Suppose that f(0)=-3 and f prime(x) ≤ 5 for all values of x then the largest value which f(2) can attain is

Question

Suppose that f(0)=-3 and f prime(x) ≤ 5 for all values of x then the largest value which f(2) can attain is (A) 7 (B) - 7 (C) 13 (D) 8

Tardigrade Admin · Accepted Answer

using LMVT in [0,2] f prime(c)=(f(2)-f(0)/2-0) text where c ∈(0,2) f prime(c)=(f(2)+3/2) But f prime( x ) ≤ 5 ( f (2)+3/2) ≤ 5 ⇒ f (2)+3 ≤ 10 ⇒ f (2) ≤ 7

Q. Suppose that $f (0) = - 3$ and $f^{'} (x) \leq 5$ for all values of $x$ then the largest value which $f (2)$ can attain is

7

- 7

13

8

using LMVT in $[0, 2]$
$f^{'} (c) = \frac{f ( 2 ) - f ( 0 )}{2 - 0} where c \in (0, 2)$
$f^{'} (c) = \frac{f ( 2 ) + 3}{2}$
But $f^{'} (x) \leq 5$
$\frac{f ( 2 ) + 3}{2} \leq 5 \Rightarrow f (2) + 3 \leq 10 \Rightarrow f (2) \leq 7$

Q. Suppose that f(0)=−3 and f′(x)≤5 for all values of x then the largest value which f(2) can attain is

7

- 7

13

8

using LMVT in [0,2] f′(c)=2−0f(2)−f(0)​ where c∈(0,2) f′(c)=2f(2)+3​ But f′(x)≤5 2f(2)+3​≤5⇒f(2)+3≤10⇒f(2)≤7

Q. Suppose that $f (0) = - 3$ and $f^{'} (x) \leq 5$ for all values of $x$ then the largest value which $f (2)$ can attain is

using LMVT in $[0, 2]$
$f^{'} (c) = \frac{f ( 2 ) - f ( 0 )}{2 - 0} where c \in (0, 2)$
$f^{'} (c) = \frac{f ( 2 ) + 3}{2}$
But $f^{'} (x) \leq 5$
$\frac{f ( 2 ) + 3}{2} \leq 5 \Rightarrow f (2) + 3 \leq 10 \Rightarrow f (2) \leq 7$