Limits, Continuity and Differentiability

⇒g(x) is decreasing in [1,2]
Now, f(1)=[‌

g(1)

n

]=[‌

60

n

],f(2)=[‌

55

n

]
For f(x) to be continuous in [1,2] its integral value should remain same in whole interval for n=9,f(1)=6,f(2)=6
(P)⟶(2)
(Q) g′(x)=(2n2−13n−15)(3x2+3)
=(2n−15)(n+1)(3x2+3)>0

for g(x) to be increasing min‌n=8
⇒Q⟶1 matching
(R) h(x)=(x2−9)n(x2+2x+3)
h(3)=0 for n>5
at n=6⇒n(3+δ)>h(3)
n(3−δ)>n(3)
⇒h(x) has local minima at x=3 for n=6
R⟶4 matching
(S) ‌‌l(x)=sin‌|x|+cos‌|x+‌

1

2

|+sin‌|x−1|+cos‌|x−‌

1

2

|
+......+sin‌|x−4|+cos‌|x−‌

7

2

|
as sin‌|x−a| is non differentiable at x=a
but cos|x−a| remains differentiable at x=a
⇒ given I(x) is non-differentiable at

x0=0,1,2,3,4‌ (5 points) ‌
(S)⟶(3) matching
(P)⟶(2)(Q)⟶(1)(R)⟶(4)(S)⟶(3)