MHT CET 2016 :: Mathematics :: General questions

• Mathematics - General questions

Which of the following equation does not represnts a pair of lines ?

Answer:

Explanation:

(a)     $x^{2}-x=0$

$\Rightarrow$      x(x-1)=0

$\Rightarrow$   x=0 and x=1

 which represent a pair of lines

(b)  xy-x=0

 $\Rightarrow$    x(y-1)=0

$\Rightarrow$      x=0, y=1

 which represent a pair of lines

(c)     $y^{2}-x+1$=0

 $\Rightarrow$    $y^{2}=(x-1)$,

 which represent a parabola

Thus , it does not represent a pair of lines

(d)   xy+x+y+1=0

 $\Rightarrow$     x(y-1)+(y+1)=0

$\Rightarrow$     (y+1)(x+1)=0

$\Rightarrow$    y=-1 and x=-1

which repesent a pair of lines

If sin x is the integrating factor (IF) of the linear differential equation $\frac{dy}{dx}+Py$=Q, then P is 

Answer:

Explanation:

We know that IF of  $\frac{dy}{dx}+Py$=Q  is

 $IF=e^{\int pdx }$

$\sin x= e^{\int pdx }$

On differentiating both sides , we get

$\cos x=e^{\int pdx }P$

cos x= sin x P

 $\Rightarrow$     P= cot x

M and N  are the mid-points of the diagonals Ac and BD  respectively of quandrilateral ABCD , then AB+AD+CB+CD is equal to

Answer:

Explanation:

Let the position vectors of A, B, C, D, M  and N are a,b, c,d  , m and n

Since , M and N are the mid-points of AC  and BD

$\therefore$   m= $\frac{a+c}{2}$, n= $\frac{b+d}{2}$

Now, AB+AD+CB+CD

=(b-a)+(d-a)+(b-c)+(d-c)

=2(b+d)-2(a+c)

=2 x 2n-2 x 2 m

=4(n-m)=4MN

The particular  solution of the differtential equation   $y(1+\log x)\frac{dx}{dy}-x\log x=0$  , when  x=e, y=e² is 

Answer:

Explanation:

Given, differential equation is  

 $y(1+\log x)\frac{dx}{dy}-x\log x=0$

 $\Rightarrow \frac{(1+\log x)dx}{x\log x}=\frac{dy}{y}$

$\Rightarrow$    $ \left(\frac{1}{x \log x}+\frac{1}{x}\right)dx=\frac{1}{y}dy$

On integrating both sides , we get

$\Rightarrow \int \left(\frac{1}{x \log x}+\frac{1}{x}\right)dx=\int \frac{1}{y}dy$

Put log x=t

 $\Rightarrow$   $\frac{1}{x}dx=dt$

$\therefore$   $ \int \frac{1}{t}dt+\int\frac{1}{x} dx=\int\frac{1}{y} dy$

$\Rightarrow$    log t+log x=log y+log c

 $\Rightarrow$   log tx=log yc

$\Rightarrow$   tx=yc

$\Rightarrow$   x log x= yc

 when x= e and y = $e^{2}$

$\therefore$    $e \log e=e^{2} c$

$\Rightarrow$    e x 1=$ e^{2}c$

$\Rightarrow$     c $\Rightarrow$  $ \frac{1}{e}$

$\therefore$       x log x= $\frac{y}{e}$

$\Rightarrow$    y= ex log x

If $a=\hat{i}+\hat{j}-2\hat{k},b=2\hat{i}-\hat{j}+\hat{k}$   and    $c=3\hat{i}-\hat{k}$ and  c=ma+nb , then m+n is equal to

Answer:

Explanation:

Given , $a=\hat{i}+\hat{j}-2\hat{k},b=2\hat{i}-\hat{j}+\hat{k}$   and    $c=3\hat{i}-\hat{k}$ and  c=ma+nb 

$\therefore$     $3\hat{i}-\hat{k}=m(\hat{i}+\hat{j}-2\hat{k})+n(2\hat{i}-\hat{j}+\hat{k})$

 $\Rightarrow$     $3\hat{i}-\hat{k}=(m+2n)\hat{i}+(m-n)\hat{j}+(-2m+n)\hat{k}$

On equating the coefficient  of $\hat{i}$, $\hat{j}$ and $\hat{k}$ , respectively

we get

    3=m+2n,0=m-n

and -1=-2m+n

 $\Rightarrow$    3=n+2n

$\Rightarrow$    n=1

$\Rightarrow$    m=1  and n=1

$\Rightarrow$     m+n=1+1=2

• Mathematics - General questions