Solution : Comparing with \(ax^2+2hxy+by^2+2gx+2fy+c\) = 0 Here a = \(\lambda\), b = 12, c = -3, f = -8, g = 5/2, h = -5 Using condition \(abc+2fgh-af^2-bg^2-ch^2\) = 0, we have \(\lambda\)(12)(-3) + 2(-8)(5/2)(-5) – \(\lambda\)(64) – 12(25/4) + 3(25) = 0 \(\implies\) -36\(\lambda\) + 200 – 64\(\lambda\) – 75 + 75 = […]
Solution : A straight line is parallel to y-axis, if its y-coefficient is zero i.e. 4 – k = 0 i.e. k = 4 Similar Questions The slope of tangent parallel to the chord joining the points (2, -3) and (3, 4) is If the line 2x + y = k passes through the point
Solution : \(m_1\) = -\(1\over 4\) \(m_2\) = -\(4\over k\) Two lines are perpendicular if \(m_1 m_2\) = -1 \(\implies\) (-\(1\over 4\))\(\times\)(-\(4\over k\)) = -1 \(\implies\) k = -1 Similar Questions If the straight line 3x + 4y + 5 – k(x + y + 3) = 0 is parallel to y-axis, then the value
If x + 4y – 5 = 0 and 4x + ky + 7 = 0 are two perpendicular lines then k is Read More »
Solution : Let the equation of line be \(x\over a\) + \(y\over b\) = 1 …..(i) This line passes through (3,4), therefore \(3\over a\) + \(4\over b\) = 1 …….(ii) It is given that a + b = 14 \(\implies\) b = 14 – a in (ii), we get \(3\over a\) + \(4\over 14 –
Solution : As given \(\bar{x}\) = 4, n = 5 and \({\sigma}^2\) = 5.2. If the remaining observations are \(x_1\), \(x_2\) then \({\sigma}^2\) = \(\sum{(x_i – \bar{x})}^2\over n\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\over 5\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2\) = 9 …..(1) Also \(\bar{x}\) = 4 \(\implies\)
Solution : Total marks obtained from three subjects out of 300 = 75 + 80 + 85 = 240 if the marks of another subject is added then the total marks obtained out of 400 is greater than 240 if marks obtained in fourth subject is 0 then minimum average marks = \(240\over 400\)\(\times\)100 =
Solution : As given \(\bar{x}\) = \(x_1 + x_2 + …. + x_n\over n\) If the mean of the series \(x_i\) + 2i, i = 1, 2, ….., n be \(\bar{X}\), then \(\bar{X}\) = \((x_1+2) + (x_2+2.2) + (x_3+2.3) + …. + (x_n + 2.n)\over n\) = \(x_1 + x_2 + …. + x_n\over n\)
Solution : By using method of differences, The \(n^{th}\) term is (2n-1)(2n+1)(2n+3) \(T_n\) = (2n-1)(2n+1)(2n+3) \(T_n\) = \(1\over 8\)(2n-1)(2n+1)(2n+3){(2n+5) – (2n-3)} = \(1\over 8\)(\(V_n\) – \(V_{n-1}\)) \(S_n\) = \({\sum}_{r=1}^{n} T_n\) = \(1\over 8\)(\(V_n\) – \(V_0\)) \(\therefore\) \(S_n\) = \((2n-1)(2n+1)(2n+3)(2n+5)\over 8\) + \(15\over 8\) = \(n(2n^3 + 8n^2 + 7n – 2)\) Similar Questions Find the
Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + …… Read More »
Solution : \(\because\) \(T_n\) = \(S_n – S_{n-1}\) = \({\sum}_{r=1}^{n} T_r\) – \({\sum}_{r=1}^{n – 1} T_r\) = \(n(n+1)(n+2)(n+3)\over 8\) – \((n-1)(n)(n+1)(n+2)\over 8\) = \(n(n+1)(n+2)\over 8\)[(n+3) – (n-1)] = \(n(n+1)(n+2)\over 8\)(4) \(T_n\) = \(n(n+1)(n+2)\over 2\) \(\implies\) \(1\over T_n\) = \(2\over n(n+1)(n+2)\) = \((n+2)-n\over n(n+1)(n+2)\) = \(1\over n(n+1)\) – \(1\over (n+1)(n+2)\) Let \(V_n\) = \(1\over n(n+1)\) \(\therefore\)
Solution : Clearly P(A) = Power set of A = set of all subsets of A = {\(\phi\), {x}, {y}, {x,y}} Similar Questions Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets A\(\times\)B having 3 or more elements is If aN = {ax : x \(\in\)
If A = {x,y}, then the power set of A is Read More »
