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Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 4 April, Shift 1 — Question 1

Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R} be a function given by f(x)={1−cos⁡2xx2,x<0,α,x=0,β1−cos⁡xx,x>0.f(x)=\begin{cases}\frac{1-\cos 2x}{x^2}, & x<0,\\\alpha, & x=0,\\\frac{\beta\sqrt{1-\cos x}}{x}, & x>0.\end{cases} where β∈R\beta \in \mathrm{R} f is continues at x=0\mathrm{x}=0 , then α2+β2 \alpha^2+\beta^2 is equal to :

  1. Option A:

    48

  2. Option B:

    12

    Correct
  3. Option C:

    3

  4. Option D:

    6

Answer: B

Step-by-step solution

f(0−)=lim⁡x→0−2sin⁡2xx2=2=αf\left(0^{-}\right)=\lim _{\mathrm{x} \rightarrow 0^{-}} \frac{2 \sin ^{2} \mathrm{x}}{\mathrm{x}^{2}}=2=\alpha

f(0+)=lim⁡x→0+β×2sin⁡x22x2=β2=2f\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

⇒β=22\Rightarrow \beta=2 \sqrt{2}

α2+β2=4+8=12\alpha^{2}+\beta^{2}=4+8=12

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 1
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Continuity

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