The function ${\begin{matrix} (- 1)^{[x]}, & x < 0 l i m n \to \infty (\frac{1}{1 + x^{n}}), & x ⩾ 0 \end{matrix}$ then the number of integral values of $x$ in $[- 3, 5]$ where $f (x)$ is discontinuous is / are
([.] denotes greatest integer function)

Open in App

Solution

Graph of $f (x)$ is
So discontinuous at four points

$f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 1 & - 3 \leq x < - 2 1 & - 2 \leq x < - 1 - 1 & - 1 \leq x < 0 1 & 0 \leq x < 1 \frac{1}{2} & x = 1 0 & x > 1 \end{matrix}$

Suggest Corrections

Similar questions

f (x) = [x] cos \frac{2 x - 1}{2} π

f (x) = [x] cos [\frac{2 x - 1}{2}] π

f (x) = [3 x^{3} - 10], []

(1, 2)

f (x) = [\frac{x^{2} + 1}{x^{2} + [| x |] + 1}]

f (x) = {\begin{matrix} [cos π x], & 0 \leq x \leq 1 | x - 1 | [x - 2], & 1 < x \leq 2 \end{matrix}

([.]

)

Join BYJU'S Learning Program