Let t denote the greatest integer ≤t and limx→0⁡x4/x=A. Then the function, fx=x^2sin⁡πx is discontinuous, when x is equal to:

Correct answer: (A) √(A+1). log_(x→0)⁡4/x-4/x=A⇒log_(x→O)⁡4-x4/x=A ⇒4-0 ⇒A=4 Hence, for √5, fx is discontinuous as it is an non-integral value.

Q49 / 75 · JEE Main 2020 (09 Jan Shift 2) · Mathematics

Let $[t]$ denote the greatest integer $\leq t$ and $\lim_{x \to 0} x [\frac{4}{x}] = A .$ Then the function, $f (x) = [x^{2}] \sin (π x)$ is discontinuous, when $x$ is equal to:

$\log_{x \to 0} (\frac{4}{x} - \{\frac{4}{x}\}) = A \Rightarrow \log_{x \to O} 4 - x \{\frac{4}{x}\} = A$

$\Rightarrow 4 - 0$

$\Rightarrow A = 4$

Hence, for $\sqrt{5}$ , $f (x)$ is discontinuous as it is an non-integral value.

Hence, the correct answer is option: 'A'.