Point (–7, 0) lies
Question: Point (7, 0) lies(a) on the negative direction of thex-axis(b) on the negative direction of they-axis(c) in the III quadrant(d) in the IV quadrant Solution: The point (7, 0) lies on the negative direction of thex-axis.Hence, the correct option is (a)....
Read More →Find the number of diagonals of
Question: Find the number of diagonals of (i) a hexagon (ii) a polygon of 16 sides. Solution: A polygon ofnsides hasnvertices. By joining any two vertices we obtain either a side or a diagonal. $\therefore$ Number of ways of selecting 2 out of $9={ }^{n} C_{2}=\frac{n(n-1)}{2}$ Out of these lines,nlines are the sides of the polygon. $\therefore$ Number of diagonals $=\frac{n(n-1)}{2}-n=\frac{n(n-3)}{2}$ (i) In a hexagon, there are 6 sides. $\therefore$ Number of diagonals $=\frac{6(6-3)}{2}=9$ (...
Read More →Point (0, –8) lies
Question: Point (0, 8) lies(a) in the II quadrant(b) in the IV quadrant(c) on thex-axis(d) on they-axis Solution: The abscissa of the point (0, 8) is zero.The point (0, 8) lies on they-axis.Hence, the correct option is (d)....
Read More →There are 10 points in a plane of which 4 are collinear.
Question: There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points. Solution: Number of straight lines formed joining the 10 points, taking 2 points at a time $={ }^{10} C_{2}=\frac{10}{2} \times \frac{9}{1}=45$ Number of straight lines formed joining the 4 points, taking 2 points at a time $={ }^{4} C_{2}=\frac{4}{2} \times \frac{3}{1}=6$ But, when 4 collinear points are joined pair wise, they give only one line. $\therefore...
Read More →Points (1, –1), (2, –2), (–3, –4), (4, –5)
Question: Points (1, 1), (2, 2), (3, 4), (4, 5)(a) all lie in the II quadrant(b) all lie in the III quadrant(c) all lie in the IV quadrant(d) do not lie in the same quadrant Solution: The point (1, 1) lies in the IV quadrant.The point (2, 2) lies in the IV quadrant.The point (3, 4) lies in the III quadrant.The point (4, 5) lies in the IV quadrant.Hence, the correct option is (d)....
Read More →The function f : R→R defined by
Question: The function $f: R \rightarrow R$ defined by $f(x)=6^{x}+6^{|x|}$ is (a) one-one and onto(b) many one and onto(c) one-one and into(d) many one and into Solution: (d) many one and intoGraph of the given function is as follows : A line parallel to X axis is cutting the graph at two different values.Therefore, for two different values of x we are getting the same value ofy.That means it is many one function . From the given graph we can see that the range is $[2, \infty)$ and $R$ is the c...
Read More →The points (–5, 3) and (3, –5) lie in the
Question: The points (5, 3) and (3, 5) lie in the(a) same quadrant(b) II and III quadrants respectively(c) II and IV quadrants respectively(d) IV and II quadrants respectively Solution: The point (5, 3) lies in the II quadrant.The point (3, 5) lies in the IV quadrant.Hence, the correct option is (c)....
Read More →A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions.
Question: A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. In how many ways can he choose the 7 questions? Solution: Required ways $={ }^{6} C_{5} \times{ }^{6} C_{2}+{ }^{6} C_{4} \times{ }^{6} C_{3}+{ }^{6} C_{3} \times{ }^{6} C_{4}+{ }^{6} C_{2} \times{ }^{6} C_{5}$ $=2\left({ }^{6} C_{5} \times{ }^{6} C_{2}+{ }^{6} C_{4} \times{ }^{6} C_{3}\...
Read More →A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions.
Question: A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. In how many ways can he choose the 7 questions? Solution: Required ways $={ }^{6} C_{5} \times{ }^{6} C_{2}+{ }^{6} C_{4} \times{ }^{6} C_{3}+{ }^{6} C_{3} \times{ }^{6} C_{4}+{ }^{6} C_{2} \times{ }^{6} C_{5}$ $=2\left({ }^{6} C_{5} \times{ }^{6} C_{2}+{ }^{6} C_{4} \times{ }^{6} C_{3}\...
Read More →The points (other than origin) for which abscissa is equal to the ordinate will lie in the quadrant
Question: The points (other than origin) for which abscissa is equal to the ordinate will lie in the quadrant(a) I only(b) I or II(c) I or III(d) II or IV Solution: (c) I or IIIExplanation:If abscissa = ordinate, there could be two possibilities.Either both are positive or both are negative. So, a point could be either (+,+), which lie in quadrant I or it could beof the type(-,-), which lie in quadrant III.Hence, the points (other then the origin) for which the abscissas are equal to the ordina...
Read More →The nth term of an A.P. is 6n + 2.
Question: The $n$th term of an A.P. is $6 n+2$. Find the common difference. Solution: In the given problem, $n$th term is given by " $a_{n}=6 n+2^{*}$. To find the common difference of the A.P., we need two consecutive terms of the A.P. So, let us find the first and the second term of the given A.P. First term $(n=1)$, $a_{1}=6(1)+2$ $=6+2$ $=8$ Second term $(n=2)$ $a_{2}=6(2)+2$ $=12+2$ $=14$ Now, the common difference of the A.P. $(d)=a_{2}-a_{1}$ $=14-8$ $=6$ Therefore, the common difference ...
Read More →In an examination, a student has to answer 4 questions out of 5 questions;
Question: In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice. Solution: A student has to answer 4 question out of 5 questions. Since questions 1 and 2 are compulsory, he/she will have to answer 2 question from the remaining 3. $\therefore$ Required number of ways $={ }^{3} C_{2}=3$...
Read More →A point both of whose coordinates are negative lies in
Question: A point both of whose coordinates are negative lies in(a) quadrant I(b) quadrant II(c) quadrant III(d) quadrant IV Solution: (c) quadrant IIIExplanation:Points of the type (-,-) lie in the third quadrant....
Read More →A student has to answer 10 questions, choosing at least 4 from each of part A and part B.
Question: A student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can the student choose 10 questions? Solution: The various possibilities for answering the 10 questions are given below: (i) 4 from part A and 6 from part B. (ii) 5 from part A and 5 from part B. (iii) 6 from part A and 4 from part B. $\therefore$ Required number of ways $={ }^{6} C_{4} \times{ }^{7} C_{6}+{ }^{6} C_{5} \times{ }...
Read More →Let f : Z→Z be given by f(x)
Question: Let $f: Z \rightarrow Z$ be given by $f(x)=\left\{\begin{array}{l}\frac{x}{2}, \text { if } x \text { is even } \\ 0, \text { if } x \text { is odd }\end{array}\right.$ Then,fis(a) onto but not one-one(b) one-one but not onto(c) one-one and onto(d) neither one-one nor onto Solution: Injectivity:Letxandybe two elements in the domain (Z), such that $f(x)=f(y)$ Case-1: Let both $x$ and $y$ be even. Then, $f(x)=f(y)$ $\Rightarrow \frac{x}{2}=\frac{y}{2}$ $\Rightarrow x=y$ Case-2: Let both ...
Read More →If a < 0 and b > 0, then the point (a, b) lies in quadrant
Question: Ifa 0 andb 0, then the point(a,b) lies in quadrant(a) IV(b) II(c) III(d) none of these Solution: Ans (b)Explanation:Points of the type (-,+) lie in the second quadrant.Hence, the pointP(a,b), wherea 0 andb 0, lie in quadrant II....
Read More →A student has to answer 10 questions, choosing at least 4 from each of part A and part B.
Question: A student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can the student choose 10 questions? Solution: The various possibilities for answering the 10 questions are given below: (i) 4 from part A and 6 from part B. (ii) 5 from part A and 5 from part B. (iii) 6 from part A and 4 from part B. $\therefore$ Required number of ways $={ }^{6} C_{4} \times{ }^{7} C_{6}+{ }^{6} C_{5} \times{ }...
Read More →If x > 0 and y < 0, then the point (x, y) lies in
Question: Ifx 0 andy 0, then the point (x,y) lies in(a) I(b) III(c) II(d) IV Solution: (d) IVExplanation:The points of the type (+,-) lie in fourth quadrant.Hence, the point (x,y), wherex 0 andy 0, lies in quadrant IV....
Read More →Find the common difference of the A.P. and write the next two terms:
Question: Find the common difference of the A.P. and write the next two terms: (i) 51, 59, 67, 75, ..(ii) 75, 67, 59, 51, ...(iii) 1.8, 2.0, 2.2, 2.4, ... (iv) $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$ (v) 119, 136, 153, 170, ... Solution: In this problem, we are given different A.P. and we need to find the common difference of the A.P., along with the next two terms. (i) $51,59,67,75, \ldots$ Here, $a_{1}=51$ $a_{2}=59$ So, common difference of the A.P. $(d)=a_{2}-a_{1}$ $=59-51$ $=8$ ...
Read More →A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII.
Question: A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the teams be constituted? Solution: A sports team of 11 students is to be constituted, choosing at least 5 students of class XI and at least 5 from class XII. Required number of ways $={ }^{20} C_{5} \times{ }^{20} C_{6}+{ }^{20} C_{6} \times{ }^{20} C_{5}=2 \times{ }^{20} C_{5} \times{ }^{20} C_{6}=2...
Read More →In which quadrant does the point (–7, –4) lie?
Question: In which quadrant does the point (7, 4) lie?(a) IV(b) II(c) III(d) None of these Solution: Points of the type (, ) lie in the III quadrant.The point (7, 4) lies in the III quadrant.Hence, the correct option is (c)....
Read More →From 4 officers and 8 jawans in how many ways can 6 be chosen
Question: From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer? Solution: (i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer. Required number of ways $={ }^{4} C_{1} \times{ }^{8} C_{5}=4 \times \frac{8 !}{5 ! 3 !}=4 \times \frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 6}=224$ (ii) From 4 officers and 8 jawans, 6 need to be chosen and at least one of them is an officer. Required n...
Read More →From 4 officers and 8 jawans in how many ways can 6 be chosen
Question: From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer? Solution: (i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer. Required number of ways $={ }^{4} C_{1} \times{ }^{8} C_{5}=4 \times \frac{8 !}{5 ! 3 !}=4 \times \frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 6}=224$ (ii) From 4 officers and 8 jawans, 6 need to be chosen and at least one of them is an officer. Required n...
Read More →Three vertices of a rectangle ABCD are A(3, 1), B(–3, 1) and C(–3, 3). Plot these points on a graph paper and find the coordinates of the fourth vertex D.
Question: Three vertices of a rectangleABCDareA(3, 1),B(3, 1) andC(3, 3). Plot these points on a graph paper and find the coordinates of the fourth vertexD. Also, find the area of rectangleABCD. Solution: LetA(3, 1),B(3, 1) andC(3, 3) be three vertices of a rectangleABCD. LetA(3, 1),B(3, 1) andC(3, 3) be three vertices of a rectangleABCD. Abscissa ofD= Abscissa ofA= 3.Ordinate ofD= Ordinate ofC= 3. coordinates ofDare (3, 3).AB= (BP+PA) = (3 + 3) units = 6 units.BC= (OQOP) = (3 1) units = 2 units...
Read More →Which of the following functions from A={x ∈ R : −1≤x≤1} to itself are bijections?
Question: Which of the following functions from $A=\{x \in R:-1 \leq x \leq 1\}$ to itself are bijections? (a) $f(x)=|x|$ (b) $f(x)=\sin \frac{\pi x}{2}$ (c) $f(x)=\sin \frac{\pi^{2} x}{4}$ (d) None of these Solution: \text { (b) } f(x)=\sin \frac{\pi x}{2} It is clear that $f(x)$ is one-one. Range of $f=\left[\sin \frac{\pi(-1)}{2}, \sin \frac{\pi(1)}{2}\right]=\left[\sin \frac{-\pi}{2}, \sin \frac{\pi}{2}\right]=[-1,1]=A=$ Co domain of $f$ $\Rightarrow f$ is onto. So, $f$ is a bijection....
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