The graph of the linear equation x − y = 0 passes through the point
Question: The graph of the linear equationxy= 0 passes through the point (a) $\left(\frac{-1}{2}, \frac{1}{2}\right)$ (b) $\left(\frac{3}{2}, \frac{-3}{2}\right)$ (c) $(0,-1)$ (d) $(1,1)$ Solution: (d) (1, 1) Given equation: $x-y=0$ or, $x=y$ When $x=1, y=1$ When $x=2, y=2 \ldots$ and so on Thus, we get the following table: Plot the points on the graph paper. Join the points and extend them in both the directions. We can see the linear equation $x-y=0$ passes through the point $(1,1)$...
Read More →The number of different words that can be made from the letters of the word INTERMEDIATE,
Question: The number of different words that can be made from the letters of the word INTERMEDIATE, such that two vowels never come together, is __________. Solution: Number of letters in INTERMEDIATE is 12. Number of vowels $(A, E, E, I, I, E)$ i.e is 6 . Number of consonants (T T . R M N D) i.e is $6 .$ $\therefore$ Total words are $\frac{12 !}{3 ! 2 ! 2 !}$ Now, number of ways of arranging 6 consonants (2 alike) is $\frac{6 !}{2 !}=6 \times 5 \times 4 \times 3$ = 360 There are 7 gaps in which...
Read More →Find the common difference and write the next four terms of each of the following arithmetic progressions:
Question: Find the common difference and write the next four terms of each of the following arithmetic progressions: (i) 1, 2, 5, 8, ...(ii) 0, 3, 6, 9, ... (iii) $-1, \frac{1}{4}, \frac{3}{2}, \ldots$ (iv) $-1, \frac{5}{6}, \frac{2}{3}, \ldots$ Solution: In the given problem, we need to find the common difference and the next four terms of the given A.P. (i) $1,-2,-5,-8, \ldots$ Here, first term $\left(a_{1}\right)=1$ Common difference $(d)=a_{2}-a_{1}$ $=-2-1$ $=-3$ Now, we need to find the ne...
Read More →The number of different words that can be made from the letters of the word INTERMEDIATE,
Question: The number of different words that can be made from the letters of the word INTERMEDIATE, such that two vowels never come together, is __________. Solution: Number of letters in INTERMEDIATE is 12. Number of vowels $(A, E, E, I, I, E)$ i.e is 6 . Number of consonants (T T . R M N D) i.e is $6 .$ $\therefore$ Total words are $\frac{12 !}{3 ! 2 ! 2 !}$ Now, number of ways of arranging 6 consonants (2 alike) is $\frac{6 !}{2 !}=6 \times 5 \times 4 \times 3$ = 360 There are 7 gaps in which...
Read More →The graph of the line y = −3 does not pass through the point
Question: The graph of the liney=3 does not pass through the point(a) (2, 3)(b) (3, 3)(c) (0, 3)(d) (3, 2) Solution: (d) (3, 2)The graph of the line y = -3 does not pass through (-3,2) since (-3,2) does not satisfy y = -3....
Read More →The number of six digit numbers, all digits of which are odd,
Question: The number of six digit numbers, all digits of which are odd, is __________. Solution: For a six-digits number; _ _ _ _ _ possible options of digit are 1, 3, 5, 7, 9 Since each place has all 5 options available number of six digit numbers, all digit of which are odd is 5 5 5 5 5 5 = 56...
Read More →The graph of the line y = 3 passes through the point
Question: The graph of the liney= 3 passes through the point(a) (3, 0)(b) (3, 2)(c) (2, 3)(d) none of these Solution: Since,the graph of the liney= 3 is parallel tox-axis at a distance of 3 units from thex-axis.Or, they-coordinate of every point on the line is always equal to 3.So, thegraph of the liney= 3 passes through the point (2, 3).Hence, the correct option is (c)....
Read More →The number of ways in which three letters can be posted in five letter boxes,
Question: The number of ways in which three letters can be posted in five letter boxes, is __________. Solution: Since each letter has 5 options each number of ways in which three letters can be posted in five letter boxes is 5 5 5 = 53 = 53...
Read More →In an examination there are three multiple choice questions and each question has four choice.
Question: In an examination there are three multiple choice questions and each question has four choice. The number of ways in which a student can fail to get all answers correct, is __________. Solution: Since each question has 4 options i.e there are 4 choices or 4 ways to answer a question Number of ways to answer 3 questions is 4 4 4 = 64 Out of 64 ways, there is only one way which has all the answer correct. So, number of ways in which a student fails to get all answer correct is 64 1 = 63 ...
Read More →The function f : [−1/2, 1/2, 1/2]→[−π/2, π/2] , defined by
Question: The function $f:[-1 / 2,1 / 2,1 / 2] \rightarrow[-\pi / 2, \pi / 2]$, defined by $f(x)=\sin ^{-1}\left(3 x-4 x^{3}\right)$, is (a) bijection(b) injection but not a surjection(c) surjection but not an injection(d) neither an injection nor a surjection Solution: $f(x)=\sin ^{-1}\left(3 x-4 x^{3}\right)$ $\Rightarrow f(x)=3 \sin ^{-1} x$ Injectivity: Let $x$ and $y$ be two elements in the domain $\left[\frac{-1}{2}, \frac{1}{2}\right]$, such that $f(x)=f(y)$ $\Rightarrow 3 \sin ^{-1} x=3 ...
Read More →The number of ways 'm' men and 'n' women (m > n) can be seated in arow so that no two women sit together is
Question: The number of ways 'm' men and 'n' women (mn) can be seated in arow so that no two women sit together is __________. Solution: mmen can be arranged inm! ways Since no two women are to be together ⇒ we havem+ 1 places for women out ofm+ 1 places, places to be taken =n i.e, the women can be seated inm+1Pn Total number of ways of seating men and women ism! (m+ 1Pn) i. e $m ! \frac{(m+1) !}{(m+1-n) !}$...
Read More →The number of permutations of n distinct objects, taken r at a time,
Question: The number of permutations ofndistinct objects, takenrat a time, when repetitions are allowed, is __________. Solution: The number of permutations ofndistinct objects, takenrat a time, when repetition is allowed is, nnn ------n(rtimes) =nr...
Read More →The number of permutations of n distinct object, taken r at a time,
Question: The number of permutations ofndistinct object, takenrat a time, when repetitions are not allowed, is __________. Solution: The number of permutations ofndistinct objects, takesrat a time, Without repetition is n(n 1) (n 2) ......... (n (r 1)) =nPr...
Read More →Solve the following
Question: If12Pr= 1320, thenr=__________. Solution: Given ${ }^{12} P_{r}=1320$ i. e $\frac{12 !}{(12-r) !}=1320$ $\Rightarrow(12-r) !=\frac{12 !}{1320}$ $\Rightarrow(12-r) !=\frac{12 \times 11 \times 10 \times 9 !}{12 \times 11 \times 10}$ $\Rightarrow(12-r) !=9 !$ $\Rightarrow 12-r=9$ $\Rightarrow r=12-9$ i. e $r=3$...
Read More →The function f : R→R defined by
Question: The function $f: R \rightarrow R$ defined by $f(x)=(x-1)(x-2)(x-3)$ is (a) one-one but not onto(b) onto but not one-one(c) both one and onto(d) neither one-one nor onto Solution: $f(x)=(x-1)(x-2)(x-3)$ Injectivity: $f(1)=(1-1)(1-2)(1-3)=0$ $f(2)=(2-1)(2-2)(2-3)=0$ $f(3)=(3-1)(3-2)(3-3)=0$ $\Rightarrow f(1)=f(2)=f(3)=0$ So, $f$ is not one-one. Surjectivity:Letybe an element in the co domainR,such that $y=f(x)$ $\Rightarrow y=(x-1)(x-2)(x-3)$ Since $y \in R$ and $x \in R, f$ is onto. So,...
Read More →Solve the following
Question: IfnP4:nP5= 1 : 2, thenn=__________. Solution: ${ }^{n} P_{4}:{ }^{n} P_{5}=1: 2$ i. e $\frac{{ }^{n} P_{4}}{{ }^{n} P_{5}}=\frac{1}{2}$ $\frac{n !}{(n-4) !} \times \frac{(n-5) !}{n !}=\frac{1}{2}$ $\frac{n !(n-5) !}{(n-4)(n-5) ! n !}=\frac{1}{2}$ $2=n-4$ i. e $n=6$...
Read More →Solve the following
Question: The value ofn 1Pr+rn 1Pr 1is __________. Solution: ${ }^{n-1} P_{r}+r^{n-1} P_{r-1}$ $=\frac{(n-1) !}{(n-1-r) !}+r \frac{(n-1) !}{(n-1-r+1) !}$ $=\frac{(n-1) !}{(n-r-1) !}+r \frac{(n-1) !}{(n-r) !}$ $=(n-1) !\left[\frac{1}{(n-r-1) !}+\frac{r}{(n-r)(n-r-1) !}\right]$ $=(n-1) !\left[\frac{n-r+r}{(n-r)(n-r-1) !}\right]$ $=\frac{(n-1) !(n-r+r)}{(n-r) !}=\frac{(n-1) ! n}{(n-r) !}={ }^{n} P_{r}$...
Read More →The graph of the line x = 3 passes through the point
Question: The graph of the linex= 3 passes through the point(a) (0, 3)(b) (2, 3)(c) (3, 2)(d) None of these Solution: (c) (3, 2)The graph of line x = 3 is a line parallel to the y-axis.Hence, its passes through (3,2), satisfying x =3....
Read More →The general term of a sequence is give by an = −4n + 15.
Question: The general term of a sequence is give by $a_{n}=-4 n+15$. Is the sequence an A.P.? If so, find its 15 th term and the common difference. Solution: In the given problem, we need to find that the given sequence is an A.P or not and then find its $15^{\text {th }}$ term and the common difference. Here, $a_{n}=-4 n+15$ Now, to find that it is an A.P or not, we will find its few terms by substituting $n=1,2,3$ So, Substitutingn= 1,we get $a_{1}=-4(1)+15$ $a_{1}=11$ Substitutingn= 2,we get ...
Read More →Solve the following
Question: The value ofn 1Pr+rn 1Pr 1is __________. Solution: ${ }^{n-1} P_{r}+r^{n-1} P_{r-1}$ $=\frac{(n-1) !}{(n-1-r) !}+r \frac{(n-1) !}{(n-1-r+1) !}$ $=\frac{(n-1) !}{(n-r-1) !}+r \frac{(n-1) !}{(n-r) !}$ $=(n-1) !\left[\frac{1}{(n-r-1) !}+\frac{r}{(n-r)(n-r-1) !}\right]$ $=(n-1) !\left[\frac{n-r+r}{(n-r)(n-r-1) !}\right]$ $=\frac{(n-1) !(n-r+r)}{(n-r) !}=\frac{(n-1) ! n}{(n-r) !}={ }^{n} P_{r}$...
Read More →The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6
Question: The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time, is (a) 432 (b) 108 (e) 36 (d) 18 Solution: Out of 3, 4, 5 and 6 If the unit place is 3 (say), then remaining three place can be filled in 3! ways. i.e 3 appears in unit place in 3! times. Similarly each digit appear in unit place 3! times So, sums of digits in unit place = 3! (3 + 4 + 5 + 6) = 3 2 (18) = 108....
Read More →The graph of the linear equation 2x + 5y = 10 meets the x-axis at the point
Question: The graph of the linear equation 2x+ 5y= 10 meets thex-axis at the point(a) (0, 2)(b) (2, 0)(c) (5, 0)(d) (0, 5) Solution: If he graph of the linear equation 2x+ 5y= 10 meets thex-axis, theny = 0. Substituting the value of $y=0$ in equation $2 x+5 y=10$, we get $2 x+5(0)=10$ $\Rightarrow 2 x=10$ $\Rightarrow x=\frac{10}{2}$ $\Rightarrow x=5$ So, the point of meeting is (5, 0).Hence, the correct option is (c)....
Read More →The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit exactly once,
Question: The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit exactly once, is (a) 120 (b) 96 (c) 24 (d) 100 Solution: Since each digit is used exactly once, for four digit number, _ _ _ _ First place has 4 options: Second place has 3 options left Thirdplace has 2 options left and Last place has 1 option Therefore, number of different four digit numbers positive = 4 3 2 1 = 24...
Read More →If a function
Question: If a function $f:[2, \infty) \rightarrow B$ defined by $f(x)=x^{2}-4 x+5$ is a bijection, then $B=$ (a) $R$ (b) $[1, \infty)$ (C) $[4, \infty)$ (d) $[5, \infty)$ Solution: Sincefis a bijection, co-domain off =range off $\Rightarrow B=$ range of $f$ Given: $f(x)=x^{2}-4 x+5$ Let $f(x)=y$ $\Rightarrow y=x^{2}-4 x+5$ $\Rightarrow x^{2}-4 x+(5-y)=0$ $\because$ Discrimant, $D=b^{2}-4 a c \geq 0$, $(-4)^{2}-4 \times 1 \times(5-y) \geq 0$ $\Rightarrow 16-20+4 y \geq 0$ $\Rightarrow 4 y \geq 4...
Read More →The number of possible outcomes when a coin is tossed 6 times,
Question: The number of possible outcomes when a coin is tossed 6 times, is (a) 36 (b) 64 (c) 12 (d) 32 Solution: Since Each coin has 2 faces Possible outcomes for each coin tossed is H, T i.e number of possible outcomes for each coin tossed is 2. Hence for 6 times tossing the coin. Hence number of positive outcomes when a coin is tossed 6 times is 26 = 2 2 2 2 2 2 = 64...
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