Solution : Let y = x + \(x^2\). Then, \((1 + x + x^2)^3\) = \((1 + y)^3\) = \(^3C_0\) + \(^3C_1 y\) + \(^3C_2 y^2\) + \(^3C_3 y^3\) = \(1 + 3y + 3y^2 + y^3\) = 1 + 3\((x + x^2)\) + 3\((x + x^2)^2\) + \((x + x^2)^3\) = \(x^6 + 3x^5 …
By using binomial theorem, expand \((1 + x + x^2)^3\). Read More »
Solution : Let a be the first term and r the common ratio of the given G.P. Then, \(a_4\) = 54 and \(a_9\) = 13122 \(\implies\) \(ar^3\) = 54 and \(ar^8\) = 13122 \(\implies\) \(ar^8\over ar^3\) = \(13122\over 54\) \(\implies\) \(r^5\) = 245 \(\implies\) r = 3 Putting r = 3 in \(ar^3\) = 54, …
If the 4th and 9th terms of a G.P. be 54 and 13122 respectively, find the G.P. Read More »
Solution : By using method of differences, The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. Clearly,these differences are in AP. Let \(T_n\) be the nth term and \(S_n\) denote the sum to n terms of the given series Then, \(S_n\) …
Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + ……. Read More »
Solution : By using method of differences, The difference between the successive terms are 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, …. Clearly,these differences are in AP. Let \(T_n\) be the nth term and \(S_n\) denote the sum to n terms of the given series Then, \(S_n\) …
Find the sum to n terms of the series : 3 + 15 + 35 + 63 + ….. Read More »
Solution : \(\int\) \(x^2 + x – 1\over x^2 – 1\) dx = \(\int\) (\(x^2 – 1\over x^2 – 1\) + \(x\over x^2 – 1\))dx = \(\int\) 1 dx + \(\int\) \(x\over x^2 – 1\)) dx Let \(x^2 – 1\) = t \(\implies\) 2x dx = dt = x + \(\int\) \(dt\over 2t\) = x …
Integrate \(x^2 + x – 1\over x^2 – 1\) with respect to x. Read More »
Solution : Let r be the radius of a sphere and \(\delta\)r be the error in measuring the radius. Then, r = 9 cm and \(\delta\)r = 0.03 cm. Let V be the volume of the sphere. Then, V = \({4\over 3}\pi r^3\) \(\implies\) \(dV\over dr\) = \(4\pi r^2\) \(\implies\) \(({dV\over dr})_{r = 9}\) = …
Solution : Let y = f(x), x = 3 and x + \(\delta x\). Then, \(\delta x\).= 0.02. For x = 3, we get y = f(3) = 45 Now, y = f(x) \(\implies\) y = \(3x^2 + 5x + 3\) \(\implies\) \(dy\over dx\) = 6x + 5 \(\implies\) \(({dy\over dx})_{x = 3}\) = 23 …
Find the approximate value of f(3.02), where f(x) = \(3x^2 + 5x + 3\). Read More »
Solution : Since a polynomial function is everywhere differentiable and so continuous also. Therefore, f(x) is continuous on [2, 3] and differentiable on (2, 3). Also, f(2) = \(2^2\) – 5 \(\times\) 2 + 6 = 0 and f(3) = \(3^2\) – 5 \(\times\) 3 + 6 = 0 \(\therefore\) f(2) = f(3) Thus, all …
Verify Rolle’s theorem for the function f(x) = \(x^2\) – 5x + 6 on the interval [2, 3]. Read More »
Solution : We are given that f(1) = f(3) = 0. \(\therefore\) \(1^3 – 6 \times 1 + a + b\) = \(3^3 – 6 \times 3^2 + 3a + b\) = 0 \(\implies\) a + b = 5 and 3a + b = 27 Solving these two equations for a and b, f'(c) is …
Solution : Let f(x) = cos x – 1, Clearly f(x) is continous on \([{\pi\over 2}, {3\pi\over 2}]\) and differentiable on \(({\pi\over 2}, {3\pi\over 2})\). Also, f\((\pi\over 2)\) = \(cos {\pi\over 2}\) – 1 = -1 = f\((3\pi\over 2)\). Thus, all the conditions of rolle’s theorem are satisfied. Consequently,there exist at least one point c …
