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

The least value of k for which the function x2+kx+1 is an increasing function in the interval 1<x<2 is


A

-4

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

-3

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

-1

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

-2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D

-2


Calculate the value of k based on given information

The given expression is x2+kx+1 where 1<x<2

Say, fx=x2+kx+1

On differentiating the above with respect to x, we get,

f'x=2x+k ...i [∵ddxxn=nxn-1]

Now, fx is strictly increasing on the interval (1,2) if,

f'x>0 for x∈(1,2)

⇒ 2x+k>0 for x∈(1,2) [using equation i]

⇒ k>-2x for x∈(1,2) ...ii

Now, since x∈(1,2)

Or, 1<x<2

⇒ 2<2x<4

⇒ -2>-2x>-4

Or, -4<-2x<-2 ...iii

Using inequalities ii and iii, it can be concluded that,

k≥-2

Then, the least value of k is -2.

Hence, option (D) is the correct option.


flag
Suggest Corrections
thumbs-up
5
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Continuity in an Interval
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon