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A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate color.

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(\(E_2\)) = \(4\over 8\) \(\times\) \(4\over 7\) …

A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate color. Read More »

From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. The probability that selection contains one of each category is

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 …

From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. The probability that selection contains one of each category is Read More »

Discuss the continuity and differentiability of the function y = f(x) defined parametrically; x = 2t – |t-1| and y = 2\(t^2\) + t|t|.

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)} = …

Discuss the continuity and differentiability of the function y = f(x) defined parametrically; x = 2t – |t-1| and y = 2\(t^2\) + t|t|. Read More »

If f(x + y) = f(x) + f(y) – 2xy – 1 for all x and y. If f'(0) exists and f'(0) = -sin\(\alpha\), then find f{f'(0)}.

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) …

If f(x + y) = f(x) + f(y) – 2xy – 1 for all x and y. If f'(0) exists and f'(0) = -sin\(\alpha\), then find f{f'(0)}. Read More »

Find the locus of middle point of the chord of the parabola \(y^2\) = 4ax which pass through a given (p, q).

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 …

Find the locus of middle point of the chord of the parabola \(y^2\) = 4ax which pass through a given (p, q). Read More »

Find the equation of the tangents to the parabola \(y^2\) = 9x which go through the point (4,10).

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 …

Find the equation of the tangents to the parabola \(y^2\) = 9x which go through the point (4,10). Read More »

Find the value of k for which the point (k-1, k) lies inside the parabola \(y^2\) = 4x.

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 »

The length of latus rectum of a parabola, whose focus is (2, 3) and directrix is the line x – 4y + 3 = 0 is

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 …

The length of latus rectum of a parabola, whose focus is (2, 3) and directrix is the line x – 4y + 3 = 0 is Read More »

If \(log_a x\) = p and \(log_b {x^2}\) = q then \(log_x \sqrt{ab}\) is equal to

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 »

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