MHT CET 2020 :: Mathematics :: General questions

• Mathematics - General questions

 The radius of the circle  passing through the points (5,7),(2,-2) and (-2,0) is

Answer:

Explanation:

 Given, points  (5,7) ,(2,-2) and (-2,0) lie on circle

 Let (h,k) be centre of circle 

$\therefore$    OA=OC

 $\therefore$  $(h+2)^{2}+k^{2}= (h-5)^{2}+(k-7)^{2}$

$\Rightarrow h^{2}+4h+4+k^{2}= h^{2}-10h+25+k^{2}-14k+49$

$\Rightarrow   14h+14k=70$

     h+k=5    .............(i)

 When OA=OB

 $\therefore$   $ (h+2)^{2}+k^{2}= (h-2)^{2}+(k+2)^{2}$

 $ \Rightarrow h^{2}+4h+4+k^{2}= h^{2}-4h+4+k^{2}+4k+4$

$ \Rightarrow 8h-4k=4$

 2h-k=1       .........(ii)

 From eqs. (i) and (ii)  we get

 h=2, k=3

 $\therefore$ Centre of circle (2,3)

 Radius of circle OA=  $\sqrt{(2+2)^{2}+3^{2}}$

  OA=  $\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5 units$

 The points of discontinuity of the function

$f(x)= \frac{1}{x-1}, if 0\leq x\leq2$

   $= \frac{x+5}{x+3}, if 2< x\leq4$

 in its domain are

Answer:

Explanation:

 We have

$f(x)=\begin{cases}\frac{1}{x-1} & if 0\leq x\leq2,\\\frac{x+5}{x+3}, & if 2<x \leq4\end{cases}$

 Clearly , f(x) is not defined at x=1

 Hence , f(x) is discontinuous at x=1

 At x=2

  $\lim_{x \rightarrow {2^{-}}}f(x)=\lim_{x \rightarrow {2^{-}}}\frac{1}{x-1}=\frac{1}{2-1} =1$

$\lim_{x \rightarrow {2^{+}}}f(x)=\lim_{x \rightarrow {2^{+}}}\frac{x+5}{x+3}=\frac{2+5}{2+3} =\frac{7}{5}$ 

$\therefore$   $\lim_{x \rightarrow {2^{-}}}f(x)\neq\lim_{x \rightarrow {2^{+}}}f(x)$

 $\therefore$     f(x) is discontinuous at x=2

 If A = {x,y,z}, B= {1,2}  , then the total number of relations from  set A to set B are

Answer:

Explanation:

 Given, A= {x,y,z}

 B= {1,2}

 n(A)=3, n(B)=2

 $\therefore$   Number of relation from set A to set B is 

2^{n(A) x n(B)} = 2^{3 x2}   =2⁶  =64

 The  symbolic  form of the following circuit is (where p,q represents  switches S₁ and S₂ closed respectively )

Answer:

Explanation:

 Here , p,q represents switches S₁ and S₂ closed respectively,

 here, S₁, S₂ , (S₁' and S₂' ) are in parallel

 and S₁'   and S₂' are series

 $\therefore$     l= $(p\vee q)\vee (\sim p\wedge \sim q)$

If   $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$, then  $\frac{dy}{dx}=$

Answer:

Explanation:

We have

$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$

 squaring both sides , we get

$\frac{x}{y}+\frac{y}{x}+2=16$

           $x^{2}+y^{2}=14xy$

 On differentiating w.r.t 'x' , we get

$2x+2y\frac{dy}{dx}=14\left(x\frac{dy}{dx}+y\right)$

 $x+y\frac{dy}{dx}=7x\frac{dy}{dx}+7y$

$\frac{dy}{dx}(y-7x)=7y-x$

$\frac{dy}{dx}=\frac{7y-x}{y-7x}$

• Mathematics - General questions