JEE 2011 :: General Questions :: Mathematics

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• General Questions - Mathematics

Let L be a normal to the parabola y² = 4x. lf L passes through the point (9, 6), then L is given by

Answer:

Explanation:

Normal to $y^{2}=4x$ is 

 y-mx-2m-m³ which passes through (9,6)

 Now,  $6=9m-2m-m^{3}$

$m^{3}-7m+6=0 \Rightarrow m=1,2,-3$

 $\therefore$ Equation of normals are

 y-x+3=0

 y+3x-33=0  and y-2x+12=0

If $f(x)=\begin{cases}-x-\frac{\pi}{2}, & x \leq -\frac{\pi}{2}\\-\cos x,&-\frac{\pi}{2}< x \leq0 \\ x-1, & 0 <x \leq1\\\log x , & x>1\end{cases}$, then

Answer:

Explanation:

$f(x)=\begin{cases}-x-\frac{\pi}{2}, & x \leq -\frac{\pi}{2}\\-\cos x,&-\frac{\pi}{2}< x \leq0 \\ x-1, & 0 <x \leq1\\\log x , & x>1\end{cases}$,

 continuity at x=-$\frac{\pi}{2}$

$f\left(-\frac{\pi}{2}\right)=-\left(-\frac{\pi}{2}\right)-\frac{\pi}{2}=0$

 RHL= $\Rightarrow$   $\lim_{h \rightarrow 0}-\cos \left(-\frac{\pi}{2}+h\right)=0$

$\therefore$  continuous at x=0

 Continuity  at x=0 $\Rightarrow$ f(0)=-1$

$RHL \Rightarrow \lim_{h \rightarrow 0}(0+h)-1=-1$

 $\therefore$  continuous  at x=0

 Continuity at x=1;f(1)=0

RHL$\Rightarrow$ $\lim_{h \rightarrow 0}\log (1+h)=0$

 $\therefore$ continuous at x=1

Here, $f(x)=\begin{cases}-1, & x \leq -\frac{\pi}{2}\\\sin x,&-\frac{\pi}{2}< x \leq0 \\ 1, & 0 <x \leq1\\\frac{1}{x} , & x>1\end{cases}$,

 Differentiable at x=0

 LHD=0, RHD=1

$\therefore$  not differentiable at x=0

 Differentiable at x=1

 LHD=1,RHD=1

 $\therefore$  Differentiable at x=1

 also,  for x$=-\frac{3}{2} \Rightarrow f(x)=-x-\frac{3}{2}$

 $\therefore$ differentiable at x= $-\frac{3}{2}$

The value of  b for which the equations $x^{2}+bx-1=0,x^{2}+x+b=0$ have  one root in common  is 

Answer:

Explanation:

If $a_{1}x^{2}+b_{1}x+c_{1}=0$

 and $a_{2}x^{2}+b_{2}x+c_{2}=0$

 Have  a comon  real root , then 

$\Rightarrow$  $(a_{1}c_{2}-a_{2}c_{1})^{2}= (b_{1}c_{2}-b_{2}c_{1})(a_{1}b_{2}-a_{2}b_{1})$

 $\because$    $x^{2}+bx-1=0$, 

                      $x^{2}+x+b$   , have a common root

$\Rightarrow$$(1+b)^{2}=(b^{2}+1)(1-b)$

$\Rightarrow$  $b^{2}+2b+1=b^{2}-b^{3}+1-b$

$\Rightarrow$ $b^{3}+3b=0\Rightarrow b(b^{2}+3)=0$

$\Rightarrow$  $b=0, \pm \sqrt{3} i$

Let $\omega  \neq 1$ be a cube root of unity and S be the set of all non-singular matrices of the form  $\begin{bmatrix}1 & a&b \\\omega & 1&c\\\omega^{2}&\omega&1 \end{bmatrix}$, where each of a,b and c is either $\omega$ or $\omega^{2}$ . Then, the number of distinct matrices in the set S is

Answer:

Explanation:

$|A| \neq 0$, as non-singular

$\begin{bmatrix}1 & a&b \\\omega & 1&c\\\omega^{2}&\omega&1 \end{bmatrix} \neq=0$

$\Rightarrow$   $1(1-c \omega)-a(\omega-c \omega^{2})+b(\omega^{2}-\omega^{2}) \neq=0$

$\Rightarrow$  $1- c \omega-a \omega+ac \omega^{2} \neq 0$

 $\Rightarrow$   $(1-c \omega)(1-a \omega) \neq 0$

 $\Rightarrow$  $a\neq \frac{1}{\omega}, c \neq \frac{1}{\omega} \Rightarrow a= \omega , c= \omega$

 and    $b \epsilon( \omega , \omega^{2}) \Rightarrow 2$ solutions 

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

Answer:

Explanation:

Equation of circle passing ihrough a point (x₁,y₁) and touching the straight
line L. is given by

 $(x-x_{1})^{2}+(y-y_{1})^{2}= \lambda L=0$

$\therefore$   Equation sf circle passing through (0, 2) and touching x = 0.

Now,  $(x-0)^{2}+(y-2)^{2}+\lambda x=0$...(i)

 Also, it passes through (-1,0)

 so, $1+4-\lambda=0 \Rightarrow  0 \Rightarrow \lambda=5$

 Eq.(i) becomes 

       $x^{2}+y^{2}-4y+4+5x=0$

 $\Rightarrow$  $x^{2}+y^{2}+5x-4y+4=0$

 For x- intercept put y=0

  $x^{2}+5x+4=0$

$\Rightarrow$ $(x+1)(x+4)=0$ 

$\therefore$   x=-1,-4

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