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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
MCQ
Let $f:R \to R$ is a function defined by $f\left( x \right) = \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi $, where $\left[ x \right]$ denotes the greatest integer function , then $ f$ is. . .
  • A
    discontinuous only at  $ x=0$ 
  • B
    discontinuous only at non-zero intergral values of $x$
  • C
    continuous only at $x=0 $
  • ✓
    continuous for every real $ x$
✓

Answer

Correct option: D.
continuous for every real $ x$
d
$f(x)=[x] \cos \left(\frac{2 x-1}{2}\right) \pi$

Doubtful points are $x=0, n \in I$

${\rm{LHL}} = \mathop {\lim }\limits_{x \to {\pi ^ - }} [x]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi $

$\Rightarrow(n-1) \cos \left(\frac{2 n-1}{2}\right) \pi$

$[x]$ is the greatest integer function.

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {n^ + }} [x]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi $

$ = n\cos \left( {\frac{{2n - 1}}{2}} \right)\pi $

Now values of the function at $x=0$ is $f(n)=0$

Since $\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(\mathrm{n})$

$f(x)=[x] \cos \left(\frac{2 x-1}{2}\right)$ is continuous for every real $x$

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