You visited us 2 times! Enjoying our articles? Unlock Full Access!

Theorems for Differentiability

In [0, 1] Lag...

Question

In [0, 1] Lagrange's mean value theorem is applicable to

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{2} - x, x < \frac{1}{2} {(\frac{1}{2} - x)}^{2} x \geq \frac{1}{2} \end{matrix}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f (x) = {\begin{matrix} \frac{s i n x}{x}, x \neq 0 1, x = 0 \end{matrix}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f(x)=x|x|

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f(x)=|x|

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $f (x) = {\begin{matrix} \frac{s i n x}{x}, x \neq 0 1, x = 0 \end{matrix}$

Suggest Corrections

Similar questions

f (x)

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{∣ ∣ x^{2} - x ∣ ∣}{x^{2} - x}, x \neq 0, 1 1, x = 0 - 1, x = 1 \end{matrix}

f (x)

f (x) = {\begin{matrix} [x] + [- x], w h e n x \neq 2 λ, w h e n x = 2 \end{matrix}

f (x)

x = 2

λ

f (x) = {\begin{matrix} x + {x} + x sin {x}; x \neq 0 0; x = 0 \end{matrix}

{x}

k

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, x \neq \frac{π}{2} 3, x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

f (x) = ⎧ ⎨ ⎩ \begin{matrix} x + 1; x < 1 λ; x = 1 x^{2} - x + 3; x > 1 \end{matrix}

x = 1

λ

Join BYJU'S Learning Program