MHT CET 2019 :: Mathematics :: General questions

• Mathematics - General questions

If P(6,10,10), Q(1,0,-5) , R (6,-10, $\lambda$)  are vertices of a triangle right angled at Q, then value of $\lambda$ is 

Answer:

Explanation:

In $\triangle$ PQR is right angled , at $\theta$

$\therefore$    PR²=PQ²+RQ²

 $\Rightarrow (6-6)^{2}+(-10-10)^{2}+(\lambda-10)^{2}+[(1-6)^{2}+(0-10)^{2}+(-5-10)^{2}]$

   = $[(1-6)^{2}+(0+10)^{2}+(-5-\lambda)^{2}]$

  $\Rightarrow $  $ 400+\lambda^{2}+100-20\lambda=(25+100+225)+(25+100+25+\lambda^{2}+10\lambda)$

 $\Rightarrow $  $ \lambda^{2}+20\lambda+500=350+150+10\lambda+\lambda^{2}$

$\Rightarrow $  $-20\lambda=10\lambda\Rightarrow30\lambda=0$

$\Rightarrow $   $  \lambda=0$

 If A is non-singular matrix and (A+l)(A-l)  =0 then A+A^-1= ......

Answer:

Explanation:

We have, A is non -singular matrix

 $\therefore$   |A|=0

 and (A+l)(A-l)=0

$\Rightarrow$ A²-l²  =0

$\Rightarrow$     A²= l² =l

 $\Rightarrow$     A.A=l $\Rightarrow$  A^-1 = A

 $\therefore$ A+A^-1 = A+A=2A

If A,B, C and D are (3,7,4) ,(5,-2,-3) ,(-4,5,6) and (1,2,3) respectively , then the volume of the parallelopiped  with AB, AC and AD as the co-terminus edges, is ........ cubic units

Answer:

Explanation:

 We have

  AB=  $(5-3)\hat{i}+(-2-7)\hat{j}+(3-4)\hat{k}$

  = $2\hat{i}-9\hat{j}-\hat{k}$

AC= $(-4-3)\hat{i}+(5-7)\hat{j}+(6-4)\hat{k}$

= $-7\hat{i}-2\hat{j}+2\hat{k}$

and    AD= $(1-3)\hat{i}+(2-7)\hat{j}+(6-4)\hat{k}$

=  $-2\hat{i}-5\hat{j}-\hat{k}$

$\therefore$   Volume of the parallelopiped with AB,AC and AD on the co-terminus edges

$[(AB$ $AC$ $AD)]=\begin{bmatrix}2 & -9 &-1 \\-7 & -2 &2\\ -2&-5&-1 \end{bmatrix}$

 =|2(2+10)+9(7+4)-1(35-4)|

=|2(12)+9(11)-1(31)|

=|(24+99-31)|

|92|=92 cubic unit

 If the function  $f(x)=\frac{(e^{kx}-1)\tan kx}{4x^{2}},x\neq0$

                                    =16                       x=0

  is continuous at x=0 , then k=...........

Answer:

Explanation:

 We have function,

$f(x)=\frac{(e^{kx}-1)\tan kx}{4x^{2}},x\neq0$

                                    =16                       x=0

is continuous at x=0

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

 $\Rightarrow$    $ \lim_{x \rightarrow 0} \frac{(e^{kx}-1)\tan kx}{4x^{2}}=16\left(\frac{0}{0}form\right)$

 $\Rightarrow$ $ \lim_{x \rightarrow 0} \frac{(e^{kx}-1)}{kx}\left(\frac{\tan kx}{kx}\right)\times\frac{k^{2}}{4}=16$

 $\Rightarrow$    $ \frac{k^{2}}{4}=16\Rightarrow k^{2}=64\Rightarrow k\pm 8$

The particular solution of the differential equation  $\log\left(\frac{dy}{dx}\right)=x$   , when x=0 , y=1 is .....

Answer:

Explanation:

 We have, differential equations,

  $\log(\frac{dy}{dx})=x\Rightarrow\frac{dy}{dx}=e^{x}$

$\Rightarrow$    $ dy=e^{x}dx$

 integrating  on both sides we get

              $\int  dy=\int e^{x}dx$

$\Rightarrow $    $ y= e^{x}+ C$  ........(i)

 On putting x=0, y=1  is Eq .(i) , we get

 1= e⁰+ C $\Rightarrow$ C=0

 Now, the particular solution of the given differential is y= e^x

• Mathematics - General questions