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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
MCQ
$y=x(x-3)^2$ decreases for the values of $x$ given by
  • ✓
    1 < x < 3
  • B
    x < 0
  • C
    $x>0$
  • D
    0 < x < $\frac{3}{2}$
✓

Answer

Correct option: A.
1 < x < 3
(a) : $y=x(x-3)^2$
$
\begin{aligned}
\Rightarrow \quad & \frac{d y}{d x}=x \cdot 2(x-3)+(x-3)^2 \\
& =(x-3)(2 x+x-3)=(x-3)(3 x-3)=3(x-3)(x-1)
\end{aligned}
$
For decreasing function, $\frac{d y}{d x}<0$
=> (x - 3) (x - 1 )< 0 = > 1 < x < 3 

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