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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Determine the values of $x$ for which the function $f(x) = x^2 - 6x + 9$ is increasing or decreasing. Also, find the coordinates of the point on the curve $y = x^2 - 6x + 9$ where the normal is parallel to the line $y = x + 5.$
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Answer

We have, $f(x) = x^2 - 6x + 9$
$\therefore f'(x) = 2x - 6$
Critical points f$'(x) = 0$
$\Rightarrow 2(x - 3) = 0 $
$\Rightarrow x = $3
Clearly, $f'(x) > 0$ if $x > 3 $
$f'(x) < 0$ if $x < 3$
Thus, $f(x) $ is increases in $(3,\infty),$ decreases in $(-\infty,3)$
II part:
The given equation of curves
$y = x^2 - 6x + 9 ....(i) $
$y = x + 5 ....(ii)$
Slope of (i) $\text{m}_1=\frac{\text{dy}}{\text{dx}}=2\text{x}-6$
Slope of (ii) $\text{m}_2=1$ Given that slope of normal to (i) is parallel to (ii)
$\therefore\ \frac{-1}{2\text{x}-6}=1$
$\Rightarrow2\text{x}-6=-1$
$\Rightarrow\text{x}=\frac{5}{2}$
From (i) $\text{y}=\frac{25}{4}-15+9$
$=\frac{25}{4}-6$ $=\frac{1}{4}$
Thus, the required point is $\Big(\frac{5}{2},\frac{1}{4}\Big).$

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