Limits, Continuity & Differentiability
Piecewise Continuity — Finding Constants
nta_pyq_2024_jan
Grade 12
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ be defined as
$$f(x)=\begin{cases}\dfrac{a-b\cos 2x}{x^2} & ,\; x<0\\x^2+cx+2 & ,\; 0\le x\le 1\\2x+1 & ,\; x>1\end{cases}$$
If $f$ is continuous everywhere in $\mathbb{R}$ and $m$ is the number of points where $f$ is NOT differentiable, then $m+a+b+c$ equals:
Step-by-Step Solution
Key Concept: For continuity at $x=0$: limit from left must exist and equal $f(0)=2$. Use $\lim_{h\to0}\frac{a-b\cos(2h)}{h^2}$ requiring $a=b$ (for limit to exist) and limit value $=2b$. For continuity at $x=1$: $1+c+2=3$, giving $c=0$. Then check differentiability at $x=0$ and $x=1$.
At $x=0$: $f(0)=2$. $\lim_{x\to0^-}\frac{a-b\cos2x}{x^2}$: for limit to exist, $a-b=0\Rightarrow a=b$; limit $=2b=2$, so $a=b=1$.
At $x=1$: $f(1^-)=1+c+2=3+c$, $f(1^+)=3$, $f(1)=1+c+2=3+c$. Continuity: $c=0$.
Differentiability at $x=0$: LHD$=0$, RHD$=0$. ✓ Differentiable everywhere. $m=0$.
$m+a+b+c=0+1+1+0=2$.
Correct Answer: 4