Let f be a twice differentiable function on 1-6- If f2-8-f-2-5-f-x-1 and f-x-4- for all x-1-6- then

F(2)=8,f'(2)=5,f'(x)≥ 1,f”(x)≥ 4 ∀ x∈(1,6) Using LMVT, f”(c_1)=f'(5) f'(2)5 2≥ 4 ∀ x∈(1,6) ⇒ f'(5)≥17 ⋯(1) nbsp; and nbsp; f'(c_2)=f(5) f(2)5 2≥ 1 ∀ x∈ ...

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Byju's Answer

Standard XII

Mathematics

Derivative of One Function w.r.t Another

Let f be a tw...

Question

Let $f$ be a twice differentiable function on $(1, 6)$ . If $f (2) = 8, f^{'} (2) = 5, f^{'} (x) \geq 1$ and $f^{''} (x) \geq 4,$ for all $x \in (1, 6),$ then

f (5) +

f^{'} (5) \geq 28

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f^{'} (5) + f^{''} (5) \leq 20

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f (5) \leq 10

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f (5) + f^{'} (5) \leq 26

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Solution

The correct option is A $f (5) +$ $f^{'} (5) \geq 28$
$f (2) = 8, f^{'} (2) = 5, f^{'} (x) \geq 1, f^{''} (x) \geq 4 \forall x \in (1, 6)$
Using LMVT,
$f^{''} (x) = \frac{f^{'} (5) - f^{'} (2)}{5 - 2} \geq 4 \forall x \in (1, 6)$
$\Rightarrow f^{'} (5) \geq 17 \dots (1)$
and
$f^{'} (x) = \frac{f (5) - f (2)}{5 - 2} \geq 1 \forall x \in (1, 6)$
$\Rightarrow f (5) \geq 11 \dots (2)$
Adding $(1)$ and $(2),$ we get
$f (5) + f^{'} (5) \geq 28$

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Let f be a twice differentiable function on (1,6). If f(2)=8,f′(2)=5,f′(x)≥1 and f′′(x)≥4, for all x∈(1,6), then

The correct option is A f(5)+f′(5)≥28f(2)=8,f′(2)=5,f′(x)≥1,f′′(x)≥4 ∀ x∈(1,6) Using LMVT, f′′(x)=f′(5)−f′(2)5−2≥4 ∀ x∈(1,6) ⇒f′(5)≥17 ⋯(1) and f′(x)=f(5)−f(2)5−2≥1 ∀ x∈(1,6) ⇒f(5)≥11 ⋯(2) Adding (1) and (2), we get f(5)+f′(5)≥28

Let $f$ be a twice differentiable function on $(1, 6)$ . If $f (2) = 8, f^{'} (2) = 5, f^{'} (x) \geq 1$ and $f^{''} (x) \geq 4,$ for all $x \in (1, 6),$ then