If a, b, c, d are four distinct positive real numbers and if 3s = a + b + c + d, then -Maths 9th

If a, b, c, d are four distinct positive real numbers and if 3s = a + b + c + d, then -Maths 9th
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If a1, a2, ....., an are distinct positive real numbers such that a1 + a2 + ..... + an = 1, then -Maths 9th

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Description : If x, y, z are three positive numbers, then the minimum value of -Maths 9th

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Description : On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Last Answer : (a) Reflexive and symmetric only(a, a) ∈ R ⇒ 1 + a . a = 1 + a2 > 0 V real numbers a ⇒ R is reflexive (a, b) ∈ R ⇒ 1 + ab > 0 ⇒ 1 + ba > 0 ⇒ (b, a) ∈ R ⇒ R is symmetricWe observe that \(\big(1,rac{1}{2}\big) ... }{2},-1\big)\) ∈ Rbut (1, - 1) ∉ R as 1 + 1 (-1) = 0 \( ot>\) 0 ⇒ R is not transitive.

If a, b, c, x, y, z are all positve real numbers, then -Maths 9th

Description : If a, b, c, x, y, z are all positve real numbers, then -Maths 9th

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If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Description : If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Last Answer : Given p,q,r are in A.P. then q=2p+r​.....(1). Now px2+qx+r=0 will have real root then q2−4pr≥0. or, 4(p+r)2​−4pr≥0 or, p2+r2−14pr≥0 or, r2−14rp+49p2≥48p2 or, (r−7p)2≥(43​p)2 or, (pr​−7)2≥(43​)2 [ Since p=0 for the given equation to be quadratic] or, ∣∣∣∣∣​pr​−7∣∣∣∣∣​≥43​.

Is ax + by + c = 0, where a, b and c are real numbers, a linear equation in two variables? Give reason. -Maths 9th

Description : Is ax + by + c = 0, where a, b and c are real numbers, a linear equation in two variables? Give reason. -Maths 9th

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Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; -Maths 9th

Description : Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; -Maths 9th

Last Answer : (c) S is an equivalence relation but R is not an equivalence relationR = {(x, y) | x, y ∈ R, x = wy, w is a rational number} Reflexive: x R x ⇒ x = wx ⇒ w = 1, (a rational number) Hence R is reflexive. Symmetric ... \(rac{r}{s}\) ⇒ \(rac{m}{n}\) S \(rac{r}{s}\) (True)∴ S is an equivalence relation.

The solution set for the inequality 2x – 10 < 3x – 15 over the set of real numbers is -Maths 9th

Description : The solution set for the inequality 2x – 10 < 3x – 15 over the set of real numbers is -Maths 9th

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The set of all real numbers x, for which x^2 – |x + 2| + x > 0, is -Maths 9th

Description : The set of all real numbers x, for which x^2 – |x + 2| + x > 0, is -Maths 9th

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The minimum value of the sum of real numbers a^(–5), a^(–4), 3a^(–3), 1, a^8 and a^10 with a > 0 is -Maths 9th

Description : The minimum value of the sum of real numbers a^(–5), a^(–4), 3a^(–3), 1, a^8 and a^10 with a > 0 is -Maths 9th

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Write the coordinates of a point on x-axis at a distance of 6 units from the origin in the positive direction of x-axis and then justify your answer. -Maths 9th

Description : Write the coordinates of a point on x-axis at a distance of 6 units from the origin in the positive direction of x-axis and then justify your answer. -Maths 9th

Last Answer : Solution :- As, any point on x-axis has coordinates (,)x0 where x is the distance from origin, so required coordinates are (6, 0).

If a^2 + b^2 + c^2 = 1, x^2 + y^2 + z^2 = 1, where a, b, c, x, y, z are positive reals then ax + by + cz is -Maths 9th

Description : If a^2 + b^2 + c^2 = 1, x^2 + y^2 + z^2 = 1, where a, b, c, x, y, z are positive reals then ax + by + cz is -Maths 9th

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If a, b, c, d are positive reals such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation -Maths 9th

Description : If a, b, c, d are positive reals such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation -Maths 9th

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There are 20 marbles in a box which are marked with distinct numbers from 1 to 20. If a marble is drawn, then find the probability that the marble bei

Description : There are 20 marbles in a box which are marked with distinct numbers from 1 to 20. If a marble is drawn, then find the probability that the marble bei

Last Answer : There are 20 marbles in a box which are marked with distinct numbers from 1 to 20. If a marble is drawn, then find the ... `2//5` C. `1//5` D. `4//5`

On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Description : On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Last Answer : (c) Symmetric only Let a ∈N. Then (a, a) ∉R as the GCD of a' and a' is a' not 2. R is not reflexive Let a, b ∈N. Then, (a, b) ∉R ⇒ GCD of a' and b' is 2 ⇒ GCD of b' and a' is 2 ⇒ (b, a) ∈R ∴ R ... , let a = 4, b = 10, c = 12 GCD of (4, 10) = 2 GCD of (10, 12) = 2 But GCD of (4, 12) = 4.

If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Description : If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Last Answer : (d) An equivalence relationWe can check the given properties as follows: Reflexive: Let (a, b) ∈ N x N. Then (a, b) ∈ N ⇒ a + b = b + a (Communtative law of Addition) ⇒ (a, b) R (b, a) ⇒ (a, b) R (a, ... , f) ⇒ (a, b) R (e, f) on N x N so R is transitive.Hence R is an equivalence relation on N N.

Show that If m > 1, then the sum of the mth powers of underline (n)even numbers is greater than n (n + 1)^m -Maths 9th

Description : Show that If m > 1, then the sum of the mth powers of underline (n)even numbers is greater than n (n + 1)^m -Maths 9th

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(A) Use two digits to make the smallest possible positve intiger.(B) Five 3s can express 37: 37=33+3+3/3Find another way to do it.(C) Use six identical dgits to make 100. (Several solutions are possible.)(D) Use five 4s to make 55.(E) Use four 9s to make 20. -Riddles

Description : (A) Use two digits to make the smallest possible positve intiger.(B) Five 3s can express 37: 37=33+3+3/3Find another way to do it.(C) Use six identical dgits to make 100. (Several solutions are possible.)(D) Use five 4s to make 55.(E) Use four 9s to make 20. -Riddles

Last Answer : (A) 1 X 1; 1/1;2/2;ect....;1-0;2-1;and many others. (B) 37=333/3X3; 37=3 X 3 X 3 + 3/.3 (C) 99 + 99/99; 55+55- 5- 5; (666-66)/6 (D) 44 + 44/4=55. (E) 9 + 99/9=20.

If r is real number such that |r| < 1 and if a = 5 (1 – r), then -Maths 9th

Description : If r is real number such that |r| < 1 and if a = 5 (1 – r), then -Maths 9th

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If x is real and x^2 + 3x + 2 > 0, x^2 – 3x – 4 < 0, then which of the following is correct? -Maths 9th

Description : If x is real and x^2 + 3x + 2 > 0, x^2 – 3x – 4 < 0, then which of the following is correct? -Maths 9th

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If the roots of the equation x^2 – 2ax + a^2 + a – 3 = 0 are real and less than 3, then which one of the following is correct ? -Maths 9th

Description : If the roots of the equation x^2 – 2ax + a^2 + a – 3 = 0 are real and less than 3, then which one of the following is correct ? -Maths 9th

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Let u = (log2x)^2 – 6(log2x) + 12, where x is a real number. Then the equation x^u = 256 has : -Maths 9th

Description : Let u = (log2x)^2 – 6(log2x) + 12, where x is a real number. Then the equation x^u = 256 has : -Maths 9th

Last Answer : ⇒ u=(log2 x)2−log2 x+12 ⇒ We can take log2 x=y ⇒ Then equation becomes y2−6y+12=u ⇒ Given that xu=256 Applying log we get, ⇒ ulog2 x=8 ∴ u=log2 x8 =y8 So our equation becomes, ⇒ y2−6y+12=y8 ⇒ ... We get y=2 ⇒ So, log2 x=2 ⇒ x=22 ∴ x=4 ∴ The given equation has exactly one solution for x

If `(2p+5q)/(2r+5s) = (4p-3q)/(4r-3s)`, then find the relation between p, q, r and s.

Description : If `(2p+5q)/(2r+5s) = (4p-3q)/(4r-3s)`, then find the relation between p, q, r and s.

Last Answer : If `(2p+5q)/(2r+5s) = (4p-3q)/(4r-3s)`, then find the relation between p, q, r and s.

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Description : If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Last Answer : We know that product of two rational number is always a rational number. Hence if a is a rational number then a2 = a x a is a rational number, a3 = 4:2 x a is a rational number. ∴ an = an-1 x a is a rational number.

Abscissa of a point is positive in -Maths 9th

Description : Abscissa of a point is positive in -Maths 9th

Last Answer : (b) Abscissa of a point is positive in I and IV quadrants.

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

The positive solutions of the equation ax + by + c = 0 always lie in the -Maths 9th

Description : The positive solutions of the equation ax + by + c = 0 always lie in the -Maths 9th

Last Answer : (a) We know that, if a line passes through the Ist quadrant, then all solution lying on the line in first quadrant must be positive because the coordinate of all points in the Ist quadrant are positive.

If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Description : If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Last Answer : We know that product of two rational number is always a rational number. Hence if a is a rational number then a2 = a x a is a rational number, a3 = 4:2 x a is a rational number. ∴ an = an-1 x a is a rational number.

Abscissa of a point is positive in -Maths 9th

Description : Abscissa of a point is positive in -Maths 9th

Last Answer : (b) Abscissa of a point is positive in I and IV quadrants.

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

The positive solutions of the equation ax + by + c = 0 always lie in the -Maths 9th

Description : The positive solutions of the equation ax + by + c = 0 always lie in the -Maths 9th

Last Answer : (a) We know that, if a line passes through the Ist quadrant, then all solution lying on the line in first quadrant must be positive because the coordinate of all points in the Ist quadrant are positive.

Show that x+a is a factor of x^n+a^n for any odd positive n -Maths 9th

Description : Show that x+a is a factor of x^n+a^n for any odd positive n -Maths 9th

Last Answer : Let f(x)=xn+an. In order to prove that x+a is a factor of f(x) for any odd positive integer n, it is sufficient to show that f(−a)=0. f(−a)=(−a)n+an=(−1)nan+an f(−a)=(−1+1)an [ n is odd positive integer ] f(−a)=0×an=0 Hence, x+a is a factor of xn+an, when n is an odd positive integer.

Show that x+a is a factor of x^n+a^n for any odd positive n -Maths 9th

Description : Show that x+a is a factor of x^n+a^n for any odd positive n -Maths 9th

Last Answer : This answer was deleted by our moderators...

In which quadrant does a point both of whose coordinates are positive lie? -Maths 9th

Description : In which quadrant does a point both of whose coordinates are positive lie? -Maths 9th

Last Answer : Solution :- I quadrant.