VITEEE 2015 :: Mathematics :: General questions

• Mathematics - General questions

 If a,b and c are three non-coplannar vectors , then (a+b-c).[(a-b) x(b-c) equals

Answer:

Explanation:

 [a+b-c}.[(a-b) x(b-c)

=(a+b-c).[a x b-a x c- bx b+ b x c]

= a.(a x b)-a.(a x c)+a.(b x c)+b.

(a x b)-b(a x c)+b. (b x c)- c.(a x b)+ c.( a x c)-c.(b x c)

= a.(b x c)-b .(a x c)-c. (a x b)

 =[a b c ] -[b a c]-[c a b]

=[a b c]+[a b c]-[a b c]

=[a b c]= a.(b x c)

 The area bounded by the curves y= cos x and y= sin x between the ordinates x=0 and   $x=\frac{3\pi}{2}$  is

Answer:

Explanation:

 Required area 

$\int_{0}^{\frac{\pi}{4}} (\cos x-\sin x)dx+\int_{\pi/4}^{5\pi/4} (\sin x-\cos x)dx$

                                                                    $\int_{5\pi/4}^{3\pi/2} (\cos x-\sin x)dx$

=$ \left[\sin x+\cos x\right]_{0}^{\pi/4}+[-cos x-sinx]^{5\pi/4}_{\pi/4}+[sin x+cos x]^{3\pi/4}_{5x/4}$

=  $(4\sqrt{2}-2)$ sq.units

 if  $a=\hat{i}-\hat{j}+2\hat{k}$   and  $b=2\hat{i}-\hat{j}+\hat{k}$ , then the angle $\theta$ between a and b s given by

Answer:

Explanation:

$\cos\theta=\frac{a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}}{\sqrt{a^{2}_{1}+a^{2}_{2}+a^{2}_{}}\sqrt{b^{2}_{1}+b^{2}_{2}+b^{2}_{3}}}$

$=\frac{1\times2+(-1)\times(-1)+2\times(1)}{\sqrt{1+1+4}\sqrt{4+1+1}}$

$\frac{2+2+2}{6}=\frac{6}{6}=1$

 So, $\theta$ = 0°  or $\theta$= 2$\pi$

 $\because sec 2\pi=1$

$\therefore$     $2\pi=\sec^{-1}(1)$

$\Rightarrow \theta  =\sec^{-1}(1)$

The normal at the point $(at_{1}^{2},2at^{}_{1})$   on the parabola meets the parabola again in the point  

$(at_{2}^{2},2at^{}_{2})$ , then 

Answer:

Explanation:

 Equation of the normal at point  $(at_{1}^{2},2at^{}_{1})$ on parabola is

 $y=-t_{1}x+2at_{1}+at^{3}_{1}$

 It also passes through $(at_{2}^{2},2at^{}_{2})$

 So,  $2at_{2}=-t_{1}(at^{2}_{2})+2at_{1}+at^{3}_{1}$

 $\Rightarrow 2t_{2}-2t_{1}=-t_{1}(t^{2}_{2}-t^{2}_{1})$

$\Rightarrow t_{1}+t_{2}=\frac{-2}{t_{1}}$

$\Rightarrow t_{2}=-t_{1}-\frac{2}{t_{1}}$

 If  $|z| \geq 3,$ then the least value of  $|z+\frac{1}{4}| $  is 

Answer:

Explanation:

$|z+\frac{1}{4}| $

= $|z-\left(\frac{-1}{4}\right)| \geq|z|-|\frac{-1}{4}|$

=  $|(-z)-\left(\frac{1}{4}\right)| \geq|3-\frac{1}{4}|=\frac{11}{4}$

$\therefore$      $|z+\frac{1}{4}| \geq\frac{11}{4}$

• Mathematics - General questions