If f - R - R is a twice differentiable function such that f-x - 0 for all x - R- and f 12-12- f1-1- then

Given, f (12)=12, f(1)=1 Let nbsp;g(x)=f(x) x, x ∈[12,1 ] Then, g(12)=g(1)=0 Using Rolle's theorem, we get g'(c)=0, c ∈(12,1) ⇒ f'(c) 1=0 ⇒ f'(c)=1 Since c ...

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Byju's Answer

Standard XIII

Mathematics

Rolle's Theorem

If f : R → R ...

Question

If $f : R \to R$ is a twice differentiable function such that $f^{''} (x) > 0$ for all $x \in R$ , and $f (\frac{1}{2}) = \frac{1}{2}, f (1) = 1, then$

f^{'} (1) \leq 0

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0 < f^{'} (1) \leq \frac{1}{2}

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\frac{1}{2} < f^{'} (1) \leq 1

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f^{'} (1) > 1

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Solution

The correct option is D $f^{'} (1) > 1$
$f (\frac{1}{2}) = \frac{1}{2}, f (1) = 1 Let g (x) = f (x) - x, x \in [\frac{1}{2}, 1] g (\frac{1}{2}) = g (1) = 0$
Using Rolle's theorem, we get
$g^{'} (c) = 0, c \in (\frac{1}{2}, 1) \Rightarrow f^{'} (c) - 1 = 0 [∵ g^{'} (x) = f^{'} (x) - 1] \Rightarrow f^{'} (c) = 1, c \in (\frac{1}{2}, 1) As f^{''} (x) > 0, \Rightarrow f^{'} (x) is increasing on R ∴ f^{'} (1) > f^{'} (c) \Rightarrow f^{'} (1) > 1$

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If f:R→R is a twice differentiable function such that f′′(x)>0 for all x∈R, and f(12)=12, f(1)=1, then

The correct option is D f′(1)>1 f(12)=12, f(1)=1Let g(x)=f(x)−x, x∈[12,1]g(12)=g(1)=0 Using Rolle's theorem, we get g′(c)=0, c∈(12,1)⇒f′(c)−1=0 [∵g′(x)=f′(x)−1]⇒f′(c)=1, c∈(12,1)As f′′(x)>0,⇒f′(x) is increasing on R∴f′(1)>f′(c)⇒f′(1)>1

If $f : R \to R$ is a twice differentiable function such that $f^{''} (x) > 0$ for all $x \in R$ , and $f (\frac{1}{2}) = \frac{1}{2}, f (1) = 1, then$