Solution : 6N = {6, 12, 18, 24, 30, …..} 8N = {8, 16, 24, 32, ….} \(\therefore\) 6N \(\cap\) 8N = {24, 48, …..} = 24N Similar Questions If A = {x,y}, then the power set of A is If A = {2, 4} and B = {3, 4, 5} then (A \(\cap\) B) […]
If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to Read More »
Solution : A = [x: x \(\in\) R,-1 < x < 1] B = [x : x \(\in\) R, x – 1 \(\le\) -1 or x – 1 \(\ge\) 1] [x: x \(\in\) R, x \(\le\) 0 or x \(\ge\) 2] \(\therefore\) A \(\cup\) B = R – D where D = [x : x
Solution : \(x^2\) = 16 \(\implies\) x = \(\pm\)4 2x = 6 \(\implies\) x = 3 There is no value of x which satisfies both the above equations. Thus, A = \(\phi\) Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then the
The set A = [x : x \(\in\) R, \(x^2\) = 16 and 2x = 6] equal Read More »
Solution : R = {(2, 4), (4, 3), (6, 2), (8, 1)} \(R^{-1}\) = {(4, 2), (3, 4), (2, 6), (1, 8)} Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to Let A =
Solution : (A \(\cap\) B) = {4} and (A \(\cup\) B) = {2, 3, 4, 5} \(\therefore\) (A \(\cap\) B) \(\times\) (A \(\cup\) B) = {(4, 2), (4, 3), (4, 4), (4, 5)} Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then
If A = {2, 4} and B = {3, 4, 5} then (A \(\cap\) B) \(\times\) (A \(\cup\) B) Read More »
Solution : Given P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2 \(\therefore\) P(A) + P(B) + P(C) = 1 then events A, B, C are exhaustive. If P(E) = Probability of introducing a new product, then as given P(E|A) = 0.7, P(E|B) = 0.6 and P(E|C) = 0.5 = 0.5 \(\times\) 0.7 +
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)
