Let f:R R be defined as : f(x)=array|c|c| a-b 2xx^2; & x1 array If f is continuous everywh…

Question

Let $f:R	o R$ be defined as :

$f(x)=\begin{array}{|c|c|} \frac{a-b\cos 2x}{{x}^{2}}; & x1 \end{array}$

If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is NOT differential then $m+a+b+c$ equals :

Accepted Answer

At $x=1,f(x)$ is continuous therefore,

$f({1}^{-})=f(1)=f({1}^{+})$

$f(1)=3+c$ .........(1)

$\begin{gathered} f({1}^{+})=\underset{h	o 0}{lim}2(1+h)+1 \ f({1}^{+})=\underset{h	o 0}{lim}3+2h=3.........(2) \end{gathered}$

from (1) and (2)

$c=0$

at $x=0,f(x)$ is continuous therefore,

$\begin{gathered} f({0}^{-})=f(0)=f({0}^{+})........(3) \ f(0)=f({0}^{+})=2...........(4) \end{gathered}$

$f({0}^{-})$has to be equal to 2

$\underset{h	o 0}{lim}\frac{a-b\cos (2h)}{{h}^{2}}$

= $\underset{h	o 0}{lim}\frac{a-b{1-\frac{4{h}^{2}}{2!}+\frac{16{h}^{4}}{4!}+\ldots }}{{h}^{2}}$

= $\underset{h	o 0}{lim}\frac{a-b+b{2{h}^{2}-\frac{2}{3}{h}^{4}\ldots }}{{h}^{2}}$

for limit to exist $a-b=0$ and limit is $2b$ from (3), (4) and (5)

$a=b=1$

checking differentiability at $x=0$

$\begin{gathered} \underset{h	o 0}{lim}\frac{\frac{1-\cos 2h}{{h}^{2}}-2}{-h} \ \underset{h	o 0}{lim}\frac{1-(1-\frac{4{h}^{2}}{2!}+\frac{16{h}^{4}}{4!}\ldots )-2{h}^{2}}{-{h}^{3}}=0 \ \underset{h	o 0}{lim}\frac{(0+h{)}^{2}+2-2}{h}=0 \end{gathered}$

Function is differentiable at every point in its domain

$	herefore m=0$

m + a + b + c = 0 + 1 + 1 + 0 = 2

Solutions · 1