Let f: [a,b]→[a,b] (where a,b∈R and a

Let g ( x ) = f ( x ) − x , x ∈ ( a , b ) g ( a ) g ( b ) = ( f ( a ) − a ) ( f ( b ) − b ) = ( b − a ) ( a − b ) = − ( b − a ) 2 < 0 By IVT, g ( c ) = 0 , C ∈ ( a , b ) ⇒ f ( c ) − C = 0 f ( c ) = C By LMVT on ( a , c ) f ' ( C 1 ) = f ( c ) − f ( a ) c − a By LMVT C 2 o n ( c , b ) f ' ( C 2 ) = f ( b ) − f ( c ) b − c f ' ( C 1 ) f ' ( C 2 ) = ( f ( c ) − f ( a ) c − a ) ( f ( b ) − f ( c ) b − c ) = c − b c − a . a − c b − c =1 | f ' ( C 1 ) | = | f ( c ) − f ( a ) c − a | = | c − b c − a | > 1 ( ∵ a < b ) | f 2 ( C 2 ) | = | f ( b ) − f ( c ) c − b | = | a − c c − b | < 1

∃ some C ∈ ( a , b ) such that | f ' ( C ) | < 1

∃ at least one C ∈ ( a , b ) such that f(C)=C

∃ some C ∈ ( a , b ) such that | f ' ( C ) | > 1

∃ C 1 , C 2 ∈ ( a , b ) , ( C 1 ≠ C 2 ) such that f ' ( C 1 ) f ' ( C 2 ) = 1

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Let f: $[a, b] \to [a, b]$ (where $a, b \in R$ and a<b) be a differentiable non-linear functions for which f(a)=b & f(b)=a then

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\exists$ some $C \in (a, b)$ such that $| f^{'} (C) | < 1$

$\exists$ at least one $C \in (a, b)$ such that f(C)=C

$\exists$ some $C \in (a, b)$ such that $| f^{'} (C) | > 1$

$\exists$ $C_{1}, C_{2} \in (a, b)$ , $(C_{1} \neq C_{2})$ such that $f^{'} (C_{1}) f^{'} (C_{2}) = 1$

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let $g (x) = f (x) - x$ , $x \in (a, b)$
$g (a) g (b) = (f (a) - a) (f (b) - b)$
$= (b - a) (a - b)$
$= - {(b - a)}^{2} < 0$
By IVT, $g (c) = 0, C \in (a, b)$
   $\Rightarrow f (c) - C = 0$
   $f (c) = C$
By LMVT on $(a, c)$
$f^{'} (C_{1}) = \frac{f (c) - f (a)}{c - a}$
By LMVT
$C_{2} o n (c, b)$
$f^{'} (C_{2}) = \frac{f (b) - f (c)}{b - c}$
$f^{'} (C_{1}) f^{'} (C_{2}) = (\frac{f (c) - f (a)}{c - a}) (\frac{f (b) - f (c)}{b - c})$
$= \frac{c - b}{c - a} . \frac{a - c}{b - c}$
=1
$| f^{'} (C_{1}) | = | \frac{f (c) - f (a)}{c - a} |$
= $| \frac{c - b}{c - a} | > 1 (∵ a < b)$
$| f^{2} (C_{2}) | = | \frac{f (b) - f (c)}{c - b} |$
   $= | \frac{a - c}{c - b} | < 1$