Possibly Æ’(x) = x2
a) Æ’"(x) = 2
b) ƒ'(x) ≠2
c) true : Using LMVT
Let f be twice diff function satisfying f(1)=1,f(2)=4,f(3)=9 then:
a)f"(x)=2
b)f'(x)=2
c)there exists at least one x E (1,3) such that f ' (x)=2
c is false.
f(x)=x2 => f'(x)=2x
2x=2 at x=1 which is not an elemnt of (1,3).
You cannot assume f(x) = x2.
Instead look at the function g(x) = f(x) - x2
We have g(1) = g(2) = g(3) = 0
What conclusions can now be drawn?
no you dont tapan..
you have to just see which of these options are possible.
oh ok!!
then using prophet suir's g(x) we can easily say
options b & c are wrong, wat is left shud be the ans
but ans is given to be c........................[7][7][7]
HELP.................
in that case the question should have been there exists x ε [1,3] such that f"(x) = 2
In that case the question is wrong. The options should be satisfied by any function satisfying the given conditions. And as sky has pointed out f(x) = x2 does not satisfy the three options.
Better check your source