Assertion (A) & Reason (B) MCQ

MCQ 11 Mark

Assertion (A): Let $A=\{2,4,6\}$ and $B=\{3,5,7,9\}$ and defined a function $f=\{(2,3),(4,5),(6,7)\}$ from $A$ to B. Then, f is not onto.
Reason (R): A function $f$ : $A \rightarrow B$ is said to be onto, if every element of $B$ is the image of some elements of $A$ under $f$.

✓
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.
D
A is false but R is true.

Answer

Correct option: A.

Both A and R are true and R is the correct explanation of A.

(a) Both A and R are true and R is the correct explanation of A.
Explanation: Assertion: Given that,
A = {2, 4, 6}
R = {3, 5, 7, 9}
and R = {(2, 3), (4, 5), (6, 7)}
Here, f(2) = 3 f(4) = 5 and f(6) = 7
It can be seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one but f is not onto as 9 \in B does not have a pre-image in A.
Hence, both Assertion and Reason are true, but Reason is not a correct explanation of Assertion

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MCQ 21 Mark

Assertion $(A)$ : If two positive numbers are such that sum is $16$ and sum of their cubes is minimum, then numbers are $8, 8.$
Reason $(R)$ : If f be a function defined on an interval $I$ and $c \in l$ and let $f$ be twice differentiable at $c,$ then $x = c$ is a point of local minima if $f\ '(c) = 0$ and $f\ "(c) > 0 $ and $f(c)$ is local minimum value of $f$.

✓
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.
D
$A$ is false but $R$ is true.

Answer

Correct option: A.

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

Let one number be $x,$ then the other number will be $(16-x)$.
Let the sum of the cubes of these numbers be denoted by $S$.
Then, $S=x^3+(16-x)^3$
On differentiating $\text{w.r.t}. x,$ we get
$\frac{d S}{d z}=3 x ^2+3(16- x )^2(-1)$
$=3 x^2-3(16- x )^2$
$\Rightarrow \frac{d^2 S}{d x^2}=6 x+6(16- x )=96$
For minima put $\frac{d S}{d x}=0$.
$\therefore 3 x^2-3(16-x)^2=0$
$\Rightarrow x^2-\left(256+x^2-32 x\right)=0$
$\Rightarrow 32 x=256$
$\Rightarrow x=8$
At $x =8,\left(\frac{d^2 S}{d x^2}\right)_{x=8}=96;0$
By second derivative test $,x = 8$ is the point of local minima of $S$.
Thus, the sum of the cubes of the numbers is the minimum when the numbers are $8$ and $16-8=8$
Hence, the required numbers are $8$ and $8$.

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Maths Assertion (A) & Reason (B) MCQ

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