JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 7)
Let ƒ be any function continuous on [a, b] and
twice differentiable on (a, b). If for all x $$ \in $$ (a, b),
ƒ'(x) > 0 and ƒ''(x) < 0, then for any c $$ \in $$ (a, b),
$${{f(c) - f(a)} \over {f(b) - f(c)}}$$ is greater than :
1
$${{b - c} \over {c - a}}$$
$${{b + a} \over {b - a}}$$
$${{c - a} \over {b - c}}$$
Explanation
_9th_January_Morning_Slot_en_7_1.png)
It is clear from graph that, slope of AC $$>$$ slope of CB
$$ \Rightarrow $$ $${{f\left( c \right) - f\left( a \right)} \over {c - a}}$$ $$>$$ $${{f\left( b \right) - f\left( c \right)} \over {b - c}}$$
$$ \Rightarrow $$ $${{f\left( c \right) - f\left( a \right)} \over {f\left( b \right) - f\left( c \right)}}$$ $$ > {{c - a} \over {b - c}}$$
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