Questions

Assertion (A) & Reason (B) MCQ

🎯

Test yourself on this topic

2 questions · timed · auto-graded

MCQ 11 Mark
Assertion (A): The function $f(x)=x^2+b x+c$ where b and c are real constants, describes onto mapping.
Reason (R): Let $A=\{1, 2, 3, \ldots, n \}$ and $B =\{ a , b \}$. Then, the number of surjections from A into B is $2^{ n }-2$.
  • 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:
Assertion: Given function is $f(x)=x^2+b x+c$
It is a quadratic equation in x.
So, we will get a parabola either downward or upward.
Hence, it is a many-one mapping and not onto mapping.
Hence, it is neither one-one nor onto mapping.
Reason: Total number of functions $=( n ( B ))^{ n ( A )}=2^{ n }$
Clearly, a function will not be onto if all elements of A map to either a or b
View full question & answer
MCQ 21 Mark
Assertion $(A):$ The absolute maximum value of the function $2 x^3-24 x$ in the interval $[1, 3]$ is $89.$
Reason $(R):$ The absolute maximum value of the function can be obtained from the value of the function at critical points and at boundary points.
  • 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.
Let $f(x)=2 x^3-24 x$
$ \Rightarrow f^{\prime}(x)=6 x^2-24=6\left(x^2-4\right)$
$=6(x+2)(x-2)$
For maxima or minima put $f^{\prime}(x)=0$.
$\Rightarrow 6(x+2)(x-2)=0$
$ \Rightarrow x=2,-2 $
We first consider the interval $[1,3]$.
So, we have to evaluate the value of $f$
at the critical point $x=2 \in[1,3]$ and at the end points of $[1,3]$.
At $x=1, f(1)=2 \times 1^3-24 \times 1=-22$
At $x=2, f(2)=2 \times 2^3-24 \times 2=-32$
At $x=3, f(3)=2 \times 3^3-24 \times 3=-18$
$\therefore$ The absolute maximum value of $f ( x )$ in the interval $[1,3]$ is $-18$ occurring at $x =3$.
Hence, Assertion is false and Reason is true.
View full question & answer
Assertion (A) & Reason (B) MCQ - Maths STD 12 Science Questions - Vidyadip