TS EAMCET 04 May 2019 Shift 2 Solved Paper

Consider the piecewise function.
f(x)={

5−3x	‌	‌ if ‌x≤‌ 5 3
x2−3x+20	‌	‌ if ‌x>‌ 5 3 .

The above function can be solved as

lim

x→‌ 5− 3

f(x)‌‌=5−3(‌

5

3

)
‌‌=0
And

lim

x→‌ 5+ 3

f(x)‌‌=(‌

5

3

)2−3(‌

5

3

)+20
‌‌=‌

160

9

Now

lim

x→‌ 5− 3

f(x)≠

lim

x→‌ 5+ 3

f(x)
Therefore, the function is discontinuousat x=‌

5

3

Hence, f(x) is derivative at x=2
‌ Consider ‌f′(x)=

lim

x→2

‌

f(x+2)−f(x)

2

lim

h→0

‌

f(x+h)−f(x)

2

=

lim

h→0

‌

(x+h)2−3(x+h)+20−x2+3x−20

2

=

lim

h→0

‌

h2+2xh−3h

h

=h+2x−3
=2x−3
As f′(2) exists, f(x) is differentiable at x=2