Assertion (A) & Reason (B) MCQ

Question 11 Mark

Assertion $(A):f(x)=2 x^3-9 x^2+12 x-3$ is increasing outside the interval $(1, 2).$
Reason $(R): f ^{\prime}( x ) < 0$ for $x \in(1,2)$

Answer

$(b)$ Both $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
Explanation: Assertion: We have, $f(x)=2 x^3-9 x^2+12 x-3$
$\Rightarrow f^{\prime}(x)=6 x^2-18 x+12$
For increasing function, $f^{\prime}(x) \geq 0$
$\therefore 6\left(x^2-3 x+2\right) \geq 0$
$\Rightarrow 6(x-2)(x-1) \geq 0$
$\Rightarrow x \leq 1 \text { and } x \geq 2 $
$\therefore f(x)$ is increasing outside the interval $(1, 2),$ therefore it is a true statement.
$\text { Reason: Now, } f ^{\prime}( x ) < 0$
$\Rightarrow 6( x -2)( x -1) < 0$
$\Rightarrow 1< x < 2$
$\therefore$ Assertion and Reason are both true but Reason is not the correct explanation of Assertion.

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Question 21 Mark

Assertion $(A):$ The function $f : R ^* \rightarrow R ^*$ defined by $f(x)=\frac{1}{x}$ is one$-$one and onto, where $R^*$ is the set of all non$-$zero real numbers.
Reason $(R):$ The function $g : N \rightarrow R ^*$ defined by $f(x)=\frac{1}{x}$ is one$-$one and onto.

Answer

$(c) A$ is true but $R$ is false.
Explanation: Assertion: It is given that$f: R^* \rightarrow R^*$ is defined by
$f(x)=\frac{1}{x}$
For one-one, $f(x)=f(y)$
$\Rightarrow \frac{1}{x}=\frac{1}{y}$
$\Rightarrow x=y$
Therefore, f is one-one.
For onto, it is clear that for $y \in R^*$,
there exists $x=\frac{1}{y} \in R^*$ (exists as $y \neq 0$ ) such that $f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y$
Therefore, $f$ is onto. Thus, the given function $( f )$ is one-one and onto.
Reason: Now, consider function $g : N \rightarrow R ^*$ defined by $g(x)=\frac{1}{x}$.
We have, $g\left(x_1\right)=g\left(x_2\right) \Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

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MATHS Assertion (A) & Reason (B) MCQ

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