JEE Main 2024 4 April Morning Shift
Exam JEE Main
Year 2024
Date 4 April
Shift Morning Shift
Subject Mathematics
Chapter Continuity and Differentiability JEE Main 2024 · 4 April · Morning Shift · Mathematics · Continuity and Differentiability
Let f: R arrow R be a function given by f(x)=\ array ll (1- 2...
Let
f : R → R f: \mathbf{R} \rightarrow \mathbf{R} f : R → R be a function given by
f(x)=\left\{ 1 − cos 2 x x 2 , x < 0 α , x = 0 , β 1 − cos x x , x > 0 \begin{array}{ll}
\frac{1-\cos 2 x}{x^2}, & x < 0 \\
\alpha, & x=0, \\
\frac{\beta \sqrt{1-\cos x}}{x}, & x>0
\end{array} x 2 1 − c o s 2 x , α , x β 1 − c o s x , x < 0 x = 0 , x > 0 \right.
where
α , β ∈ R \alpha, \beta \in \mathbf{R} α , β ∈ R . If
f f f is continuous at
x = 0 x=0 x = 0 , then
α 2 + β 2 \alpha^2+\beta^2 α 2 + β 2 is equal to :
Step-by-step solution
f ( 0 − ) = lim x → 0 − 2 sin 2 x x 2 = 2 = α f ( 0 + ) = lim x → 0 + β × 2 sin x 2 2 x 2 = β 2 = 2 ⇒ β = 2 2 α 2 + β 2 = 4 + 8 = 12 \begin{aligned} & f\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 x}{x^2}=2=\alpha \\ & f\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \beta \times \sqrt{2} \frac{\sin \frac{x}{2}}{2 \frac{x}{2}}=\frac{\beta}{\sqrt{2}}=2 \\ & \Rightarrow \beta=2 \sqrt{2} \\ & \alpha^2+\beta^2=4+8=12\end{aligned} f ( 0 − ) = x → 0 − lim x 2 2 sin 2 x = 2 = α f ( 0 + ) = x → 0 + lim β × 2 2 2 x sin 2 x = 2 β = 2 ⇒ β = 2 2 α 2 + β 2 = 4 + 8 = 12
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