tricky busters JEE MAINS
Calculus
Level
2
the set of values of a for which the function f(x)=x^3 + (a+2)x^2 +3ax +5 is one one is [a,b] then find the value of a+b.
The answer is 5.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
f(x) =x^3+(a+2)x^2 +3ax +5
the derivative of the function is 3x^2 +2(a+2)x +3a clearly it is a continuous function
thus for the function to be one one it should not have any turning points thus 4(a+2)^2 -36a <=0 thus a is bounded between 1 and 4 thus 1+4=5.