JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 18)
If the function f given by f(x) = x3 – 3(a – 2)x2 + 3ax + 7, for some a$$ \in $$R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, $${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$$ is :
$$-$$ 7
5
7
6
Explanation
f '(x) = 3x2 $$-$$ 6(a $$-$$ 2)x + 3a
f '(x) $$ \ge $$ 0 $$\forall $$ x $$ \in $$ (0, 1]
f '(x) $$ \le $$ 0 $$\forall $$ x $$ \in $$ [1, 5)
$$ \Rightarrow $$ f '(x) = 0 at x = 1 $$ \Rightarrow $$ a = 5
f(x) $$-$$ 14 = (x $$-$$ 1)2 (x $$-$$ 7)
$${{f(x) - 14} \over {{{\left( {x - 1} \right)}^2}}} = x - 7$$
f '(x) $$ \ge $$ 0 $$\forall $$ x $$ \in $$ (0, 1]
f '(x) $$ \le $$ 0 $$\forall $$ x $$ \in $$ [1, 5)
$$ \Rightarrow $$ f '(x) = 0 at x = 1 $$ \Rightarrow $$ a = 5
f(x) $$-$$ 14 = (x $$-$$ 1)2 (x $$-$$ 7)
$${{f(x) - 14} \over {{{\left( {x - 1} \right)}^2}}} = x - 7$$
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