For positive real numbers a, b, c, the least value of a^(logb – logc) + b^(logc – loga) + c^(loga – logb) is -Maths 9th

For positive real numbers a, b, c, the least value of a^(logb – logc) + b^(logc – loga) + c^(loga – logb) is -Maths 9th
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Last Answer : (a) x + y + z + 2 = xyz.x = logabc ⇒ ax = bc ⇒ ax + 1 = abc ⇒ a = (abc)1/(x +1) Similarly, b = (abc)1/(y + 1), c = (abc)1/(z + 1)∴ abc = \((abc)^{rac{1}{x+1}+rac{1}{y+1}+rac{1}{z+1}}\)⇒ \({rac{1}{x+1}+ ... + xz + x + z + 1 + xy + y + x + 1 = xyz + xy + yz + zx + x + y + z + 1⇒ x + y + z + 2 = xyz.

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Last Answer : (b) \(rac{1}{{1-log_az}}\) y = \(rac{1}{a^{1-log_ax}}\) = \(a^{-(1-log_ax)}\)⇒ logay = \(rac{1}{{1-log_ax}}\) and loga z = \(rac{1}{{1-log_ay}}\)∴ logaz = \(rac{1}{1-\bigg(rac{1}{1-log_ax}\bigg)}\) = \( ... x = \(rac{1}{{1-log_az}}\)⇒ x = \(a^{rac{1}{1-log_az}}\) = ak ⇒ k = \(rac{1}{{1-log_az}}\).

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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).

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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.

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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

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

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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).

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Last Answer : As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

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