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Write the points of non-differentiability of
Concept: undefined >> undefined
Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.
Concept: undefined >> undefined
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Let \[f\left( x \right) = \left( x + \left| x \right| \right) \left| x \right|\]
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Let f (x) = |x| and g (x) = |x3|, then
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The function f (x) = sin−1 (cos x) is
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The set of points where the function f (x) = x |x| is differentiable is
Concept: undefined >> undefined
The function f (x) = e−|x| is
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If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text { is }\]
Concept: undefined >> undefined
If \[f\left( x \right) = x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . ,\]
then at x = 0, f (x)
Concept: undefined >> undefined
If \[f\left( x \right) = \left| \log_e x \right|, \text { then}\]
Concept: undefined >> undefined
If \[f\left( x \right) = \left| \log_e |x| \right|\]
Concept: undefined >> undefined
Let \[f\left( x \right) = \begin{cases}\frac{1}{\left| x \right|} & for \left| x \right| \geq 1 \\ a x^2 + b & for \left| x \right| < 1\end{cases}\] If f (x) is continuous and differentiable at any point, then
Concept: undefined >> undefined
Let f (x) = |sin x|. Then,
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The function f (x) = |cos x| is
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If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is
Concept: undefined >> undefined
If \[f\left( x \right) = \begin{cases}\frac{1}{1 + e^{1/x}} & , x \neq 0 \\ 0 & , x = 0\end{cases}\] then f (x) is
Concept: undefined >> undefined
The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is
Concept: undefined >> undefined
Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is
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Give a condition that three vectors \[\vec{a}\], \[\vec{b}\] and \[\vec{c}\] form the three sides of a triangle. What are the other possibilities?
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Prove that a necessary and sufficient condition for three vectors \[\vec{a}\], \[\vec{b}\], \[\vec{c}\] to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that \[l \vec{a} + m \vec{b} + n \vec{c} = \vec{0} .\]
Concept: undefined >> undefined
