Let f be differentiable for all x- If f1 - -2 and f -x-2 for all x-1-6- then which of the following cannot be the value of f6-

By Lagrange's Mean Value Theorem, there exists c ∈ (1, 6) such that f'(c)=f(6) f(1)/6 1 f(6) + 2/5≥2 (∵ f'(x) ≥ 2 for all x∈[1,6] nbsp; ) f(6)+2≥10 or f(6) ...

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Standard XII

Mathematics

LaGrange's Mean Value theorem

Let f be diff...

Question

Let f be differentiable for all x. If $f (1) = - 2$ and $f' (x) \geq 2$ for all $x \in (1, 6]$ , then which of the following cannot be the value of f(6)?

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Solution

The correct option is D 7
By Lagrange's Mean Value Theorem, there exists $c \in (1, 6)$ such that
$f^{'} (c) = \frac{f (6) - f (1)}{6 - 1}$
$\frac{f (6) + 2}{5} \geq 2 (∵ f^{'} (x) \geq 2 for all x \in [1, 6])$
$f (6) + 2 \geq 10$ or $f (6) \geq 8$
$∴$ 6 and 7 cannot be the value of f(6)

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Let f be differentiable for all x. If f(1)=−2 and f '(x)≥2 for all x∈(1,6], then which of the following cannot be the value of f(6)?

The correct option is D 7By Lagrange's Mean Value Theorem, there exists c∈(1,6) such that f′(c)=f(6)−f(1)6−1 f(6) + 25≥2(∵f′(x)≥2 for all x∈[1,6]) f(6)+2≥10 or f(6)≥8 ∴ 6 and 7 cannot be the value of f(6)

Let f be differentiable for all x. If $f (1) = - 2$ and $f' (x) \geq 2$ for all $x \in (1, 6]$ , then which of the following cannot be the value of f(6)?

The correct option is D 7
By Lagrange's Mean Value Theorem, there exists $c \in (1, 6)$ such that
$f^{'} (c) = \frac{f (6) - f (1)}{6 - 1}$
$\frac{f (6) + 2}{5} \geq 2 (∵ f^{'} (x) \geq 2 for all x \in [1, 6])$
$f (6) + 2 \geq 10$ or $f (6) \geq 8$
$∴$ 6 and 7 cannot be the value of f(6)