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If the function $f$ given by $f(x)=x^{3}-3\left ( a-2 \right )x^{2}+3ax+7,$ for some $a\equiv R$ is increasing in $(0,1]$ and decreasing in $[1,5),$ then a root of the equation, $\frac{f(x)-14}{\left ( x-1 \right )^{2}}=0\: \left ( x\neq 1 \right )$ is :
Option 1)
$5$
Option 2)
$6$
Option 3)
$7$
Option 4)
$-7$

Answers (1)

best_answer

Condition for increasing functions -

For increasing function tangents drawn at any point on it makes an acute slope with positive x-axis.

$M_{T}=tan\theta\geq 0$

$\therefore \:\:\:\frac{dy}{dx}=f'(x)\geq 0\:\:for\:\:x\epsilon (a,b)$

- wherein

Where f(x) is continuous and differentiable for (a,b)

Condition for Decreasing function -

The tangents drawn at any point on it make an obtuse angle with positive direction of the x-axis.

$M_{T}=tan\theta\leq 0$

$\therefore \:\:\frac{dy}{dx}\leq 0\:\:\:\:\:x\epsilon (a,b)$

- wherein

y=f(x) is continuous and differentiable in (a,b)

${f}'\left ( x \right )=\: \: 3\left ( x^{2}-2(a-2)x+a \right )\\$

${f}'\left ( 1 \right )=0\\\\$

$\Rightarrow a=5$

$\frac{f(x)-14}{(x-1)^{2}}=\frac{x^{3}-ax^{2}+15x-7}{(x-1)^{2}}=x-7=0\\\\$

$\Rightarrow x=7$

Option 1)

$5$

Option 2)

$6$

Option 3)
$7$
Option 4)

$-7$

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