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  • y = x (x - 3)2 decreases for the values of x given by : A. 1 < x < 3 B. x < 0 C. x > 0

y = x (x - 3)^2 decreases for the values of x given by :
A. 1 < x < 3
B. x < 0
C. x > 0
D. 0<x<\frac{3}{2}

Answers (1)

Given y = x (x - 3)\textsuperscript{2}

\\ { \Rightarrow y=x(x\textsuperscript{2}-6x+9)}\\ { \Rightarrow y=x\textsuperscript{3}-6x\textsuperscript{2}+9x}\\

Apply first derivative and get
\frac{d y}{d x}=\frac{d\left(x^{3}-6 x^{2}+9 x\right)}{d x}$
Apply sum rule of differentiation and get
\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(6 \mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(9 \mathrm{x})}{\mathrm{dx}}$
Apply power rule and get
\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}-12 \mathrm{x}+9$ \\$\Rightarrow \frac{d y}{d x}=3\left(x^{2}-4 x+3\right)$
Now, split middle term and get
\\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3\left(\mathrm{x}^{2}-3 \mathrm{x}-\mathrm{x}+3\right)$ \\$\Rightarrow \frac{d y}{d x}=3(x(x-3)-1(x-3))$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3((\mathrm{x}-3)(\mathrm{x}-1))$

Now, \frac{\mathrm{d} y}{\mathrm{dx}}=0 gives us

x=1, 3

The points divide this real number line into three intervals

\\ {(- \infty , 1), (1,3) and (3, \infty )}\\ {(i)$ in the interval (- \infty , 1), f'(x)>0}\\ { \therefore \text{ f(x) is increasing in} (- \infty ,1)}\\ {(ii)$ \text{in the interval} (1,3), f'(x) \leq 0}\\ { \therefore $f(x) is decreasing in (1,3)}\\ {(iii) in the interval (3, \infty ), f'(x)>0}\\ { \therefore \text{f(x) is increasing in} (3, \infty )}\\

So, the interval on which the function decreases is (1,3) i.e., 1<x<3

So the correct answer is option A.

Posted by

infoexpert22

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