Limits, Continuity and Differentiability Part 3

At x=1,f(x) is continuous therefore,
‌f(1)=f(1)=f(1+)
‌f(1)=3+c ......(1)
‌f(1+)=

lim

h→0

2(1+h)+1
‌f(1+)=

lim

h→0

3+2h=3 ......(2)
from (1) & (2)
c=0
at x=0,f(x) is continuous therefore,
f(0−)=f(0)=f(0+) ......(3)
f(0)=f(0+)=2 ......(4)
f(0−) has to be equal to 2

lim

h→0

‌

a−b‌cos(2h)

h2

‌

lim

h→0

‌

a−b{1− 4h2 2! + 16h4 4! +...}

h2

‌

lim

h→0

‌

a−b+b{2h2− 2 3 h4...}

h2

for limit to exist a−b=0 and limit is 2b ........(5)
from (3), (4) & (5)
a=b=1
checking differentiability at x=0
‌‌ LHD : ‌

lim

h→0

‌

‌ 1−cos‌2‌h h2 −2

−h

‌

lim

h→0

‌

1−(1− 4h2 2! + 16h4 4! ...)−2h2

−h3

=0
‌ RHD : ‌

lim

h→0

‌

(0+h)2+2−2

h

=0
Function is differentiable at every point in its domain
‌∴m=0
‌m+a+b+c =0+1+1+0=2