Question
If $f(x)=0$ be a quadratic equation such that $f(-\pi)=f(\pi)=0$ and $f\left(\frac{\pi}{2}\right)=-\frac{3 \pi^{2}}{4}$, then $\lim _{x \rightarrow-\pi} \frac{f(x)}{\sin (\sin x)}$ is equal to(a) 0(b) $\pi$(c) $2 \pi$(d) None of these
Step 1
Step 1: Since \( f(x) = 0 \) is a quadratic equation with roots at \( x = -\pi \) and \( x = \pi \), we can express \( f(x) \) in the form: \[ f(x) = k(x + \pi)(x - \pi) = k(x^2 - \pi^2) \] for some constant \( k \). Show more…
Show all steps
Your feedback will help us improve your experience
Harmender Singh Yadav and 75 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If $\mathrm{f}(\mathrm{x})=0$ be a quadratic equation such that $\mathrm{f}(-\pi)=\mathrm{f}(\pi)=0$ and $\mathrm{f}\left(\frac{\pi}{2}\right)=-\frac{3 \pi^{2}}{4}$, then $\lim _{\mathrm{x} \rightarrow-\pi} \frac{\mathrm{f}(\mathrm{x})}{\sin (\sin \mathrm{x})}$ is equal to (A) 0 (B) $\pi$ (C) $2 \pi$ (D) None of these
If $f(x)=\sin x, \quad x \neq n \pi$ $=2, \quad x=n \pi$ where $n \in Z$ and $g(x)=x^{2}+1, \quad x \neq 2$ $=3$, $x=2$. then $\lim _{x \rightarrow 0} g[f(x)]$ is (A) 1 (B) 0 (C) 3 (D) Does not exist
If $f(x)=\sqrt{\frac{x-\sin x}{x+\cos ^{2} x}}$, then $\lim _{\mathrm{x} \rightarrow \infty} f(x)$ is (a) 0 (b) $\infty$ (c) 1 (d) none of these
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD