MHT CET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If $\alpha$ and $\beta$ are roots of the equation x² +5|x|-6=0, then the value of $|tan^{-1}\alpha-\tan^{-1}\beta|=$ is 

Answer:

Explanation:

Given $\alpha$ and $\beta$  be the roots of the equation

 x² +5|x|-6=0,

 Now   |x|² +5|x|-6=0

 |x|²+6|x|-|x|-6=0

                    [by factorisation]

|x|(|x|+6)-1(|x|+6)=0

(|x|+6)(|x|-1)=0

|x|=-6 or |x|=1

(since, modulus cannot be giving negative values)

$\therefore$   $|x|=1\Rightarrow x=\pm1$

So,   $\alpha$ =1 and $\beta$  =-1 $\therefore$    Now, |$\tan ^{-1} \alpha - \tan ^{-1} \beta$|=|$\tan ^{-1} 1 - \tan ^{-1} -1|$

= $|\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)|=|\frac{\pi}{4}+\frac{\pi}{4}|=|\frac{\pi}{2}|$

If vector r with dc's l,m, n is equally inclined to the coordinate axes, then the total number of such vectors is 

Answer:

Explanation:

 Given vector  r with  direction cosines l,m,n is equally inclined to the coordinate axes,

$\therefore$     l=m=n ........(i)

$\because$      l²+m²+n²=1

         l²+l²+l²=1  [from E.q,(i)]

 $\Rightarrow$      3l² =l$\rightarrow$ l²=$\frac{1}{3}$

 $\Rightarrow$    $l=\pm\frac{1}{\sqrt{3}}$

$\therefore$     l=m=n= $\pm\frac{1}{\sqrt{3}}$

 Now vector =   $r=|r|\left(\pm \frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}\pm\frac{1}{\sqrt{3}}\hat{k}\right)$

 Since, each has 2 choices i.e, l=m=n= $\frac{1}{\sqrt{3}}$

$\therefore$   Total number of such vectors =2³=8

Let $\Box $ PQRS  be a quadrilateral , if M and N are the mid-points of the sides PQ and RS respectively , then PS+QR=

Answer:

Explanation:

Let p,q, r, s , m and n be the position vectors of P.Q,R,S,M and N be respectively.

Given ,M and N are the mid-points of PQ and RS respectively of quadrilateral PQRS

 $m=\frac{p+q}{2}$  and $n=\frac{p+s}{2}$   .......(i)

Now, PS QR = (PV of S- PV of P)+(PV of R-PV of Q)

 = s+p+r-q

 =(r+s)-(p+q)

=2n-2m     [from Eq.(i)]

=2(n-m)=2MN

If the origin and the points P(2,3,4) , Q(1,2,3) and R(x,y,z)  are coplanar, then

Answer:

Explanation:

Let O (0,0,0) be the origin ,

It is given that O(0,0,0) , P(2,3,4), Q(1,2,3) and R(x,y,z) are coplannar

$\therefore$  [OR OP OQ]

$\begin{bmatrix}x-0 & y-0&z-0 \\2-0 & 3-0 &4-0\\1-0 &2-0 &3-0 \end{bmatrix}=0$

   $\begin{bmatrix}x & y&z \\2 & 3 &4\\1 &2 &3 \end{bmatrix}=0$

  x(9-8)-y(6-4)+z(4-3)=0

$\Rightarrow$     x-2y+z=0

Which of the following statement pattern is a tautology?

Answer:

Explanation:

The truth table is given as 

$\therefore$ $(q\rightarrow p)\vee (\sim p\leftrightarrow q)$ statement pattern is a tautology

• Mathematics - General questions