Mathemerize

Angle between asymptotes of hyperbola xy=8 is

Solution : Since given hyperbola xy = 8 is rectangular hyperbola. And eccentricity of rectangular hyperbola is \(\sqrt{2}\) Angle between asymptotes of hyperbola is \(2sec^{-1}(e)\) \(\implies\) \(\theta\) = \(2sec^{-1}(\sqrt{2})\) \(\implies\) \(\theta\) = \(2sec^{-1}(sec 45)\) \(\implies\) \(\theta\) = 2(45) = 90 Similar Questions Find the normal to the hyperbola \(x^2\over 16\) – \(y^2\over 9\) = 1 …

Angle between asymptotes of hyperbola xy=8 is Read More »

The sum of the slopes of the tangent of the parabola \(y^2\)=4ax drawn from the point (2,3) is

Solution : The equation of tangent to the parabola \(y^2\) = 4ax is y = mx + \(a\over m\). Since it is drawn from point (2,3) Therefore it lies on tangent y = mx + \(a\over m\). \(\implies\) 3 = 2m + \(a\over m\) \(\implies\) 3m = 2\(m^2\) + a \(\implies\) 2\(m^2\) – 3m + …

The sum of the slopes of the tangent of the parabola \(y^2\)=4ax drawn from the point (2,3) is Read More »

If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then

Leave a Comment / Maths Questions, Series Questions / By mathemerize

Solution : Since, x, y and z are in AP \(\therefore\) 2y = x + z Also, \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are in AP \(\therefore\) 2\(tan^{-1}y\) = \(tan^{-1}x\) + \(tan^{-1}z\) \(\implies\) \(tan^{-1}({2y\over {1 – y^2}})\) = \(tan^{-1}({x + z\over {1 – xz}})\) \(\implies\) \(x + z\over {1 – y^2}\) = \(x + z\over {1 – …

If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then Read More »

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is

Leave a Comment / Maths Questions, Series Questions / By mathemerize

Solution : 0.7 + 0.77 + 0.777 + …… + upto 20 terms = \(7\over 10\) + \(77\over 10^2\) + \(777\over 10^3\) + ….. + upto 20 terms = 7[ \(1\over 10\) + \(11\over 10^2\) + \(111\over 10^3\) + ….. + upto 20 terms ] = \(7\over 9\)[ \(9\over 10\) + \(99\over 100\) + \(999\over …

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is Read More »

Three positive integers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then the common ratio of GP is

Leave a Comment / Maths Questions, Series Questions / By mathemerize

Solution : Let a, ar, \(ar^2\) are in GP (r > 1) According to the question, a, 2ar, \(ar^2\) in AP. \(\implies\) 4ar = a + \(ar^2\) \(\implies\) \(r^2\) – 4r + 1 = 0 \(\implies\) r = \(2 \pm \sqrt{3}\) Hence, r = \(2 + \sqrt{3}\) [ \(\because\) AP is increasing] Similar Questions …

Three positive integers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then the common ratio of GP is Read More »

If \((10)^9\) + \(2(11)^1(10)^8\) + \(3(11)^2(10)^7\) + …… + \(10(11)^9\) = \(K(10)^9\), then k is equal to

Leave a Comment / Maths Questions, Series Questions / By mathemerize

Solution : \(K(10)^9\) = \((10)^9\) + \(2(11)^1(10)^8\) + \(3(11)^2(10)^7\) + …… + \(10(11)^9\) K = 1 + 2\(({11\over 10})\) + 3\(({11\over 10})^2\) + ….. + 10\(({11\over 10})^9\) ……(i) \(({11\over 10})\)K = 1\(({11\over 10})\) + 2\(({11\over 10})^2\) + 3\(({11\over 10})^3\) + ….. + 10\(({11\over 10})^{10}\) …..(ii) On subtracting equation (ii) from (i), …

If \((10)^9\) + \(2(11)^1(10)^8\) + \(3(11)^2(10)^7\) + …… + \(10(11)^9\) = \(K(10)^9\), then k is equal to Read More »

A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is 2/5, then p is equal to

Leave a Comment / Maths Questions, Probability Questions / By mathemerize

Solution : Let the probability of getting a head be p and not getting a head be q. Since, head appears first time in an even throw 2 or 4 or 6. \(\therefore\) \(2\over 5\) = qp + \(q^3\)p + \(q^5\)p + …… \(\implies\) \(2\over 5\) = \(qp\over {1 – q^2}\) Since q = 1- …

A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is 2/5, then p is equal to Read More »

The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series India’s second win occurs at the third test is

Leave a Comment / Maths Questions, Probability Questions / By mathemerize

Solution : Let \(A_1\), \(A_2\), \(A_3\) be the events of match winning in first, second and third matches respectively and whose probabilities are P(\(A_1\)) = P(\(A_2\)) = P(\(A_2\)) = \(1\over 2\) \(\therefore\) Required Probability = P(\(A_1\))P(\(A_2’\))P(\(A_3\)) + P(\(A_1’\))P(\(A_2\))P(\(A_3\)) = \(({1\over 2})^3\) + \(({1\over 2})^3\) = \(1\over 8\) + \(1\over 8\) = \(1\over 4\) Similar Questions …

The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series India’s second win occurs at the third test is Read More »

A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is

Leave a Comment / Maths Questions, Probability Questions / By mathemerize

Solution : Probability of getting success, p = \(1\over 6\) and probability of failure, q = \(5\over 6\) Now, we must get 2 sixes in seven throws, so probability is \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\) and the probability that 8th throw is \(1\over 6\). \(\therefore\) Required Probability = \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\)\(1\over 6\) = \(^7C_2\times 5^5\over {6^8}\) Similar …

A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is Read More »

If A and B are two mutually exclusive events, then

Leave a Comment / Maths Questions, Probability Questions / By mathemerize

Question : If A and B are two mutually exclusive events, then (a) P(A) < P(B’) (b) P(A) > P(B’) (c) P(A) < P(B) (d) None of these Solution : Since, A and B are two mutually exclusive events. \(\therefore\) A \(\cap\) B = \(\phi\) \(\implies\) Either A \(\subseteq\) B’ or B \(\subseteq\) A’ …

If A and B are two mutually exclusive events, then Read More »