मराठी

The Function F (X) = |Cos X| is

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प्रश्न

The function f (x) = |cos x| is

पर्याय

  • everywhere continuous and differentiable

  • everywhere continuous but not differentiable at (2n + 1) π/2, n ∈ Z

  • neither continuous nor differentiable at (2n + 1) π/2, n ∈ Z

  • none of these

MCQ
थोडक्यात उत्तर
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उत्तर

We have,

\[f\left( x \right) = \left| \cos x \right|\]

`⇒ f(x) ={(cosx , 2npile x < (4n +1)_2^pi),(0, x= (4n +1)_2^pi),(-cos x, (4n +1)_2^pi < x< (4n +3)^_2^pi),(0,x = (4n + 3)_2^pi),(cos x, (4n +3)_2^pi < x le (2n +2)pi):}`

\[\text { When, x is in first quadrant, i . e . 2n}\pi \leq x < \left( 4n + 1 \right)\frac{\pi}{2} , \text { we have} \]

\[ f\left( x \right) = \text { cos x which being a trigonometrical function is continuous and differentiable in}  \left( 2n\pi, \left( 4n + 1 \right)\frac{\pi}{2} \right)\]

\[\text { When, x is in second quadrant or in third quadrant, i . e }. , \left( 4n + 1 \right)\frac{\pi}{2} < x < \left( 4n + 3 \right)\frac{\pi}{2} , we have\]

\[ f\left( x \right) = - \text { cos x which being a trigonometrical function is continuous and differentiable in } \left( \left( 4n + 1 \right)\frac{\pi}{2}, \left( 4n + 3 \right)\frac{\pi}{2} \right)\]

\[\text { When, x is in fourth quadrant, i . e}  . , \left( 4n + 3 \right)\frac{\pi}{2} < x \leq \left( 2n + 2 \right)\pi ,\text {  we have }\]

\[ f\left( x \right) =\text{cos x which being a trigonometrical function is continuous and differentiable in } \left( \left( 4n + 3 \right)\frac{\pi}{2}, \left( 2n + 2 \right)\pi \right)\]

\[\text { Thus possible point of non - differentiability of } f\left( x \right)\text {  are x} = \left( 4n + 1 \right)\frac{\pi}{2}, \left( 4n + 3 \right)\frac{\pi}{2}\]

\[\text { Now, LHD } \left[ \text { at x }= \left( 4n + 1 \right)\frac{\pi}{2} \right] = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]

\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{\cos x - 0}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]

\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{- \sin x}{1 - 0} \left[\text {  By L'Hospital rule } \right]\]

\[ = - 1\]

\[\text { And RHD } \left( \text { at x } = \left( 4n + 1 \right)\frac{\pi}{2} \right) = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]

\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{- \cos x - 0}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]

\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{\sin x}{1 - 0} \left[ \text { By L'Hospital rule}  \right]\]

\[ = 1\

\[ \therefore \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} f\left( x \right)\]

\[\text { So }f\left( x \right)\text { is not differentiable at x }= \left( 4n + 1 \right)\frac{\pi}{2}\]

\[\text { Now, LHD}  \left[\text { at x } = \left( 4n + 3 \right)\frac{\pi}{2} \right] = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]

\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} \frac{- \cos x - 0}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]

\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} \frac{\sin x}{1 - 0} \left[\text {  By L'Hospital rule }\right]\]

\[ = 1\]

\[\text { And RHD } \left( \text { at x } = \left( 4n + 3 \right)\frac{\pi}{2} \right) = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]

\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{\cos x - 0}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]

\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{- \sin x}{1 - 0} \left[\text {  By L'Hospital rule} \right]\]

\[ = - 1\

\[ \therefore \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} f\left( x \right)\]

\[\text { So} f\left( x \right) \text { is not differentiable at x}  = \left( 4n + 3 \right)\frac{\pi}{2}\]

\[\text { Therefore} , f\left( x \right)\text {  is neither differentiable at }\left( 4n + 1 \right)\frac{\pi}{2} \text { nor at } \left( 4n + 3 \right)\frac{\pi}{2}\]

\[\text { i . e } . f\left( x \right) \text { is not differentiable at odd multiples of } \frac{\pi}{2}\]

\[\text { i . e .} f\left( x \right) \text { is not differentiable at x }= \left( 2n + 1 \right)\frac{\pi}{2}\]

\[\text { Therefore, f(x) is everywhere continuous but not differentiable at } \left( 2n + 1 \right)\frac{\pi}{2} .\]

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पाठ 9: Differentiability - Exercise 10.4 [पृष्ठ १८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 9 Differentiability
Exercise 10.4 | Q 7 | पृष्ठ १८

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