MHT CET 2018 :: Mathematics :: General questions

• Mathematics - General questions

The equation of the line passing through the point (-3,1)  and bisecting the angle between  coordinate axes  is 

Answer:

Explanation:

 Since, the line bisects the angle between coordinate axes,

Now, equation of the line passing through the point (-3,1) and making angle $\theta$ = 135° with positive direction of X-axis , is

 y-1=m(x+3)

$\Rightarrow$   y -1= $\tan 135^{0} (x+3) $

                                              [ $\because m= \tan \theta$  and $\theta =135^{0}$]

$\Rightarrow$     y-1=-1(x+3)

$\Rightarrow$    y-1= -x-3

$\Rightarrow$   x+y+2=0

 Hence, option (a) is correct.

If the slope of one of the lines given by ax²+2hxy+by² =0 is  two times the other, then

Answer:

Explanation:

 We have . ax²+2hxy+by²=0

 Let slope of one line is m

$\therefore$ Slope  of another line is 2m,

 We know that,

$m_{1}+m_{2}=-\frac{2h}{b}$

 and    $m_{1}m_{2}=\frac{a}{b}$

$\therefore$    $m+2m=-\frac{2h}{b}$

 $\Rightarrow$    $3m=-\frac{2h}{b}$  ......(i)

  and   $m(2m)=\frac{a}{b}$

 $\Rightarrow$     $2m^{2}=\frac{a}{b}$  ......(ii)

 On eliminating  m, we get

$2\left(\frac{-2h}{3b}\right)^{2}=\frac{a}{b}$

 $\Rightarrow$       $8h^{2}=9ab$

If points P(4,5,x) , Q(3,y,4) and R (5,8,0) are collinear , then the value of x+y is 

Answer:

Explanation:

Given P(4,5,x), Q (3,y,4) and R(5,8,0) are collinear

 $\therefore$     $\overrightarrow{PQ}=\lambda \overrightarrow{QR}$

$\Rightarrow$   $ \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$

 = $(3\hat{i}+y\hat{j}+4\hat{k})-(4\hat{i}+5\hat{j}+x\hat{k})=-\hat{i}+(y-5)\hat{j}+(4-x)\hat{k}$

 and 

$\overrightarrow{QR}=\overrightarrow{OR}-\overrightarrow{OQ}=(5\hat{i}+8\hat{j})-(3\hat{i}+y\hat{j}+4\hat{k})= 2\hat{i}+(8-y)\hat{j}-4\hat{k}$

 $\therefore$    $-\hat{i}+(y-5)\hat{j}+(4-x)\hat{k}$

 = $\lambda[2\hat{i}+(8-y)\hat{j}-4\hat{k}]$

 On equating  the component of vector both sides, we get,

$\frac{-1}{2}=\lambda,\frac{y-5}{8-y}=\lambda,\frac{4-x}{-4}=\lambda$

 On putting the value of $\lambda$ , we get

$(y-5)=(8-y)(-\frac{1}{2})$

 and   $(4-x)=-4(-\frac{1}{2})$

 $\Rightarrow$   2y-10=y-8

    and 4-x=2

$\Rightarrow$   y=2

 and x=2

 $\therefore$   x+y=2+2=4

The maximum  value of 2x+y  subject to 3x+5y $\leq$ 26  and 5x+3y $\leq$ 30, x $\geq$ 0, y $\geq$ o , is 

Answer:

Explanation:

We have ,

 Maximize   Z= 2x+y

  Subject to constraints

$3x+5y\leq 26$

 $5x+3\leq30$

$x\geq 0, y\geq0$

 The graph of in equality  are

Maximum value of z=12

If f(x) = $\frac{e^{x^{2}}-\cos x}{x^{2}}$  , for x≠ o, is continuous  at x=0, then of f(0) is

Answer:

Explanation:

We have,

f(x) = $\frac{e^{x^{2}}-\cos x}{x^{2}}$   is continuous at x=0

 $\therefore$     $\lim_{x \rightarrow 0}f(x)=f(0)$

 $\Rightarrow$   $f(0)=\lim_{x \rightarrow0}\frac{e^{x^{2}}-\cos x}{x^{2}}$

 $\Rightarrow$    $f(0)=\lim_{x \rightarrow0}\frac{2xe^{x^{2}}+\sin x}{2x}$

                                    [ apply L' Hosptial rule]

 $\Rightarrow$    $f(0)=\lim_{x \rightarrow0}\frac{2xe^{x^{2}}}{2x}+\lim_{x \rightarrow 0}\frac{\sin x}{2x}=1+\frac{1}{2}=\frac{3}{2}$

• Mathematics - General questions