Solution : \(E_1\) : Event that first drawn ball is red, second is blue and so on. \(E_2\) : Event that first drawn ball is blue, second is red and so on. \(\therefore\) P(\(E_1\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) and \(\therefore\) P(\(E_2\)) = \(4\over 8\) \(\times\) \(4\over 7\) …
Solution : n(S) = \(^{10}C_4\) = 210 n(E)= \(^5C_2 \times ^3C_1 \times ^2C_1\) + \(^5C_1 \times ^3C_2 \times ^2C_1\) + \(^5C_1 \times ^3C_1 \times ^2C_2\) = 105 \(\therefore\) P(E) = \(105\over 210\) = \(1\over 2\) Similar Questions How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no …
Solution : Here x = 2t – |t-1| and y = 2\(t^2\) + t|t|. Now when t < 0; x = 2t – {-(t-1)} = 3t – 1 and y = 2\(t^2\) – \(t^2\) = \(t^2\) \(\implies\) = y = \({1\over 9}{(x+1)}^2\) = when 0 \(\le\) t < 1 x = 2t – {-(t-1)} = …
Solution : f'(x) = \(\displaystyle{\lim_{h \to 0}}\) \(f(x+h) – f(x)\over h\) = \(\displaystyle{\lim_{h \to 0}}\) \({f(x)+f(h)-2xh-1} – f(x)\over h\) = \(\displaystyle{\lim_{h \to 0}}\) -2x + \(\displaystyle{\lim_{h \to 0}}\) \(f(h)-1\over h\) = -2x + \(\displaystyle{\lim_{h \to 0}}\) \(f(h) – f(0)\over h\) [Putting x = 0 = y in the given relation we find f(0) = f(0) …
Solution : Let P(h,k) be the mid point of chord of the parabola \(y^2\) = 4ax, so equation of chord is yk – 2a(x+h) = \(k^2\) – 4ah. Since it passes through (p,q) \(\therefore\) qk – 2a(p+h) = \(k^2\) – 4ah \(\therefore\) Required locus is \(y^2\) – 2ax – qy + 2ap = 0 Similar …
Solution : Equation of tangent to the parabola \(y^2\) = 9x is y = mx + \(9\over 4m\) Since it passes through (4,10) \(\therefore\) 10 = 4m + \(9\over 4m\) \(\implies\) 16\(m^2\) – 40m + 9 = 0 m = \(1\over 4\), \(9\over 4\) \(\therefore\) Equation of tangent’s are y = \(x\over 4\) + 9 …
Solution : \(\because\) Point (k-1, k) lies inside the parabola \(y^2\) = 4x. \(\therefore\) \({y_1}^2 – 4ax_1\) < 0 \(\implies\) \(k^2\) – 4(k-1) < 0 \(\implies\) \(k^2\) – 4k + 4 < 0 \((k-2)^2\) < 0 \(\implies\) k \(\in\) \(\phi\) Similar Questions The slope of the line touching both the parabolas \(y^2\) = 4x and …
Find the value of k for which the point (k-1, k) lies inside the parabola \(y^2\) = 4x. Read More »
Solution : The length of latus rectum = 2 x perpendicular from focus to the directrix = 2 x |\({2-4(3)+3}\over {\sqrt{1+16}}\)| = \(14\over \sqrt{17}\) Similar Questions The slope of the line touching both the parabolas \(y^2\) = 4x and \(x^2\) = -32 is Find the locus of middle point of the chord of the parabola …
Solution : We have \(2^{x+2}\) > \(2^{-2/x}\) Since the base 2 > 1, we have x + 2 > \(-2\over x\) (the sign of inequality is retained) Now, x + 2 + \(-2\over x\) > 0 \(\implies\) \({x^2 + 2x + 2}\over x\) > 0 \(\implies\) \(({x+1})^2 + 1\over x\) > 0 \(\implies\) x \(\in\) …
Solve for x : \(2^{x+2}\) > \(({1\over 4})^{1/x}\) Read More »
Solution : \(log_a x\) = p \(\implies\) \(a^p\) = x \(\implies\) a = \(x^{1/p}\) Similarly \(b^q\) = \(x^2\) \(\implies\) b = \(x^{2/q}\) Now, \(log_x \sqrt{ab}\) = \(log_x \sqrt{x^{1/p}x^{2/q}}\) = \(log_x x^{({1\over p}+{2\over q}){1\over 2}}\) = \(1\over {2p}\) + \(1\over q\). Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Evaluate the …
If \(log_a x\) = p and \(log_b {x^2}\) = q then \(log_x \sqrt{ab}\) is equal to Read More »
