Maths

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man’s eye is 30° (Ignore man’s height). After sailing for 20 seconds, towards the base of the tower (which is the level of water), the boat has reached a point B, where the angel of depression is 45°. Then the time taken (in second) by the boat from B to reach the base of the tower is:

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3 months ago

All possible values of $\in [0, 2 π]$ for which sin 2q + tan 2q > 0 lie in:

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sin 2qθ+ tan 2θ> 0

$\frac{2 t a n θ}{1 + t a n^{2} θ} + \frac{2 t a n θ}{1 - t a n^{2} θ} > 0$

Let tan q = x

$\frac{2 x}{1 + x^{2}} + \frac{2 x}{1 - x^{2}} > 0$

$t a n θ < - 1 o r 0 < t a n θ < 1$

$θ \in (0, \frac{π}{4}) \cup (\frac{π}{2}, \frac{3 π}{4}) \cup (π, \frac{5 π}{4}) \cup (\frac{3 π}{2}, \frac{7 π}{4})$

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Let the lines (2 – i) z = (2 + i) $\bar{z} a n d (2 + i) z + (i - 2) \bar{z} - 4 i = 0,$ (here i² = -1) be normal to a circle C. If the line iz + $\bar{z} + 1 + i = 0$ is tangent to this circle C, then its radius is:

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(2 – i) z = (2 + i) $\bar{z}$ , put z = x + iy

$y = \frac{x}{2} . . . . . . . . (i)$

(ii) $(2 + i) z + (i - 2) \bar{z} - 4 i = 0$

x + 2y = 2

(iii) $i z + \bar{z} + 1 + i = 0$

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

$x = 1 . y = \frac{1}{2} \Rightarrow c e n t r e (1, \frac{1}{2})$

Perpendicular distance of point $(1, \frac{1}{2})$ from x – y + 1 = 0 is p = r $\Rightarrow | \frac{1 - \frac{1}{2} + 1}{\sqrt[]{2}} | = r$

$r = \frac{3}{2 \sqrt[]{2}}$

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If $0 < θ, ? < \frac{π}{2}, x = \sum_{n = 0}^{\infty} c o s^{2 n} θ, y = \sum_{}^{} s i n^{2 n} ? a n d z = \sum_{n = 0}^{\infty} c o s^{2 n} θ . s i n^{2 n} ? t h e n :$

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$x = \sum_{n = 0}^{\infty} c o s^{2 n} θ = c o s^{0} θ + c o s^{2} θ + c o s^{4} θ + . . . . . = 1 + c o s^{2} θ + c o s^{4} θ + . . . . .$

a = 1, r = cos² θ

$x = S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - c o s^{2} θ} = \frac{1}{s i n^{2} θ}$

Similarly, y = $\frac{1}{c o s^{2} θ} \Rightarrow \frac{1}{y} = c o s^{2} θ$

$z = \frac{x y}{x y - 1} \Rightarrow x y z - z = x y . . . . . . . . . . . (i)$

Also, $\frac{1}{x} + \frac{1}{y} = 1 \Rightarrow x + y = x y . . . . . . . . . . (i i)$

(i) & (ii) ->xyz = xy + z -> (x + y) z = xy + z

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When a missile is fired from a ship, the probability that it is intercepted is $\frac{1}{3}$ and the probability that the missile hits the target, given that it is not intercepted, is $\frac{3}{4} .$ If three missiles are fired independently from the ship, then probability that all three hit the target, is:

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Let

A : Missile hit the target

B : Missile intercepted

P (B) = $\frac{1}{3} P (A / \bar{B}) = \frac{3}{4}$

$P (\bar{B}) = \frac{2}{3}$

$\Rightarrow P (\bar{B} \cap A) = \frac{3}{4} * \frac{2}{3} = \frac{1}{2}$

Required Probability = $\frac{2}{3} * \frac{3}{4} * \frac{2}{3} * \frac{3}{4} * \frac{2}{3} * \frac{3}{4} = \frac{1}{8}$

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$\underset{n \to \infty}{l i m} (1 + \frac{1 + \frac{1}{2} + . . . . . . . . . . + \frac{1}{n}}{n^{2}}) i s e q u a l t o :$

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Contributor-Level 10

$\underset{n \to \infty}{l i m} {(1 + \frac{1 + \frac{1}{2} + . . . . . . . + \frac{1}{n}}{n^{2}})}^{n}$ limit is in the form of $1^{\infty}$

$l = e x p (\underset{n \to \infty}{l i m} \frac{1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n}}{n^{2}})$

$0 \leq 1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n} \leq 1 + \frac{1}{\sqrt[]{2}} + \frac{1}{\sqrt[]{3}} + . . . . + \frac{1}{\sqrt[]{n}} \leq 2 \sqrt[]{n} - 1$

Taking limit $(n \to \infty)$

$l$ = exp (0) (from sandwich)

$l = 1$

Second Method :

1 + \frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{n} \geq l n (n + 1) . . . . . . . (i)

1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n} \leq 1 + \int_{1}^{2} \frac{1}{2} d x \leq 1 + l n n . . . . . . . . . . (i i)

From (i) & (ii)

$l n (n + 1) \leq 1 + \frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{n} \leq 1 + I n n, \forall n \in N, n \geq 2$

As $\underset{n \to \infty}{l i m} \frac{l n (n + 1)}{n} = 0$

and $\underset{n \to \infty}{l i m} \frac{1 + l n (n)}{n} = 0$

$∴$ from sandwich theorem

$\underset{n \to \infty}{l i m} \frac{1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n}}{n} = 0$

$∴ e^{0} = 1$

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3 months ago

A tangent is drawn to the parabola y² = 6x which is perpendicular to the 2x + y = 1. Which of the following points does NOT lie on it?

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Contributor-Level 10

2x + y = 1

$m_{1} = \frac{- 2}{1} = - 2 a n d m_{1} m_{2} = - 1$

$- 2 . m_{2} = - 1$

$m_{2} = \frac{1}{2}$

y² = 6x

y² = 4 (3/2) x

$y = m x + \frac{a}{m}$

$y = \frac{1}{2} x + \frac{\frac{3}{2}}{\frac{1}{2}}$

$y = \frac{x}{2} + 3$

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3 months ago

The statement A -> (B -> A) is equivalent to:

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Contributor-Level 10

$A \to (B \to A)$

$\sim A \lor (B \to A)$

$\sim A \lor (\sim B \lor A)$

$(\sim A \lor B) \lor (\sim A \lor A) = t A \to (A \cup B)$

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3 months ago

A plane passes through the points A(1,2,3) B(2,3,1) and C(2,4,2). If O is the origin and P is (2, -1,1), then the projection of on this plane is of length

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Payal Gupta

Contributor-Level 10

$\vec{O} P = (2, - 1, 1)$

Normal vector to the plane

$= \vec{A} B * A \vec{C} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & - 2 \\ 1 & 2 & - 1 \end{matrix} | = (3, - 1, 1) = \vec{n}$

Projection of $\vec{O} P o n \vec{n} i s | \frac{\vec{O} P . \vec{n}}{| \vec{n} |} |$

PN = $\frac{6 + 1 + 1}{\sqrt[]{11}} = \frac{8}{11}$

Projection of OP on plane = ON = $\sqrt[]{O P^{2} - P N^{2}} = \sqrt[]{6 - \frac{64}{11}} = \sqrt[]{\frac{2}{11}}$

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3 months ago

If the curves, $\frac{x^{2}}{a} + \frac{y^{2}}{b} = 1 a n d \frac{x^{2}}{c} + \frac{y^{2}}{d} = 1$ intersect each other at an angle of 90°, then which of the following relations is TRUE?

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