(b) None of these Explanation: $4 \notin A$ $\{4\} \not \subset A$ $B \not \subset A$ Therefore, we can say that none of these options satisfy the given relation.
(b) $\{x: x>1, x \in R\}$ Explanation: $6 x -1>5$ $\begin{array}{l}\Rightarrow 6 x-1+1>5+1 \\ \Rightarrow 6 x>6 \\ \Rightarrow x>1\end{array}$ Hence the solution set is $\{x: x>1, x \in R\}$
(a) $\frac{1}{2} n(n+1)$ Explanation: We know that $\frac{C_r}{C_{r 1}}=\frac{n-r+1}{r}$, Substituting $r =1,2,3, \ldots, n$, we obtain $\frac{C_1}{C_0}+2 \cdot \frac{C_2}{C_1}+3 \cdot \frac{C_3}{C_2}+\ldots+n \cdot \frac{C_n}{C_{n-1}}= n +( n -1)+( n -2)+\ldots+1=\frac{1}{2} n(n+1)$.
Let $F_1$ be the set of parallelograms, $F_2$ the set of rectangles, $F_3$ the set of rhombuses, $F_4$ the set of squares and $F_5$ the set of trapeziums in a plane. Then $F _1$ may be equal to
A
$F_2 \cap F_3$
B
$F _3 \cap F_4$
C
$F _2 \cup F_5$
D
$F _2 \cup F_3 \cup F_4 \cup F_1$
Answer
(d) $F_2 \cup F_3 \cup F_4 \cup F_1$ Explanation: We know that Every rectangle, square and rhombus is a parallelogram But, no trapezium is a paralleogrm Thus, $F_1=F_2 \cup F_3 \cup F_4 \cup F_1$
If ${ }^{20} C_{r+1}={ }^{20} C_{r-1}$, then r is equal to
A
19
B
10
C
12
D
11
Answer
(b) 10 Explanation: $r +1+ r -1=20\left[\because{ }^n C_x={ }^n C_y \Rightarrow n = x + y\right.$ or $\left.x = y \right]$ $\begin{array}{l}\Rightarrow 2 r =20 \\ \Rightarrow r =10\end{array}$
A line passes through P (1, 2) such that its intercept between the axes is bisected at P. The equation of the line is
A
x + 2y = 5
B
2x + y – 4 = 0
C
x + y – 3 = 0
D
x – y + 1 = 0
Answer
(b) 2x + y – 4 = 0 Explanation: We know that the equation of a line making intercepts a and b with x-axis and y-axis, respectively, is given by $\frac{x}{a}+\frac{y}{b}=1$ which give a = 2 and b = 4. Thus, now we have to find the required equation of the line is given by $\frac{x}{2}+\frac{y}{4}=1$ or $2 x+y-4=0$
If $f(x)=\frac{x^n-a^n}{x-a}$ for some constant, a, then $f ^{\prime}(a)$ is equal to
A
$1 / 2$
B
does not exist
C
1
D
$0$
Answer
(b) does not exist Explanation: Given $f(x)=\frac{x^n-a^n}{x-a}$ $f^{\prime}(x)=\frac{(x-a)\left(n \cdot x^{n-1}\right)-\left(x^n-a^n\right) \cdot 1}{(x-a)^2}$ $\therefore \quad f^{\prime}(a)=\frac{(a-a)\left(n \cdot a^{n-1}\right)-\left(a^n-a^n\right)}{(a-a)^2}$ So $f^{\prime}(a)=\frac{0}{0}=$ does not exist
Distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is
A
$\sqrt{\alpha^2+\gamma^2}$
B
$|\beta|+|\gamma|$
C
$|\beta|$
D
$\beta$
Answer
(a) $\sqrt{\alpha^2+\gamma^2}$ Explanation: The foot of perpendicular from point $P(\alpha, \beta, \gamma)$ on y -axis is $Q(0, \beta, 0)$ $\therefore$ Required distance, $P Q=\sqrt{(\alpha-0)^2+(\beta-\beta)^2+(\gamma-0)^2}=\sqrt{\alpha^2+\gamma^2}$
Number of relations that can be defined on the set A = {a, b, c, d} is
A
24
B
$4^4$
C
16
D
$2^{16}$
Answer
(d) $2^{16}$ Explanation: No. of elements in the set $A =4$. Therefore, the no. of elements in $A \times A=4 \times 4=16$. As, the no. of relations in $A \times A=$ no. of subsets of $A \times A=2^{16}$.