f(x)=x3sin3x+αsinx−βcos3x is continuous at x=0 x→0lim=x33x−∣3(3x)3+…+α(x−∣3x3…)−β(1−⌊2(3x)2…)=f(0) $\begin{aligned}
& \lim _{x \rightarrow 0}= \frac{-\beta+x(3+\alpha)+\frac{9 \beta x^2}{\underline{|2}}+\left(\frac{-27}{\underline{|3}}-\frac{\alpha}{\underline{|3}}\right) x^3 \ldots}{x^3}=f(0) \
& \text { for exist } \
& \beta=0,3+\alpha=0,-\frac{27}{\underline{|3}}-\frac{\alpha}{\underline{|3}}=f(0)
\end{aligned}\begin{aligned} & \alpha=-3,-\frac{27}{6}-\frac{(-3)}{6}=f(0) \ & f(0)=\frac{-27+3}{6}=-4\end{aligned}$
If the function f(x)= sin 3 x+α sin x-β cos 3 x x^3, x ∈ R, is continuous at x=0, then f(0) is equal to :
JEE Main 2024 — Mathematics Calculus
If the function f(x)=x3sin3x+αsinx−βcos3x,x∈R, is continuous at x=0, then f(0) is equal to :
Held on 5 Apr 2024 · Verified 6 Jul 2026.
Options
- A
2
- B
-2
- C
4
- D
-4
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