2 Marks Questions

Question 12 Marks

In what ratio is the line joining the points $(2, 3)$ and $(4, -5)$ divided by the line passing through the points $(6, 8)$ and $(-3, -2).$

Answer

Therefore required equation of the line joining the points $(6, 8)$ and $(-3, -2)$ is
$y-8=\frac{-2-8}{-3-6}(x-6)$
$\Rightarrow 10 x-9 y+12=0$
Suppose $10x - 9y + 12 = 0$ divide the line joining the points $(2, 3)$ and $(4, -5)$ at point $P$ in the ratio $k:1$
$\therefore P \equiv\left(\frac{4 k+2}{k+1}, \frac{-5 k+3}{k+1}\right)$
$P$ lies on the line $10x - 9y + 12 = 0$
$\therefore 10\left(\frac{4 k+2}{k+1}\right)-9\left(\frac{-5 k+3}{k+1}\right)+12=0$
$\Rightarrow 40 k+20+45 k-27+12 k+12=0$
$\Rightarrow 97 k+5=0$
$\Rightarrow k=\frac{-5}{97}$

View full question & answer→

Question 22 Marks

If $A=\{a, b, c, d, e\}, B=\{a, c, e, g\}$ and $C=\{b, e, f, g\},$ verify that : $(A \cap B) \cap C=A \cap(B \cap C)$

Answer

Suppose $x$ be any element of $(A \cap B) \cap C$
$\Rightarrow x \in(A \cap B) \text { } x \in C$
$\Rightarrow x \in A \text { } x \in B \text { } x \in C$
$\Rightarrow x \in A \text { } x \in(B \cap C)$
$\Rightarrow x \in A \cap(B \cap C)$
$\Rightarrow(A \cap B) \cap C \subset A \cap(B \cap C) \ldots( i )$
Now, suppose $x$ be an element of $A \cap(B \cap C)$
Then, $x \in A$ $(B \cap C)$
$\Rightarrow x \in A \text { } x \in B \text { } x \in C$
$\Rightarrow x \in(A \cap B) \text { } x \in C$
$\Rightarrow x \in(A \cap B) \cap C$
$\Rightarrow A \cap(B \cap C) \subset(A \cap B) \cap C \ldots \ldots(\text { ii) }$
Using $(i)$ $(ii),$ we have $(A \cap B) \cap C=A \cap(B \cap C)$
Hence, proved.

View full question & answer→

Question 32 Marks

An integer is chosen at random from the numbers ranging from $1$ to $50$. What is the probability that the integer chosen is a multiple of $2$ or $3$ or $10$ ?

Answer

We have to find the probability that the integer is chosen is a multiple of $2$ or $3$ or $10$
Out of 50 integers an integer can be chosen in ${ }^{50} C _1$ ways.
Total number of elementary events $={ }^{50} C _1=50$
Consider the following events:
$A =$ Getting a multiple of $2, B =$ Getting a multiple of $3$ and $, C =$ Getting a multiple of $10$
Clearly $, A = \{2, 4,..., 50\}, B = \{3, 6,..., 48\}, C = \{10, 20,..., 50\}$
$A \ n\ B = \{6,12,..., 48\}, B\ n \ C = \{30\}, A\ n\ C = \{10, 20,..., 50\}$ and $, A\ n\ B\ n\ C = \{30\}$
$\therefore P(A)=\frac{25}{50}, P(B)=\frac{16}{50}, P(C)=\frac{5}{50}, P(A \cap B)=\frac{8}{50}, P(B \cap C)=\frac{1}{50}$
$P(A \cap C)=\frac{5}{50}$ and $ P(A \cap B \cap C)=\frac{1}{50}$
$\text { Required probability }= P ( A \cap B \cap C )$
$= P ( A )+ P ( B )+ P ( C )- P ( A \cap B )- P ( A \cap C )- P ( B \cap A )+ P ( A \cap B \cap C )$
$=\frac{25}{50}+\frac{16}{50}+\frac{5}{50}-\frac{8}{50}-\frac{1}{50}-\frac{5}{50}+\frac{1}{50}=\frac{33}{50}$

View full question & answer→

Question 42 Marks

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be a black card (i.e., a club or, a spade).

Answer

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52. Let C be the event ‘card drawn is black card’.
Since total number of black cards = 26
So, $P ( C )=\frac{26}{52}=\frac{1}{2}$
Thus, probability of a black card $=\frac{1}{2}$

View full question & answer→

Question 52 Marks

Differentiate the function with respect to $x:\left(3 x^2-x+1\right)^4$.

Answer

To Find : $\frac{d}{d x}\left(3 x^2- x +1\right)^4$
Formula used: $\frac{d}{d x}\left(y^n\right)=n y^{ n -1} \times \frac{d y}{d x}$
Let $y=\left(3 x^2-x+1\right)^4$
So, by using the above formula, we have
$\frac{d}{d x}\left(3 x^2- x +1\right)^4=4\left(3 x ^2- x +1\right)^3 \times \frac{d}{d x}\left(3 x ^2- x +1\right)=4\left(3 x ^2- x +1\right)^3(6 x -1)$

View full question & answer→

Question 62 Marks

Find the domain and range of the function $f(x)=1-|x-2|$

Answer

According to the question, $f(x)=1-|x-2|$
We observe that $f(x)$ is defined for all $x \in R$. Therefore, Domain $(f)=R$
$\therefore 0 \leq|x-2|<\infty$ for all $x \in R$
$\Rightarrow-\infty<-|x-2| \leq 0$ for all $x \in R$
$\Rightarrow-\infty<1-|x-2| \leq 1$ for all $x \in R$
$\Rightarrow-\infty < f(x) \leq 1$ for all $x \in R$
$\therefore$ Range $(f)=(-\infty, 1]$

View full question & answer→

Question 72 Marks

f, g and h are three functions defined from R to R as follows:
i. $f(x)=x^2$
ii. $g(x)=x^2+1$
iii. $h(x)=\sin x$
Then, find the range of each function.

Answer

i. Given: $f: R \rightarrow R$ such that $f(x)=x^2$
since the value of $x$ is squared, $f(x)$ will always be equal or greater than 0
$\therefore$ the range is $[0, \infty)
ii. Given: g: R \rightarrow R$ such that $g(x)=x^2+1$
since, the value of x is squared and also adding with 1, g(x) will always be equal or greater than 1
$\therefore$ Range of $g(x)=[1, \infty)$
iii. Given: $h: R \rightarrow R$ such that $h(x)=\sin x$
We know that, sin (x) always lies between -1 to 1
$\therefore$ Range of $h(x)=(-1,1)$

View full question & answer→

Maths 2 Marks Questions

Test yourself on this topic