JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 9)
The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = $$\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}$$ is :
$$-$$ 222
$$-$$ 122
$$122$$
222
Explanation
S = {x $$ \in $$ R, x2 + 30 $$-$$ 11x $$ \le $$ 0}
= {x $$ \in $$ R, 5 $$ \le $$ x $$ \le $$ 6}
Now f(x) = 3x3 $$-$$ 18x2 + 27x $$-$$ 40
$$ \Rightarrow $$ f '(x) = 9(x $$-$$ 1)(x $$-$$ 3),
which is positive in [5, 6]
$$ \Rightarrow $$ f(x) increasing in [5, 6]
Hence maximum value = f(6) = 122
= {x $$ \in $$ R, 5 $$ \le $$ x $$ \le $$ 6}
Now f(x) = 3x3 $$-$$ 18x2 + 27x $$-$$ 40
$$ \Rightarrow $$ f '(x) = 9(x $$-$$ 1)(x $$-$$ 3),
which is positive in [5, 6]
$$ \Rightarrow $$ f(x) increasing in [5, 6]
Hence maximum value = f(6) = 122
Comments (0)
