Question

2.
Let f(x) =
ax' + bx +C, X < 1
2ax + b,
XZ1
If f(x) is differentiable at x = 1, then
(1) Minimum of fis a, when a > 0
(2) Maximum of fis a, when a < 0
(3) fincreases, when x and a have same sign
(4) f increases, when 0 < x < 1​

Answer 1

Answer:

Correct option is

B

f(6)≥8

Given that f(1)=−2 and f  

′

(x)≥2 ∀x∈[1,6]  

Lagrange's mean value theorem states that if f(x) be continuous on [a,b] and differentiable on (a,b) then there exists some c between a and b such that f  

′

(c)=  

b−a

f(b)−f(a)

​

 

Given that f is differentiable for all x. Therefore, lagrange's mean value theorem can be applied.

Therefore, f  

′

(c)=  

6−1

f(6)−f(1)

​

≥2 (Since, [a,b]=[1,6])

⟹f(6)−f(1)≥2(5)

⟹f(6)−(−2)≥10

⟹f(6)≥10−2

⟹f(6)≥8