JEE 2012 :: General Questions :: Mathematics

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 The ellipse   $E_{1}:\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ is  inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E₂ passing through the point (0, 4)circumscribes the rectangle R. The eccentricity of the ellipse $E_{2}$ is 

Answer:

Explanation:

Concept involved equation of an ellipse is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$    (a >b)

 Ecentricity  $e^{2}=1-\frac{b^{2}}{a^{2}}$       (a >b)

Description of Situation As ellipse circumscribes the rectangle, then it must
pass through all four vertices.

Sol. Let the equation of Ellipse  $E_{2}$ be 

 $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , where a<b and b=4

 Also, it passes through (3,2) 

 $\Rightarrow$   $\frac{9}{a^{2}}+\frac{4}{b^{2}}=1$                     $(\because$ b=4)

 $\Rightarrow$  $\frac{9}{a^{2}}+\frac{1}{4}=1 $ or  $a^{2}=12$

 Eccentricity of $E_{2}$

 $\Rightarrow$   $e^{2}=1-\frac{a^{2}}{b^{2}}=1-\frac{12}{16}=\frac{1}{4}$     $( \because $ a <b)

 $\therefore$  e=$\frac{1}{2}$

Let P=$[a_{ij}]$ be 3x3 matrix and let Q=$[b_{ij}]$,  where $b_{ij}=2^{i+j}a_{ij}$ for  $1 \leq i, j \leq 3$ . If the determinant of P is 2, then the determinant of the matrix Q is 

Answer:

Explanation:

Concept Involved it is a simple question on scalar  multipication  i.e, 

$\begin{bmatrix}ka_{1} & ka_{2}&ka_{3} \\b_{1} & b_{2}&b_{3}\\c_{1}&c_{2}&c_{3} \end{bmatrix}=k \begin{bmatrix}a_{1} & a_{2}&a_{3} \\b_{1} & b_{2}&b_{3}\\c_{1}&c_{2}&c_{3} \end{bmatrix}$

Description of Situation Construction of Matrix

i.e, if A=$ [a_{ij}]_{3\times 3}= \begin{bmatrix}a_{11} & a_{12}&a_{13} \\a_{21} & a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{bmatrix}$

 sol. here , P=$[a_{ij}]_{3\times 3}= \begin{bmatrix}a_{11} & a_{12}&a_{13} \\a_{21} & a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{bmatrix}$

 $Q=[b_{ij}]_{3x3}= \begin{bmatrix}b_{11} & b_{12}&b_{13} \\b_{21} & b_{22}&b_{23}\\b_{31}&b_{32}&b_{33} \end{bmatrix}$

 where    $b_{ij}=2^{i+j}a_{ij}$

 $\therefore$   $|Q|= \begin{bmatrix}4a_{11} & 8a_{12}&16a_{13} \\8a_{21} &16 a_{22}&32a_{23}\\16a_{31}&32a_{32}&64a_{33} \end{bmatrix}$

   =$4 \times 8 \times 16  \begin{bmatrix}a_{11} & a_{12}&a_{13} \\2a_{21} & 2a_{22}&2a_{23}\\4a_{31}&4a_{32}&4a_{33} \end{bmatrix}$

=$2^{9} \times 2 \times 4  \begin{bmatrix}a_{11} & a_{12}&a_{13} \\a_{21} & a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{bmatrix}$

 =$2^{12}.|P|=2^{12}.2=2^{13}$

The locus of the end point of the chord of contact of tangents drawn from points lying on the straight line 4x-5y=20 to the circle $x^{2}+y^{2}=9$ is 

Answer:

Explanation:

Concept Involved if 

 $S: ax^{2}+2hxy+by^{2}+2gx+2fy+c$

then equation of chord bisected at P (x₁,y₁) is T=S₁

 or $axx_{1}+h(xy_{1}+yx_{1})+byy_{1}+g(x+x_{1})+f(y+y_{1})+C$

       =$ax_{1}^{2}+2hx_{1}y_{1}+by_{1}^{2}+2gx_{1}+2fy_{1}+C$

Description of Situation As equation of
chord of contact is T= 0

Sol. Here, equation of chord of contact wrt  P is 

$x\lambda+y\left(\frac{4\lambda-20}{5}\right)=9$

 $5 \lambda x+(4 \lambda-20)y=45$  .........(i)

and equation of chord bisected at the point Q (h, k) is

 $xh+yk-9=h^{2}+k^{2}-9$

$\Rightarrow$  $xh+ky=h^{2}+k^{2}$.....(ii)

From Eqs.(i) and (ii)   We get

 $\frac{5 \lambda}{h}=\frac{4 \lambda-20}{k}=\frac{45}{h^{2}+k^{2}}$

$\therefore$  $\lambda=\frac{20h}{4h-5k}$ and   $\lambda=\frac{9h}{h^{2}+k^{2}}$

$\Rightarrow$   $\frac{20h}{4h-5k}=\frac{9h}{h^{2}+k^{2}}$

 or   $20(h^{2}+k^{2})=9(4h-5k)$

or     $20(x^{2}+y^{2})=36x-45y$

The function $f:[0,3] \rightarrow [1,29],$ defined by $f(x)=2x^{3}-15$

         $x^{2}+36x+1$, is 

Answer:

Explanation:

Concept Involved To check nature of the function

(i) One-One To check one-one, we must check whether

 $f'(x) >0$ or $f'(x) <0$ in given domain

(ii) Onto To check onto, we must check
Range = Co-domain

Description of Situation To find range in given domain [a,b] put f'(x)=0 and find $x= \alpha_{1},\alpha_{2},......\alpha_{n}\epsilon[a,b]$

 Now find, 

  {$f(a),f(\alpha_{1}),f(\alpha_{2}),.....f(\alpha_{n}),f(b)}$}

its the greatest and least gives you range.

 $f:[0,3] \rightarrow [1,29]$

     $f(x)=2x^{3}-15x^{2}+36x+1$

 $\therefore$  $f'(x)=6x^{2}-30x+36$

 =$6(x^{2}-5x+6)=6(x-2)(x-3)$

For given domain [0,3] ,f(x) is increasing  as well as decreasing

$\Rightarrow$ many-one

 Now, put $f'(x)=0 \Rightarrow x=2,3$

 Thus, for range f(0)=1

  f(2)=29,f(3)=28

 $\Rightarrow$   Range  $\epsilon[1,29]$ $ \therefore$  onto

if  $\lim_{x \rightarrow \infty}\left(\frac{x^{2}-x+1}{x+1}-ax-b\right)=4$ , then

Answer:

Explanation:

Concept involved   $( \frac{\infty}{\infty})$ form

 $\lim_{x \rightarrow \infty}\frac{a_{0}x^{n}+a_{1}x^{n-1}+.....+a_{n}}{b_{0}x^{m}+b_{1}x^{m-1}+.....+b_{m}}$

=$\begin{cases}0 ,& if n=m\\\frac{a_{0}}{b_{0}}, & if n <m \\+\infty, &if n>m and a_{0}b_{0}>0\\-\infty&if n>m and a_{0}b_{0}<0\end{cases}$

Description of Situation As to make degree of numerator equal to degree of denominator.

Sol. $\lim_{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-ax-b\right)=4$

$\Rightarrow$   $\lim_{x \rightarrow \infty}\frac{x^{2}+x+1-ax^{2}-ax-bx-b}{x+1}=4$

$\Rightarrow$  $\lim_{x \rightarrow \infty}\frac{x^{2}(1-a)+x(1-a-b)+(1-b)}{x+1}=4$

 Here, we make degree of N^r = degree of D^r

$\therefore$   1-a=0

 and  $\lim_{x \rightarrow \infty}\frac{x(1-a-b)+(1-b)}{x+1}=4$

$\Rightarrow$    1-a-b=4

$\Rightarrow$   b=-4   $[ \because (-a)=0]$

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