1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For all real values of x, the minimum value of 1âˆ’x+x21+x+x2 is

A
0
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
1
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
3
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
13
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
E
2
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

The correct option is A 13Let's find the first derivative of this function (2x−1)(x2+x+1)−(2x+1)(x2−x+1)(x2+x+1)2Putting this to zero, we get x2−1=0Which gives us x=±1Since we can see that the double derivative is quite difficult to solve, so lets put the value of x in main function and see the result. So for x=1 the function is 13 and for x=−1 function is 2So minimum value is 13 at x=1

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Second Derivative Test Failure Condition
MATHEMATICS
Watch in App
Join BYJU'S Learning Program