JEE 2018 :: Mathematics (2018) :: Solved questions (2018)

• Mathematics (2018) - Solved questions (2018)

The boolean expression $\sim (p\vee q) \vee (\sim p \wedge q)$ is equivalent to 

Answer:

Explanation:

Key Idea De- Morgan's and distributive law

   We have  $\sim (p \vee q) \vee(\sim p\wedge q)$

   =  $(\sim p\wedge\sim q)\vee (\sim p \wedge q)$  [ $\therefore$ By De-Morgan's law]

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

   ≡$\sim p \wedge (\sim q\vee q)   $         [ By distributive law]

   ≡   $\sim p \wedge t   $      [ $\sim q \vee q=t $]

    ≡ $\sim p$

PQR is a triangular park with PQ=PR=200 m. A TV tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P,Q and R are respectively 45°,30° and 30° ,then the height of the tower (in m) is 

Answer:

Explanation:

Let height of tower TM be h

In            $\triangle PMT.\tan45^{o}=\frac{TM}{PM}$

$\Rightarrow$           $1=\frac{h}{PM}$

 $\Rightarrow$    PM=h

In $\triangle TQM, \tan30^{o}=\frac{h}{QM^{2}}$

   $QM=\sqrt{3}h$

In $\triangle PMQ $   $PM^{2}+QM^{2} =PQ^{2}$

                               $h^{2}+(\sqrt{3h})^{2} =200^{2}$

$\Rightarrow$    $4h^{2}=(200)^{2}$

$\Rightarrow$   h=100 m

Tangents are drawn to the hyperbola  4x²-y²=36 at  the points P and Q . If these tangents intersect at the point T ( 0,3), then the area ( in sq units ) of $\triangle PTQ$ is

Answer:

Explanation:

Tangents are drawn to the hyperbola 4x²-y²=36 at the point P and Q.

Tangent intersects at point T ( 0,3).

Clearly , PQ is chord of contact.

$\therefore$   Equation  of PQ is -3y=36

$\Rightarrow$   y=-12

Solving the curve 4x²-y²=36 and y=-12.

we get      x = $\pm 3\sqrt{5}$

Area of $\triangle PQT=\frac{1}{2}\times PQ \times ST= \frac{1}{2}(6\sqrt{5}\times 15)$

= $45\sqrt{5}$

Tangent and normal are drawn at P ( 16,16) on the parabola y²=16x, which intersect the axis  of the parabola at A and B , respectively . If C is centre of the circle through the points P,A and B and $\angle$CPB= θ, then the value of $\tan\theta$ is

Answer:

Explanation:

Equation of tangent and normal to the curve y²=16x at ( 16,-16) is x-2y+16= 0 and 2x+y-48=0 respectively,

C is the centre of circle passing through PAB

i.e.                C= ( 4,0)

Slope of  $PC= \frac{16-0}{16-4}=\frac{16}{12}=\frac{4}{3}=m_{1}$

Slope of  $PB= \frac{16-0}{16-24}=\frac{16}{-8}=-2=m_{2}$

                               $\tan\theta=\mid \frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\mid$

$\Rightarrow$                $\tan\theta=\mid \frac{\frac{4}{3}+2}{1-( \frac{4}{3})( 2)}\mid$

$\Rightarrow$      $\tan\theta=2$

If the tangent at ( 1,7) to the curve  $x^{2}=y-6$ touches the circle  $x^{2}+y^{2}+16x+12y+c=0$, then the value of c is

Answer:

Explanation:

Key Idea Equation of tangent to the curve

x²=4ay  at ( x₁,y₁) is $xx_{1}=4a( \frac{y+y_{1}}{2})$

 Tagent to the curve  $x^{2}=y-6$ at ( 1,7)

           $x=\frac{y+7}{2}-6$

$\Rightarrow$      2x-y+5=0            .........( i)

Equation of circle is

$x^{2}+y^{2}+16x+12y+c=0$

Centre ( -8,-6)

$r=\sqrt{8^{2}+6^{2}-c}=\sqrt{100-c}$

Since , line 2x-y+5=0 also touches the circle.

$\therefore$     $\sqrt{100-c}= \mid \frac{2( -8)-( -6)+5}{\sqrt{2^{2}+1^{2}}}\mid$

$\therefore$     $\sqrt{100-c}= \mid \frac{-16+6+5}{\sqrt{5^{}}}\mid$

$\therefore$    $\sqrt{100-c}= \mid -\sqrt{5}\mid$

$\therefore$    100-c=5 $\therefore$    c=95

• Mathematics (2018) - Solved questions (2018)