Skip to main content

Posts

The ranges and heights for two projectiles projected ....

The ranges and heights for two projectiles projected with the same initial velocity at angles $42^\circ $ and $48^\circ $ with the horizontal are $R_1\,,R_2$ and $H_1\,,H_2$ respectively. Choose the correct option: (A) $R_1 < R_2 $ and $H_1 < H_2 $ (B) $R_1 > R_2 $ and $H_1 = H_2 $ (C) $R_1 = R_2 $ and $H_1 = H_2 $ (D) $R_1 = R_2 $ and $H_1 < H_2 $ Solution $R=\frac {u^2 sin 2\theta }{g} $, $H=\frac {u^2 sin^2 \theta }{2g}$  $\frac {R_1}{R_2} = \frac {sin 2 \times 42^\circ}{sin 2\times 48^\circ} = \frac {cos 6^\circ}{cos 6^\circ} = 1 $ $\frac {H_1}{H_2}= \frac {sin^2 42^\circ }{sin^2 48^\circ } = tan^2 42^\circ < 1 $ Answer: (D)

The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)=?$

The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is: (A) $2\sqrt 6$ (B) $\sqrt {26}$ (C) $2\sqrt 5 $ (D) $4\sqrt 2 $ Solution We have, $3x-z+4=0$ or $z=3x+4$ & $3y-2z-1=0$ or $3y-2(3x+4)-1=0$ or $y=2x+3$ Any point $(x,y,z)$ on the line $ \equiv (t,2t + 3,3t + 4)$ Let $d$ be the distance between $(2,-1,6)$ & $ (t,2t + 3,3t + 4)$ Then, $d^2=(t-2)^2+(2t+3+1)^2+(3t+4-6)^2=14t^2+24$ Minimum $d$ = Required answer = $\sqrt {24}=2\sqrt 6$ when t = 0. Answer: (A)

$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$

$cos^{-1}cos(-5)+$

$sin^{-1}sin6$-

$tan^{-1}tan12 =? $

$cos^{-1}(cos(-5))+sin^{-1}(sin(6))-tan^{-1}(tan(12)) $ is equal to:  (The inverse trigonometric functions take the principal values) (A) $3\pi - 11$ (B) $3\pi + 1 $ (C) $4\pi - 11$ (D) $4\pi - 9$ Solution The given expression can be rewritten as, $cos^{-1}cos5+sin^{-1}sin 6-tan^{-1} tan12 $ Considering the principal values the given expression can be further rewritten as, $cos^{-1}cos(2\pi - 5)+sin^{-1}sin (-(2\pi- 6))-tan^{-1} tan(-(4\pi -12)) $ Or $(2\pi - 5)+ (-(2\pi- 6))-(-(4\pi-12)) = 4\pi -11 $ Answer: (C)

A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1 $ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$

Planes $x-2y-2z+1=0$ & $2x-3y-6z+1=0$ ....

Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane P. Then which of the following points lies on P? (A) $(0,2,-4)$ (B) $(4,0,-2)$ (C) $(-2,0,-\frac {1}{2})$ (D) $(3,1,-\frac {1}{2})$ Solution Bisectors are given by, $\frac{{x - 2y - 2z + 1}}{3} =  \pm \frac{{2x - 3y - 6z + 1}}{7}$ $ \Rightarrow 7x - 14y - 14z + 7 =  \pm (6x - 9y - 18z + 3)$ Hence, $x-5y+4z+4=0$ & $13x-23y-32z+10=0$ Let $\theta $ be the angle between $x-2y-2z+1=0$ & $x-5y+4z+4=0$ $\cos \theta  = \frac{{|1 + 10 - 8|}}{{3 \times \sqrt {42} }} = \frac{1}{{\sqrt {42} }} < \frac{1}{{\sqrt 2 }}$ $\theta > 45^\circ $ So, $x-5y+4z+4=0$ is the obtuse angle bisector. $\therefore $ Acute angle bisector $P \equiv 13x - 23y - 32z + 10 = 0$ The point $(-2,0,-\frac {1}{2})$ satisfies P. Answer: (C)