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

Using LMVT in [0,2] 2 − 0 f ( 2 ) − f ( 0 ) = f ′ ( c ) , where c ∈ ( 0 , 2 ) 2 f ( 2 ) + 3 ≤ 5 ⇒ f ( 2 ) ≤ 7

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Mathematics3D-Coordinate GeometryMiscellaneous/MixedMedium2 minPYQ_2020

MathematicsMediumsingle choice

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

Answer:

Solution:

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

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

⏱ 2mℹ️ Source: PYQ_2020

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