JEE 2012 :: General Questions :: Mathematics

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• General Questions - Mathematics

Let$ f:(-1,1)\rightarrow R$ be such  that  $f(\cos 4 \theta)=\frac{2}{2 -\sec^{2} \theta}$  for 

$\theta \epsilon\left(0,\frac{\pi}{4}\right)\cup\left(\frac{\pi}{4},\frac{\pi}{2}\right)$ then, the value (s) of $f(\frac{1}{3})$ is/are 

Answer:

Explanation:

$f(\cos 4 \theta)=\frac{2}{2-\sec^{2} \theta}$ .......(i)

 at , $\cos 4 \theta=\frac{1}{3}$

 $\Rightarrow$   $2 \cos^{2} 2 \theta-1=\frac{1}{3}$

$\Rightarrow$   $\cos^{2} 2 \theta=\frac{2}{3}$

$\Rightarrow$    $ \cos 2 \theta= \pm \sqrt{\frac{2}{3}}$ ........(ii)

 $\therefore$     $f(\cos 4 \theta)=\frac{2 \cos^{2} \theta}{2 \cos^{2} \theta-1}$

                           =$ \frac{1+ \cos 2 \theta}{\cos 2 \theta}$

 $\Rightarrow$   $f\left(\frac{1}{3}\right)=1 \pm\sqrt{\frac{3}{2}}$ from eq (ii)

If the adjoint of a 3x 3 matrix P is 

$\begin{bmatrix}1 & 4&4 \\2 & 1&7\\1&1&3 \end{bmatrix}$, then the possible value(s) of the determinant of P is,/are

Answer:

Explanation:

 Concept involved  if $|A_{n xn}|=\triangle$,  then   $|adj A|= \triangle ^{n-1}$

So., Here,$adj  P_{3x3}$=$\begin{bmatrix}1 & 4&4 \\2 & 1&7\\1&1&3 \end{bmatrix}$

 $\Rightarrow$    $|adj P|=|P|^{2}$

 $\therefore$   $|adj P|$= $\begin{bmatrix}1 & 4&4 \\2 & 1&7\\1&1&3 \end{bmatrix}$

  =1(3-7)-4(6-7)+4(2-1)

=-4+4+4=4

|P|= $\pm 2$

If the straight lines $\frac{x-1}{2}=\frac{y+1}{K}=\frac{z}{2}$  and  $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane (s) containing  these two lines is/are 

Answer:

Explanation:

Concept Involved If the straight lines are coplanar. They the should lie in same plane.
Description of Situation If straight lines are coplanar.

 $\Rightarrow$  $\begin{bmatrix}x_{2}-x_{1} & y_{2}-y_{1}&z_{2}-z_{1} \\a_{1} & b_{1}&c_{1}\\a_{2}& b_{2}&c_{2} \end{bmatrix}=0$

 Sol; Since  

$\frac{x-1}{2}=\frac{y+1}{K}=\frac{z}{2}$

 and  $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,

 $\Rightarrow$   $\begin{bmatrix}2& 0&0 \\2 & K&2\\5&2&K \end{bmatrix}=0\Rightarrow K^{2}=4$

 $\Rightarrow$   $K=\pm 2$

 $\therefore$  $n_{1}=b_{1} \times d_{1}=6j-6k$ , for k=2

For every integer n, let  $a_{n}$  and $b_{n}$ be real numbers. Let  function $f:R \rightarrow R$ be given by 

$f(x)=\begin{cases}a_{n}+\sin \pi x & for x\epsilon[2n,2n+1] \\b_{n}+\cos \pi x & for x \epsilon (2n-1,2n)\end{cases}$

for all integers n, 

If f is continuous ,  then which of the following hold9s) for all n?

Answer:

Explanation:

$f(2n)=a_{n},f(2n^{+})=a_{n}$

 $f(2n^{-})=b_{n}+1$

$ \Rightarrow$   $a_{n}-b_{n}=1$

 $f(2n+1)=a_{n}$

 $f((2n+1)^{-})=a_{n}$

    $f((2n+1)^{+})=b_{n+1}-1$

$\Rightarrow$   $a_{n}=b_{n+1}-1$ or $a_{n}-b_{n+1}=-1$

 or     $a_{n-1}-b_{n}=-1$

 If $f(x)= \int_{0}^{x} e^{t^{2}}(t-2)(t-3) dt,\forall x \epsilon (0,\infty),then$

Answer:

Explanation:

 Concept  involved

 use of Newton Leibnitz formula

 $\frac{d}{dx}\left\{\int_{f(x)}^{g(x)} \phi (t) dt\right\}=\phi(g(x)).g'(x)-\phi (f(x)).f'(x)$

 Sol. Here, $f(x) =\int_{0}^{x} e^{t^{2}} (t-2)(t-3) dt$

 $\Rightarrow$ $f'(x)=e^{x^{2}} (x-2)(x-3)$

$\therefore$  maximum at x=2

 minimum at x=3

 decreasing on (2,3) 

 Also, f'(x)=0 has two roots x=2  and x=3

 i.e, $f'(2)=f'(3)=0$

 Thus, by rolle's theorm

 $f"(c)=0 $ must have atleast one root $\epsilon (2,3)$

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