What the mind can conceive and believe, it can achieve.– Napoleon Hill
Lecture - 3 Chapter 5 Continuity and Differentiability
In this lecture, I am discussing about how to check continuity and discontinuity in a graph of a function through limits. I am using my own assignments to explain NCERT Exercise 5.1 Questions as well as Extra (HOTS) question for explanation and CBSE Board exam point of view.
Questions Discussed in this lecture:
12. f(x)=\begin{cases} \frac{\sin (a+1)x+\sin x}{x}, & \text{ if } x<0 \\ c, & \text{ if } x=0 \\ \frac{\sqrt{x+bx^{2} } -\sqrt{x} }{bx^{\frac{3}{2} } }, & \text{ if } x>0 \end{cases}
13. f(x)=\begin{cases} \frac{1-\cos kx}{x\sin x}, & \text{ if } x\ne 0 \\ \frac{1}{2}, & \text{ if } x=0 \end{cases}
14. f(x)=\begin{cases} \frac{x-4}{|x-4|} +a, & \text{ if } x<4 \\ a+b, & \text{ if } x=4 \\ \frac{x-4}{|x-4|} +b, & \text{ if } x>4 \end{cases}
15. f(x)=\begin{cases} \frac{1-\sin ^{3} x}{3\cos ^{2} x}, & \text{ if } x<\frac{\pi }{2} \\ a, & \text{ if } x=\frac{\pi }{2} \\ \frac{b, (1-\sin x)}{(\pi -2x)^{2} }, & \text{ if } x>\frac{\pi }{2} \end{cases}
16. f(x)=\begin{cases} \frac{\sqrt{1+px} -\sqrt{1-px} }{x}, & \text{ if } -1\le x<0 \\ \frac{2x+1}{x-2}, & \text{ if } 0\le x\le 1 \end{cases}
17. f(x)=\begin{cases} a \sin \frac{\pi }{2} (x+1), & \text{ if } x\le 0 \\ \frac{\tan x-\sin x}{x^{3} }, & \text{ if } x>0 \end{cases}
18. f(x)=\begin{cases} \frac{\log (1+3x)-\log (1-2x)}{x}, & \text{ if } x\ne 0 \\ k, & \text{ if } x=0 \end{cases}
19. f(x)=\begin{cases} \frac{1-\cos 10x}{x^{2} }, & \text{ if } x<0 \\ a, & \text{ if } x=0 \\ \frac{\sqrt{x} }{\sqrt{625+\sqrt{x} } -25}, & \text{ if } x>0 \end{cases}