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

• Mathematics (2018) - Solved questions (2018)

Let the orthocentre and centroid of a triangle be  A ( -3,5) and B ( 3,3) respectively. If C is the circumcentre of this triangle, then the radius of  the circle having line segment AC as diameter , is

Answer:

Explanation:

Key Idea    Orthocentre, centroid and circumcentre are collinear and centroid divide orthocentre and circumcentre in 2:1 ratio.

We have orthocentre and centroid of a triangle be A ( -3,5) and ( 3,3) respectively and  C circumcentre

Clearly , $AB= \sqrt{(3+3)^{2}+( 3-5)^{2} }=\sqrt{36+4}=2\sqrt{10}$

We know that, AB:BC=2:1

$\therefore$       BC= $\sqrt{10}$

Now,

AC=AB+BC=$2\sqrt{10}+\sqrt{10}=3\sqrt{10}$

Since, AC is a diameter of circle

$\therefore$    $r=\frac{AC}{2}$

$\Rightarrow$      $r=\frac{3\sqrt{10}}{2}=3\sqrt{\frac{5}{2}}$

A straight line through a fixed point ( 2,3) intersects the coordinate axes at distinct  point P and Q, If O is the origin and the rectangle OPRQ is completed, then the locus of R is 

Answer:

Explanation:

Equation of line PQ is  $\frac{x}{\alpha}+\frac{y}{\beta}=1$

 Since this line is passes through fixed point ( 2,3)

$\therefore$       $\frac{2}{\alpha}+\frac{3}{\beta}=1$

$\therefore$    Locus of R is  $2\beta +3\alpha=\alpha\beta$

i.e.       2y+3x=xy

$\Rightarrow$   3x+2y=xy

Let $f( x)=x^{2}+\frac{1}{x^{2}}$ and  $g( x)=x^{}-\frac{1}{x^{'}}$ x $\in$ R - {-1,0,1}. If $h( x)=\frac{f( x)}{g( x)}$ , then the local minimum value of h(x) is

Answer:

Explanation:

We have 

    $f( x)=x^{2}+\frac{1}{x^{2}}$ and   $g( x)=x^{}-\frac{1}{x}$

    $\Rightarrow$       $h( x)=\frac{g( x)}{h( x)}$

$\therefore$      $h( x)=\frac{x^{2}+\frac{1}{x^{2}}}{x-\frac{1}{x}}=\frac{( x-\frac{1}{x})^{2}+2}{x-\frac{1}{x}}$

$\therefore$     $h( x)=( x-\frac{1}{x})+\frac{2}{x-\frac{1}{x}}$

                      $x-\frac{1}{x}>0,$

                   $( x-\frac{1}{x})+\frac{2}{x-\frac{1}{x}}\in [2\sqrt{2},\infty]$

                 $x-\frac{1}{x}<0$

                $( x-\frac{1}{x})+\frac{2}{x-\frac{1}{x}}\in ( -\infty,2\sqrt{2})$

$\therefore$      Local minimum value is  $2\sqrt{2}$

If the curves $y^{2}=6x,9x^{2}+by^{2}=16$  intersect each other at right angles, then the value of b is

Answer:

Explanation:

 We have  y²=6x

$\Rightarrow$      $2y\frac{\text{d}y}{\text{d}x}=6\Rightarrow \frac{\text{d}y}{\text{d}x}=\frac{3}{y}$

       Slope of tangent at ( x₁,y₁) is m₁  = $m_{1}=\frac{3}{y_{1}}$

      Also   ,  $9x^{2}+by^{2}=16$

$\Rightarrow$    $18x+2by\frac{\text{d}y}{\text{d}x}=0  \Rightarrow \frac{\text{d}y}{\text{d}x}=\frac{-9x}{by}$

      Slope of tangent at ( x₁,y₁) is $m_{2}=\frac{-9x_{1}}{by_{1}}$

Since these are intersection at right angle

$\therefore$                 $m_{1}m_{2}=-1\Rightarrow\frac{27x_{1}}{by^{2}_{1}}=1$

$\Rightarrow$               $\frac{27x_{1}}{6bx^{}_{1}}=1$     [$\because$   $y_{1}^{2}=6x_{1}$]

$\Rightarrow$              $b=\frac{9}{2}$

The sum of the coefficients of all odd degree terms in the expansion is

$(x+\sqrt{x^{3}-1})^{5}+(x-\sqrt{x^{3}-1} )^{5},(x>1)is$

Answer:

Explanation:

Key Idea =  $(a+b)^{n}+(a-b)^{n}$

    = $2(^{n}C_{0}a^{n}+^{n}C_{2}a^{n-2}b^{2}+^{n}C_{4}a^{n-4}b^{4}.....)$

We have,

           $(x+\sqrt{x^{3}-1})^{5}+(x-\sqrt{x^{3}-1})^{5},x>1$

           = $2(^{5}C_{0}x^{5}+^{5}C_{2}x^{3}(\sqrt{x^{3}-1})^{2}+^{5}C_{4}x(\sqrt{x^{3}-1})^{4})$

           = $2(x^{5}+10x^{3}(x^{3}-1)+5x(x^{3}-1)^{2})$

           = $2(x^{5}+10x^{6}-10x^{3}+5x^{7}- 10x^4+5x)$

  Sum of coefficients of all odd degree terms is 2 (1-10+5+5)=2

• Mathematics (2018) - Solved questions (2018)