} = \lim_{\text{h} \to {0}^{+}}\dfrac{h^{\text{p} + 1}\cos^{- 1}\dfrac{1}{h} - 0}{h}
= \mathop{\text{l}\text{i}\text{m}}_{\text{h} \to {0}^{+}}{\text{h}}^{\text{p}}{\text{cos}}^{- 1}\dfrac{1}{\text{h}} =00 if \text{p}>0> 0
∴∴\text{f} \left(\text{x}\right) is differentiable at \text{x} =00 if \text{p}> 0
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