Question

# Let f:R→R be a differentiable function satisfying f(x+y)=f(x)+f(y)+xy for all x,y∈R and limh→01hf(h)=3. Which of the following statements is (are) CORRECT?

A
f is a linear function
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2f(1)=[limx0(1+2x)1/x], where [.] denotes the greatest integer function
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
The minimum value of f is 92
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
f(1)=4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct options are B 2f(1)=[limx→0(1+2x)1/x], where [.] denotes the greatest integer function C The minimum value of f is −92 D f′(1)=4f(x+y)=f(x)+f(y)+xy for all x,y∈R ⋯(1) f′(x)=limh→0f(x+h)−f(x)h =limh→0f(x)+f(h)+xh−f(x)h =limh→0f(h)+xhh =limh→0f(h)h+limh→0x =3+x ∴f(x)=3x+x22+C Putting x=y=0 in (1), we get f(0)=2f(0)⇒f(0)=0 So, C=0 Hence, f(x)=3x+x22 limx→0(1+2x)1/x=e2 ∴[limx→0(1+2x)1/x]=7 Also, 2f(1)=7 Putting f′(x)=0, we get x=−3 f′′(x)=1>0 ∴fmin=f(−3)=−92 f′(1)=3+1=4

Suggest Corrections
0
Related Videos
Derivative of Simple Functions
MATHEMATICS
Watch in App