JEE Main 2024 5 April Morning Shift
Exam JEE Main
Year 2024
Date 5 April
Shift Morning Shift
Subject Mathematics
Chapter Continuity and Differentiability JEE Main 2024 · 5 April · Morning Shift · Mathematics · Continuity and Differentiability
If the function f(x)=( 3 x+α x-β 3 x)/(x^3), x ∈ R , is continuous...
If the function
f ( x ) = sin 3 x + α sin x − β cos 3 x x 3 , x ∈ R f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}, x \in \mathbf{R} f ( x ) = x 3 s i n 3 x + α s i n x − β c o s 3 x , x ∈ R , is continuous at
x = 0 x=0 x = 0 , then
f ( 0 ) f(0) f ( 0 ) is equal to :
Step-by-step solution
f ( x ) = sin 3 x + α sin x − β cos 3 x x 3 f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3} f ( x ) = x 3 s i n 3 x + α s i n x − β c o s 3 x
is continuous at
x = 0 lim x → 0 = 3 x − ( 3 x ) 3 ∣ 3 ‾ + … + α ( x − x 3 ∣ 3 ‾ … ) − β ( 1 − ( 3 x ) 2 ⌊ 2 … ) x 3 = f ( 0 ) \mathrm{x}=0 \lim _{x \rightarrow 0}=\frac{3 x-\frac{(3 x)^3}{\underline{|3}}+\ldots+\alpha\left(x-\frac{x^3}{\underline{|3}} \ldots\right)-\beta\left(1-\frac{(3 x)^2}{\lfloor 2} \ldots\right)}{x^3}=f(0) x = 0 lim x → 0 = x 3 3 x − ∣3 ( 3 x ) 3 + … + α ( x − ∣3 x 3 … ) − β ( 1 − ⌊ 2 ( 3 x ) 2 … ) = f ( 0 ) lim x → 0 = − β + x ( 3 + α ) + 9 β x 2 ∣ 2 ‾ + ( − 27 ∣ 3 ‾ − α ∣ 3 ‾ ) x 3 … x 3 = f ( 0 ) for exist β = 0 , 3 + α = 0 , − 27 ∣ 3 ‾ − α ∣ 3 ‾ = 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{\text{ for } exist } \\
& \beta=0,3+\alpha=0,-\frac{27}{\underline{|3}}-\frac{\alpha}{\underline{|^{3}}}=f(0)
\end{aligned} x → 0 lim = x 3 − β + x ( 3 + α ) + ∣2 9 β x 2 + ( ∣ 3 − 27 − ∣ 3 α ) x 3 … = f ( 0 ) for exist β = 0 , 3 + α = 0 , − ∣3 27 − ∣ 3 α = f ( 0 ) α = − 3 , − 27 6 − ( − 3 ) 6 = f ( 0 ) f ( 0 ) = − 27 + 3 6 = − 4 \begin{aligned} & \alpha=-3,-\frac{27}{6}-\frac{(-3)}{6}=f(0) \\ & f(0)=\frac{-27+3}{6}=-4\end{aligned} α = − 3 , − 6 27 − 6 ( − 3 ) = f ( 0 ) f ( 0 ) = 6 − 27 + 3 = − 4 $$
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