wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let f:RRbe a function such that f(x)=x3+x2f'(1)+xf''(2)+f'''(3)for all xR. Then f(2)=


A

-4

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

30

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

-2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

-2


Explanation of correct answer :

Determining the value of f(2) :

We have, f(x)=x3+x2f'(1)+xf''(2)+f'''(3)

Differentiating the equation:

f'(x)=3x2+2xf'(1)+f''(2)(1)

Differentiating equation (1),

f''(x)=6x+2f'(1)(2)

Differentiating equation (2),

f'''(x)=6(3)

Substituting x=3, we get,

f'''(3)=6(A)

Substituting x=1 in equation in 1.

f'(1)=312+2×1f'(1)+f''(2)=3+2f'(1)+f''(2)(i)

Substituting x=2in equation (2).

f''(2)=2×6+2f'(1)=12+2f'(1)(ii)

Now substituting the value of f''(2) in equation 1.

-f'(1)=3+12+2f'(1)Substituting(ii)in(i)-3f'(1)=15f'(1)=-5(B)

Now, substituting the value of f'(1) in equation (ii).

f''(2)=2×6+2(-5)=2(C)

Now, putting the values of A,B,C in f(x)=x3+x2f'(1)+xf''(2)+f'''(3).

f(2)=(2)3+22×-5+22+6=8-20+4+6=-2

Hence, the value of f(2) is -2.

Therefore, the correct answer is option (C).


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Parametric Differentiation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon