????

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

10 Answers

11
rkrish ·


Possibly Æ’(x) = x2

a) Æ’"(x) = 2
b) ƒ'(x) ≠2
c) true : Using LMVT

1
skygirl ·

c is false.

f(x)=x2 => f'(x)=2x

2x=2 at x=1 which is not an elemnt of (1,3).

341
Hari Shankar ·

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?

21
tapanmast Vora ·

BTW wat is the questn heree???

do v hav to findthe functtn f ??

62
Lokesh Verma ·

no you dont tapan..

you have to just see which of these options are possible.

21
tapanmast Vora ·

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

24
eureka123 ·

but ans is given to be c........................[7][7][7]

HELP
.................

341
Hari Shankar ·

in that case the question should have been there exists x ε [1,3] such that f"(x) = 2

24
eureka123 ·

maybe sir thats misprint..............but it is 2005 question.....

341
Hari Shankar ·

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

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