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JEE Mathematics Solved Question Paper 2013

Multiple Choice Questions

21.

For any two independent events E₁ and E₂ in a space S, $P [(E_{1} \cup E_{2}) \cap (E_{1} \cap E_{2})]$ is equal to

$\leq \frac{1}{4}$
$> \frac{1}{4}$
$\geq \frac{1}{2}$
$> \frac{1}{2}$

22.

The resultant of P and Q is R. If Q is doubled, R is also doubled and if Q is reversed, R is again doubled. Then, P² : Q² : R² gven by

2 2 3
3 2 2
2 3 2
2 3 1

23.
Forces ofmagnitudes 3, P, 5, 10 and Q are respectively acting along the sides AB, BC, CD, AD and the diagonal CA of a rectangle ABCD, where AB = 4m and BC = 3m. Ifthe resultant is a single force along the other diagonal BD, then P, Q and the resultant are

$4, 10 \frac{5}{12}, 12 \frac{11}{12}$

5, 6, 7

$3 \frac{1}{2}, 8, 9 \frac{1}{2}$

None of the above

$4, 10 \frac{5}{12}, 12 \frac{11}{12}$

ABCD is a rectangle in which AB = 4m and BC =3m

Then, $\tan (θ) = \frac{BC}{AB} = \frac{3}{4}$

The forces 3, P, 5, 10 and Q newtons have the resultant R Newton as shown in the figure.

$Rcos (θ) = Qcos (θ) + 5 - 3 = Qcos (θ) + 2 . . . (i) and Rsin (θ) = P + 10 - Qsin (θ) . . . (ii) and Q ABsin (θ) + 5 BC = 10 . AB [∵ \sin (θ) = \frac{3}{5} and \cos (θ) = \frac{4}{5}] \Rightarrow \frac{12}{5} Q = 40 - 15 = 25 \Rightarrow Q = \frac{125}{12} = 10 \frac{5}{12} newton Then, find R from Eq . (i) and P from Eq . (ii) \frac{4 R}{5} = \frac{125}{12} \times \frac{4}{5} + 2 \Rightarrow 4 R = \frac{125}{3} + 10 = \frac{155}{3} \Rightarrow R = \frac{155}{12} = 12 \frac{11}{12} and \frac{155}{12} \times \frac{3}{5} = P + 10 - \frac{125}{12} \times \frac{3}{5} \Rightarrow \frac{155}{4} = 5 P + 50 - \frac{125}{4} \Rightarrow 5 P = \frac{155}{4} + \frac{125}{4} - 50 \Rightarrow 5 P = \frac{280}{4} - 50 = 70 - 50 = 20 \Rightarrow P = 4 Hence, option (a) 4, 10 \frac{5}{12}, 12 \frac{11}{12} is correct .$