MHT CET 2016 :: Mathematics :: General questions

• Mathematics - General questions

 If p: Every square is a rectangle .

q:  Every rhombus is kite , then truth values of $p\rightarrow q$    and  $ p\leftrightarrow q$ are _ and _ respectively

Answer:

Explanation:

Here , statement p is true (T)  and statement q is false (F)

$\therefore$     $p\rightarrow q=T\rightarrow F= F and p\leftrightarrow q=T\leftrightarrow F=F$

If G(g) , H(h) and P(p)  are centroid orthocentre and circumcentre of a triangle and xp+yh+zg=0, then (x,y,z) is equal to 

Answer:

Explanation:

 We know that, orthocentre , centroid and circumcentre of a triangle are collinear and centroid  divides orthocentre and circumcentre in the ratio 2:1

 By using internally division. 

$\frac{2p+1h}{2+1}=g$

$\Rightarrow$                2p+h-3g=0

 But it is given xp+yh+zg=0

$\therefore$        x=2, y=1 and z=-3

If line joining parts A and B having position vectors  6a-4b+4c and -4c respectively and the line joining the points C and D  having position vectors -a-2b-3c and a+2b-5c intersect , then point of intersection is 

Answer:

Explanation:

Coordinate of points A and B  are (6,-4,4)  and (0,0,-4)  and coordinate of points C and D are (-1,-2,-3) and (1,2,-5)

Now , equation of line passing through (0,0,-4) and (6,-4,4) is

   $\frac{x-0}{6}=\frac{y-0}{-4}=\frac{z+4}{4+4}$=k   [say]

 $\Rightarrow$        x=6k, y=-4k

      and z=8k-4       ........(i)

 Again , equation of line passing through (-1,-2,-3) and (1,2,-5) is

  $\frac{x+1}{1+1}=\frac{y+2}{2+2}=\frac{z+3}{-5+3}$

 $\Rightarrow$       $\frac{x+1}{2}=\frac{y+2}{4}=\frac{3+3}{-2}$  ........(ii)

 Since, two lines are intersect , therefore point (6k,-4k,8k-4) satisfy Eq.(ii) , we get

   $\frac{6k+1}{2}=\frac{-4k+2}{4}= \frac{8k-4+3}{-2}$

 $\Rightarrow$     6k+1=-2k+1=-(8k-1)

$\therefore$    6k+1=-2k+1

$\Rightarrow$     8k=0

$\Rightarrow$   k=0

$\therefore$  x=6 x 0, y=-4 x0

 and z=8 x0-4

 $\Rightarrow$  x=0 , y=0 and z=-4

 Which is equal to the B coordinate

Principal solutions of the equation $\sin 2x+\cos 2x$=0 , where $\pi <x< 2\pi$ are 

Answer:

Explanation:

Given equation is $\sin 2x+\cos 2x$=0

$\Rightarrow$  sin 2x=-cos 2x

$\Rightarrow$ tan 2x=-1

   $[\because \pi <x<2 \pi\Rightarrow2 \pi <2x <4\pi]$

  $\Rightarrow$       2x=  $2\pi+\frac{3 \pi}{4},2 \pi +\left( \frac{3 \pi}{2}+\frac{\pi}{4}\right)$

 $\Rightarrow$       $2x= \frac{11 \pi}{8}, \frac{15 \pi}{4}$

 $\Rightarrow$     x= $\frac{11\pi}{8},\frac{15 \pi}{8}$

Probability of guessing correctly atleast 7 out of 10 answers  in a 'True' or 'False' test is equal to 

Answer:

Explanation:

In each  true and false question , probability of guessing  correctly  is p= $\frac{1}{2}$ and probability  of not guessing correctly , q= $\frac{1}{2}$

 Here, n=0

 $\therefore$   ,The probability of guessing atleast 7 correctly = P $(X\geq7)$

 =P(X=7)+P(X=8)+P(X=9)+P(X=10)

= $^{10}C_{7}\left( \frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{3}+^{10}C_{8}\left( \frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{2}+^{10}C_{9}\left( \frac{1}{2}\right)^{9}\left(\frac{1}{2}\right)^{1}+^{10}C_{10}\left( \frac{1}{2}\right)^{10}$

         $[\because P(x=r)=^{n}C_{r}p^{r}q^{n-r}]$

=   $120\times(\frac{1}{2})^{10}+45\left(\frac{1}{2}\right)^{10}+10\left( \frac{1}{2}\right)^{10}+1\left( \frac{1}{2}\right)^{10}$

 =$\frac{120+45+10+1}{2^{10}}=\frac{176}{1024}=\frac{11}{64}$

• Mathematics - General questions