Assertion (A) & Reason (B) MCQ

MCQ 11 Mark

Assertion (A): Let $A =\{1,5,8,9\}, B =\{4,6\}$ and $f =\{(1,4),(5,6),(8,4),(9,6)\}$, then f is a bijective function.
Reason (R): Let $A=\{1,5,8,9\}, B=\{4,6\}$ and $f=\{(1,4),(5,6),(8,4),(9,6)\}$, then $f$ is a surjective function.

A
Both A and R are true and R is the correct explanation of A.
B
Both A and R are true but R is not the correct explanation of A.
C
A is true but R is false.
✓
A is false but R is true

Answer

Correct option: D.

A is false but R is true

(d) A is false but R is true.
Explanation: We have, $A=\{1,5,8,9\}, B=\{4,6\}$ and $f=\{(1,4),(5,6),(8,4),(9,6)\}$
So, all elements of B has a domain element on A or we can say elements 1 and 8 5 and 9 have some range 4 6, respectively.
Therefore, $f : A \rightarrow B$ is a surjective function not one to one function.
Also, for a bijective function, f must be both one to one onto.

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MCQ 21 Mark

Assertion $(A)$ : If manufacturer can sell $x$ items at a price of $₹ \left(5-\frac{x}{100}\right)$ each. The cost price of $x$ items is $₹ \left(\frac{x}{5}+500\right)$
Then, the number of items he should sell to earn maximum profit is $240$ items.
Reason $(R)$ : The profit for selling $x$ items is given by $\frac{24}{5} x-\frac{x^2}{100}-300$

A
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
B
Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
✓
$A$ is true but $R$ is false.
D
$A$ is false but $R$ is true.

Answer

Correct option: C.

$A$ is true but $R$ is false.

Let $S(x)$ be the selling price of $x$ items and let $C(x)$ be the cost price of $x$ items.
Then, we have
$S(x)=\left(5-\frac{x}{100}\right) x=5 x-\frac{x^2}{100}$ and $C(x)=\frac{x}{5}+500$
Thus, the profit function $P(x)$ is given by
$ P ( x )= S ( x )- C ( x )=5 x -\frac{x^2}{100}-\frac{x}{5}-500$
$\text { i.e. } P ( x )=\frac{24}{5} x-\frac{x^2}{100}-500$
On differentiating both sides $\text{w.r.t.} \ x,$ we get
$P ^{\prime}( x )=\frac{24}{5}-\frac{x}{50}$
Now, $P ^{\prime}( x )=0$ gives $x =240$.
Also, $P ^{\prime}( x )=\frac{-1}{50}$.
So, $P ^{\prime}(240)=\frac{-1}{50}<0$
Thus, $x = 240$ is a point of maxima.
Hence, the manufacturer can earn maximum profit, if he sells $240$ items.

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Maths Assertion (A) & Reason (B) MCQ

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