Let the function- f - -7- 0- - R be continuous on -7- 0- and differentiable on -7- 0- If f -7 - -3 and f-x - 2- for all x - -7- 0- then for all such functions f- f-1 - f0 lies in the interval-

F( 7) = 3 and f'(x) ≤ 2 Applying LMVT in [−7,0], we get f( 7) f(0) 7 = f'(c) ≤ 2 ⇒ 3 f(0) 7≤ 2 ⇒ f(0) + 3 ≤ 14 nbsp; ⇒ f(0) ≤ 11 Applying LMVT in [−7, ...

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Byju's Answer

Standard XII

Mathematics

LaGrange's Mean Value theorem

Let the funct...

Question

Let the function, $f : [- 7, 0] \to R$ be continuous on $[- 7, 0]$ and differentiable on $(- 7, 0)$ . If $f (- 7) = - 3$ and $f^{'} (x) \leq 2,$ for all $x \in (- 7, 0),$ then for all such functions $f, f (- 1) + f (0)$ lies in the interval:

[- 3, 11]

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[- 6, 20]

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(- \infty, 20]

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(- \infty, 11]

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Solution

The correct option is C $(- \infty, 20]$
$f (- 7) = - 3$ and $f^{'} (x) \leq 2$
Applying LMVT in $[- 7, 0]$ , we get
$\frac{f (- 7) - f (0)}{- 7} = f^{'} (c) \leq 2 \Rightarrow \frac{- 3 - f (0)}{- 7} \leq 2 \Rightarrow f (0) + 3 \leq 14 \Rightarrow f (0) \leq 11$

Applying LMVT in $[- 7, - 1]$ , we get
$\frac{f (- 7) - f (- 1)}{- 7 + 1} = f^{'} (c) \leq 2 \Rightarrow \frac{- 3 - f (- 1)}{- 6} \leq 2 \Rightarrow f (- 1) + 3 \leq 12 \Rightarrow f (- 1) \leq 9$

Therefore, $f (- 1) + f (0) \leq 20$

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Let the function, f:[−7,0]→R be continuous on [−7,0] and differentiable on (−7,0). If f(−7)=−3 and f′(x)≤2, for all x∈(−7,0), then for all such functions f, f(−1)+f(0) lies in the interval:

Let the function, $f : [- 7, 0] \to R$ be continuous on $[- 7, 0]$ and differentiable on $(- 7, 0)$ . If $f (- 7) = - 3$ and $f^{'} (x) \leq 2,$ for all $x \in (- 7, 0),$ then for all such functions $f, f (- 1) + f (0)$ lies in the interval: