Skip to content
EXERCISE 8.1 · Q2

Q.Examine the continuity of f(x)=sin⁡xf(x) = \sin x, for x≤π4x \le \dfrac{\pi}{4}, =cos⁡x= \cos x, for x>π4x > \dfrac{\pi}{4}, at x=π4x = \dfrac{\pi}{4}.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
3% · 2/73 Questions
✓ Free question

Here f(x)=sin⁡xf(x)=\sin x for x≤π/4x\le\pi/4 and f(x)=cos⁡xf(x)=\cos x for x>π/4x>\pi/4. Since x=π/4x=\pi/4 falls in the first branch, f(π/4)=sin⁡(π/4)=22f(\pi/4)=\sin(\pi/4)=\dfrac{\sqrt2}{2}.

Left-hand limit: lim⁡x→(π/4)−f(x)=lim⁡x→π/4sin⁡x=sin⁡(π/4)=22\displaystyle\lim_{x\to(\pi/4)^-} f(x)=\lim_{x\to\pi/4}\sin x=\sin(\pi/4)=\frac{\sqrt2}{2}.

Right-hand limit: lim⁡x→(π/4)+f(x)=lim⁡x→π/4cos⁡x=cos⁡(π/4)=22\displaystyle\lim_{x\to(\pi/4)^+} f(x)=\lim_{x\to\pi/4}\cos x=\cos(\pi/4)=\frac{\sqrt2}{2}.

Both one-sided limits equal 22\dfrac{\sqrt2}{2}, so lim⁡x→π/4f(x)=22=f(π/4)\displaystyle\lim_{x\to\pi/4} f(x)=\dfrac{\sqrt2}{2}=f(\pi/4).

✓Final answer

f(x)f(x) is continuous at x=π/4x=\pi/4; f(π/4)=22f(\pi/4)=\dfrac{\sqrt2}{2}.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.