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Observe the following statements
A : $f(x)=2 x^3-9 x^2+12 x-3$ is increasing outside the interval $(1,2)$
$\mathrm{R}: f^{\prime}(x) < 0$ for $x \in(1,2)$.
Then, which of the following is true?
Options:
A : $f(x)=2 x^3-9 x^2+12 x-3$ is increasing outside the interval $(1,2)$
$\mathrm{R}: f^{\prime}(x) < 0$ for $x \in(1,2)$.
Then, which of the following is true?
Solution:
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Verified Answer
The correct answer is:
Both A and R are true, and R is not the correct
reason for A
reason for A
Statement A
For increasing function, $f^{\prime}(x)>0$
$\begin{array}{lcrl}
& \therefore & & 6\left(x^2-3 x+2\right)>0 \\
\Rightarrow & & 6(x-2)(x-1)>0 \\
\Rightarrow & & x < 1 \text { and } x>2
\end{array}$
$\therefore f(x)$ is increasing outside the interval $(1,2)$, therefore it is true statement.
From Eq. (ii)
$f^{\prime}(x)=6 x^2-18 x+12$
for decreasing
$\begin{array}{lcc}
& & f^{\prime}(x) < 0 \\
\Rightarrow & & 6(x-2)(x-1) < 0 \\
\Rightarrow & x>1 \text { and } x < 2 \\
\therefore & f(x) \text { is decreasing in }(1,2) .
\end{array}$
$\therefore \quad A$ and $R$ are both true, but $R$ is not the correct reason.$
For increasing function, $f^{\prime}(x)>0$
$\begin{array}{lcrl}
& \therefore & & 6\left(x^2-3 x+2\right)>0 \\
\Rightarrow & & 6(x-2)(x-1)>0 \\
\Rightarrow & & x < 1 \text { and } x>2
\end{array}$
$\therefore f(x)$ is increasing outside the interval $(1,2)$, therefore it is true statement.
From Eq. (ii)
$f^{\prime}(x)=6 x^2-18 x+12$
for decreasing
$\begin{array}{lcc}
& & f^{\prime}(x) < 0 \\
\Rightarrow & & 6(x-2)(x-1) < 0 \\
\Rightarrow & x>1 \text { and } x < 2 \\
\therefore & f(x) \text { is decreasing in }(1,2) .
\end{array}$
$\therefore \quad A$ and $R$ are both true, but $R$ is not the correct reason.$
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