TG EAMCET 2 May 2025 Shift 2 Paper

Given,g(x)={

K√x+1,‌	0≤x≤3
mx+2,‌	3<x≤5

is differentiable
Since, g(x) is differentiable therefore it is continuous.
at x=3 ‌

lim

x⟶3−

f(x)=

lim

x⟶3+

f(x)
⇒

lim

h⟶0

f(3−h)=

lim

h⟶0

f(3+h)
⇒

lim

h⟶0

(K√(3−h+1))=

lim

h⟶0

(m(3+h)+2)
⇒K√4=m(3)+2
⇒2K=3m+2
for 0≤x≤3 g′(x)=K‌

1

2√x+1

And for 3
g′(x)=m
for differentiability at x=3
‌Lg′(3)=Rg′(3)
‌⇒‌

K

2√4

=m
⇒K=4m
On Solving Eqs. (i) and (ii), we get
‌2(4m)‌=3m+2
⇒‌8m‌=3m+2
⇒‌5m‌=2⇒m=‌

2

5

And ‌‌K=‌

8

5

∴‌‌K+m=‌

8

5

+‌

2

5

=‌

10

5

=2