If f : R → R is a twice differentiable function such that f ′ ′ x > 0 for all x ∈ R and f 1 2 = 1 2 , f 1 = 1 then

Using LMVT on f x for x ∈ 1 2 , 1 f 1 - f 1 2 1 - 1 2 = f ' c , where c ∈ 1 2 , 1 1 - 1 2 1 2 = f ' c ⇒ f ' c = 1 , where c ∈ 1 2 , 1 ∵ f ″ x > 0 ∀ x ∈ R ∴ f ' x is an increasing function ∀ x ∈ R ∴ f ′ 1 > 1

Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Mathematics3D-Coordinate GeometryMiscellaneous/MixedEasy2 minPYQ_2017

MathematicsEasysingle choice

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

Options:

Answer:

Solution:

Using LMVT on $f (x)$ for $x \in [\frac{1}{2}, 1]$

$\frac{f (1) - f (\frac{1}{2})}{1 - \frac{1}{2}} = f^{'} (c),$ where $c \in (\frac{1}{2}, 1)$

$\frac{1 - \frac{1}{2}}{\frac{1}{2}} = f^{'} (c) \Rightarrow f^{'} (c) = 1,$ where $c \in (\frac{1}{2}, 1)$

$∵ f^{″} (x) > 0 \forall x \in R$

$∴ f^{'} (x)$ is an increasing function $\forall x \in R$

$∴ f^{'} (1) > 1$

Stream:JEE_ADVSubject:MathematicsTopic:3D-Coordinate GeometrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2017

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