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The inverse of the function f : R→{x ∈ R : x < 1} given by

Question: The inverse of the function $f: R \rightarrow\{x \in R: x1\}$ given by $f(x)=\frac{e^{z}-e^{-x}}{e^{x}+e^{-x}}$ is (a) $\frac{1}{2} \log \frac{1+x}{1-x}$ (b) $\frac{1}{2} \log \frac{2+x}{2-x}$ (c) $\frac{1}{2} \log \frac{1-x}{1+x}$ (d) none of these Solution: Let $f^{-1}(x)=y \quad \ldots(1)$ $\Rightarrow f(y)=x$ $\Rightarrow \frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=x$ $\Rightarrow \frac{e^{-y}\left(e^{2 y}-1\right)}{e^{-y}\left(e^{2 y}+1\right)}=x$ $\Rightarrow\left(e^{2 y}-1\right)=x\left(e...

The point whose ordinate is 3 and which lies on the y-axis is

Question: The point whose ordinate is 3 and which lies on they-axis is(a) (3, 0)(b) (0, 3)(c) (3, 3)(d) (1, 3) Solution: The ordinate of a point is they-coordinate of the point. So, they-coordinate of the point is 3.Also, any point on they-axis has coordinates in the form (0,y).Thus, the point whose ordinate is 3 and which lies on they-axis is (0, 3).Hence, the correct answer is option (b)....

We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can the selection be made?

Question: We wish to select 6 persons from 8, but if the personAis chosen, thenBmust be chosen. In how many ways can the selection be made? Solution: 6 people are to be selected from 8. There are two case. (i) When A is selected, then B must be chosen. $\therefore$ Number of ways $={ }^{6} C_{4}=15$ (ii) When A is not chosen: Number of ways $={ }^{7} C_{6}=7$ Total number of ways = 15 + 7 = 22...

The point at which the two coordinate axes meet is called

Question: The point at which the two coordinate axes meet is called(a) the abscissa(b) the ordinate(c) the origin(d) the quadrant Solution: (c) the originExplanation: The point at which two axes meet is called as the origin....

Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king?

Question: Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king? Solution: There are 4 kings in the deck of cards. So, we are left with 48 cards out of 52. $\therefore$ Required combination $={ }^{48} C_{1} \times{ }^{4} C_{4}+{ }^{48} C_{2} \times{ }^{4} C_{3}+{ }^{48} C_{3} \times{ }^{4} C_{2}+{ }^{48} C_{4} \times{ }^{4} C_{1}$ $=48+4512+103776+778320$ $=886656$...

Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king?

Question: Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king? Solution: There are 4 kings in the deck of cards. So, we are left with 48 cards out of 52. $\therefore$ Required combination $={ }^{48} C_{1} \times{ }^{4} C_{4}+{ }^{48} C_{2} \times{ }^{4} C_{3}+{ }^{48} C_{3} \times{ }^{4} C_{2}+{ }^{48} C_{4} \times{ }^{4} C_{1}$ $=48+4512+103776+778320$ $=886656$...

Abscissa of a point is positive in

Question: Abscissa of a point is positive in(a) I and II quadrants(b) I and IV quadrants(c) I quadrant only(d) II quadrant only Solution: (b) I and IV quadrantsExplanation:If abscissa of a point is positive, then the ordinate could be either positive or negative.It means that the type of any point can be either (+,+) or (+, -).Points of the type (+,+) lie in quadrant I, whereas points of the type (+,-) lie in quadrant IV....

If g (f (x))=|sin x| and f (g (x))

Question: If $g(f(x))=|\sin x|$ and $f(g(x))=(\sin \sqrt{x})^{2}$, then (a) $f(x)=\sin ^{2} x, g(x)=\sqrt{x}$ (b) $f(x)=\sin x, g(x)=|x|$ (c) $f(x)=x^{2}, g(x)=\sin \sqrt{x}$ (d) $f$ and $g$ cannot be determined. Solution: If we solve it by the trial-and-error method, we can see that (a) satisfies the given condition.From (a): $f(x)=\sin ^{2} x$ and $g(x)=\sqrt{x}$ $\Rightarrow f(g(x))=f(\sqrt{x})=\sin ^{2} \sqrt{x}=(\sin \sqrt{x})^{2}$ So, the answer is (a)....

Find the number of

Question: Find the number of (i) diagonals (ii) triangles formed in a decagon. Solution: A decagon has 10 sides. (i) Number of diagonals $=\frac{n(n-3)}{2}=\frac{10(10-3)}{2}=35$ (ii) Number of triangles (i.e. 3 sides are to be selected) $={ }^{10} C_{3}=\frac{10}{3} \times \frac{9}{2} \times \frac{8}{1}=120$...

The perpendicular distance of the point A(3, 4) from the y-axis is

Question: The perpendicular distance of the pointA(3, 4) from they-axis is(a) 3(b) 4(c) 5(d) 7 Solution: The perpendicular distance of a point from they-axis is equal to thex-coordinate of the point. Perpendicular distance of the point A(3, 4) from they-axis =x-coordinate of A(3, 4) = 3 Hence, the correct answer is option (a)....

If A(–2, 3) and B(–3, 5) are two given points then (abscissa of A) – (abscissa of B) = ?

Question: IfA(2, 3) andB(3, 5) are two given points then (abscissa ofA) (abscissa ofB) = ?(a) 2(b) 1(c) 1(d) 2 Solution: The given points areA(2, 3) andB(3, 5).Abscissa ofA =x-coordinate of A =2Abscissa ofB =x-coordinate of B =3Abscissa ofAAbscissa ofB =2 (3) =2 + 3 = 1Hence, the correct answer is option (b)....

If f : R→R is given by f(x)=3x−5,

Question: If $f: R \rightarrow R$ is given by $f(x)=3 x-5$, then $f^{-1}(x)$ (a) is given by $\frac{1}{3 x-5}$ (b) is given by $\frac{x+5}{3}$ (c) does not exist because $f$ is not one-one (d) does not exist because $f$ is not onto Solution: Clearly, $f$ is a bijection. So, $f^{-1}$ exists. Let $f^{-1}(x)=y$ ...(1) $\Rightarrow f(y)=x$ $\Rightarrow 3 y-5=x$ $\Rightarrow 3 y=x+5$ $\Rightarrow y=\frac{x+5}{3}$ $\Rightarrow f^{-1}(x)=\frac{x+5}{3} \quad[$ from $(1)]$ So, the answer is (b)....

A committee of 3 persons is to be constituted from a group of 2 men and 3 women.

Question: A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women? Solution: A committee of 3 people is to be constituted from a group of 2 men and 3 women. $\therefore$ Number of ways $={ }^{2} C_{0} \times{ }^{3} C_{3}+{ }^{2} C_{1} \times{ }^{3} C_{2}+{ }^{2} C_{2} \times{ }^{3} C_{1}=1+2 \times 3+3 \times 1=10$ Number of committees consisting of 1 man and 2 women $={ }^...

If O(0, 0), A(3, 0), B(3, 4), C(0, 4) are four given points then the figure OABC is a

Question: IfO(0, 0),A(3, 0),B(3, 4),C(0, 4) are four given points then the figureOABCis a(a) square(b) rectangle(c) trapezium(d) rhombus Solution: The point O(0, 0) is the origin.A(3, 0) lies on the positive direction ofx-axis.B(3, 4) lies in the Ist quadrant.C(0, 4) lies on the positive direction ofy-axis.The points O(0, 0), A(3, 0), B(3, 4) and C(0, 4) can be plotted on the Cartesian plane as follows: Here, the figure OABC is a rectangle.Hence, the correct answer is option (b)....

Find:

Question: Find: (i) $10^{\text {th }}$ term of the A.P. $1,4,7,10, \ldots$ (ii) $18^{\text {th }}$ term of the A.P. $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2} \ldots$ (iii) $n^{\text {th }}$ term of the A.P. $13,8,3,-2, \ldots$ (iv) $10^{\text {th }}$ term of the A.P. $-40,-15,10,35, \ldots$ (v) 8th term of the A.P. 117, 104, 91, 78,... (vi) 11 th term of the A.P. $10.0,10.5,11.0,11.5, \ldots$ (vii) 9 th term of the A.P. $\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \ldots$ Solution: In this probl...

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has

Question: A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl? (ii) at least one boy and one girl? (iii) at least 3 girls? Solution: A group consists of 4 girls and 7 boys. Out of them, 5 are to be selected to form a team. (i) If the team has no girls, then the number of ways of selecting 5 members $={ }^{7} C_{5}=\frac{7 !}{5 ! 2 !}=\frac{7 \times 6}{2}=21$ (ii) If the team has at least 1 boy and 1 girl, then the number of way...

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has

Question: A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl? (ii) at least one boy and one girl? (iii) at least 3 girls? Solution: A group consists of 4 girls and 7 boys. Out of them, 5 are to be selected to form a team. (i) If the team has no girls, then the number of ways of selecting 5 members $={ }^{7} C_{5}=\frac{7 !}{5 ! 2 !}=\frac{7 \times 6}{2}=21$ (ii) If the team has at least 1 boy and 1 girl, then the number of way...

If the y-coordinate of a point is zero then this point always lies

Question: If they-coordinate of a point is zero then this point always lies(a) on they-axis(b) on thex-axis(c) in the I quadrant(d) in the IV quadrant Solution: The coordinates of a point on thex-axis are of the form (x, 0) and that of the point on they-axis is of the form (0,y).Thus, if they-coordinate of a point is zero, then this point always lies on thex-axis.Hence, the correct answer is option (b)....

In a village, there are 87 families of which 52 families have at most 2 children.

Question: In a village, there are 87 families of which 52 families have at most 2 children. In a rural development programme, 20 families are to be helped chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made? Solution: 52 families have at most 2 children, while 35 families have more than 2 children. The selection of 20 families of which at least 18 families must have 2 children can be made in the ways given below. (i) 18 famil...

The ordinate of every point on the x-axis is

Question: The ordinate of every point on thex-axis is(a) 1(b) 1(c) 0(d) any real number Solution: The ordinate of every point on thex-axis is 0.Hence, the correct option is (c)....

In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected?

Question: In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected? Solution: 5 persons are to be selected out of 6 men and 4 women. At least, one woman has to be selected in all cases. Required number of ways $={ }^{4} C_{1} \times{ }^{6} C_{4}+{ }^{4} C_{2} \times{ }^{6} C_{3}+{ }^{4} C_{3} \times{ }^{6} C_{2}+{ }^{4} C_{4} \times{ }^{6} C_{1}$ $=60+120+60+6$ $=246$...

In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected?

Question: In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected? Solution: 5 persons are to be selected out of 6 men and 4 women. At least, one woman has to be selected in all cases. Required number of ways $={ }^{4} C_{1} \times{ }^{6} C_{4}+{ }^{4} C_{2} \times{ }^{6} C_{3}+{ }^{4} C_{3} \times{ }^{6} C_{2}+{ }^{4} C_{4} \times{ }^{6} C_{1}$ $=60+120+60+6$ $=246$...

The point which lies on the y-axis at a distance of 5 units in the negative direction of the y-axis is

Question: The point which lies on they-axis at a distance of 5 units in the negative direction of they-axis is(a) (5, 0)(b) (0, 5)(c) (5, 0)(d) (0, 5) Solution: The point which lies on they-axis at a distance of 5 units in the negative direction of they-axis is (0, 5). Hence, the correct option is (b)....

How many triangles can be obtained by joining 12 points,

Question: How many triangles can be obtained by joining 12 points, five of which are collinear? Solution: Out of 12 points, 5 points are collinear and 3 points are required to form a triangle. Required ways $={ }^{12} C_{3}-{ }^{5} C_{3}$ $=\frac{12}{3} \times \frac{11}{2} \times \frac{10}{1}-\frac{5}{3} \times \frac{4}{2} \times \frac{3}{1}$ $=220-10$ $=210$...

How many triangles can be obtained by joining 12 points,

Question: How many triangles can be obtained by joining 12 points, five of which are collinear? Solution: Out of 12 points, 5 points are collinear and 3 points are required to form a triangle. Required ways $={ }^{12} C_{3}-{ }^{5} C_{3}$ $=\frac{12}{3} \times \frac{11}{2} \times \frac{10}{1}-\frac{5}{3} \times \frac{4}{2} \times \frac{3}{1}$ $=220-10$ $=210$...

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