JEE 2013 :: Main 2013 :: Mathematics 2013

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• Main 2013 - Mathematics 2013

  The x- coordinate of the incentre of the triangle that has the coordinates of mid-points of its sides as (0,1),(1,1) and (1,0) is

Answer:

Explanation:

 Given midpoints  of a triangle are (0,1),(1,1)  and (1,0) . Plotting these points on graph paper and make a triangle. 

 So the sides of a triangle will be 2,2 and  $ \sqrt{2^{2}+2^{2}}$  $i.e 2\sqrt{2}$

 x- coordinate of incentre

  =$\frac{2\times0+2\sqrt{2}.0+2.2}{2+2+2\sqrt{2}}$

  $=\frac{2}{2+\sqrt{2}}\times\frac{2-\sqrt{2}}{2-\sqrt{2}}=2-\sqrt{2}$

Consider

statement I

    $(p\wedge\sim q)\wedge(\sim p\wedge q)$ is a fallacy.

 Statement II

 $(p\rightarrow q)\leftrightarrow(\sim q \rightarrow \sim p)$ is a tautology

Answer:

Explanation:

statement II

  $(p\rightarrow q)\leftrightarrow (\sim q\rightarrow \sim p)$

$\equiv (p\rightarrow q)\leftrightarrow (p\rightarrow q)$

 which is always true, so statement II is true.

  Statement I       $(p\wedge \sim q)\wedge (\sim p \wedge q)$

 $\equiv p \wedge \sim q \wedge \sim p \wedge q$

  $\equiv p \wedge \sim p \wedge \sim q \wedge q$

  $\equiv f\wedge f\equiv f$

 Hence, it is a fallacy statement

 So, statement I is true

 if x, y and z are  in AP and tan^-1x, tan^-1y and tan^-1 z are also in AP, then

Answer:

Explanation:

 Since x,y and z are in AP

 $\therefore$              2y=x+z

 also  tan^-1x, tan^-1y and tan^-1 z  are in AP

 $\therefore$  2 tan^-1y=tan^-1x= tan^-1 (z)

$    \Rightarrow$    $ \tan^{-1}\left(\frac{2y}{1-y^{2}}\right)=\tan^{-1} \left(\frac{x+z}{1-xz}\right)$

$    \Rightarrow$     $ \left(\frac{x+z}{1-y^{2}}\right)= \left(\frac{x+z}{1-xz}\right)\Rightarrow y^{2}=xz$  Since x,y and z are in AP as well as in GP

    $\therefore$   x=y=z

The circle passing through (1,-2) and touching the axis of x at (3,0) also passes through the point 

Answer:

Explanation:

 Let the equation of circle be 

$(x-3)^{2}+(y-0)^{2}+\lambda y=0$

 As it passes  through  (1,-2)

 $\therefore$   $(1-3)^{2}+(-2)^{2}+\lambda (-2)=0$

 $\Rightarrow$    $ 4+4-2\lambda=0\Rightarrow\lambda=4$

 $\therefore$    $(x-3)^{2}+y^{2}+4y=0$   By hit and trial method, we see  that point(5,-2) satisfies equation of circle.

If the equations x²+2x+3=0  and  ax²+bx+c=0 , a,b,c ε R , have a common root , then a:b:c is 

Answer:

Explanation:

Given equation  are 

$x^{2}+2x+3=0$...............(i)

and $ax^{2}+bx+c=0$   ..........(ii)

 Since, Eq.(i0 has imaginary roots so, Eq.(ii)   will also have both roots same as  Eq.(i)

Thus,   $\frac{a}{1}=\frac{b}{2}=\frac{c}{3}$

 Hence, a:b:c is 1:2:3

• Main 2013 - Physics 2013
• Main 2013 - Chemistry 2013
• Main 2013 - Mathematics 2013