JEE 2011 :: General Questions :: Mathematics

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If $\lim_{x \rightarrow0}[1+x \log (1+b^{2})]^{1/x}=2 b\sin^{2} \theta  b>0$ and $\theta\epsilon (-\pi, \pi]$ , then the value of $\theta$ is 

Answer:

Explanation:

 Here,  $\lim_{x \rightarrow0}[1+x \log (1+b^{2})]^{1/x}$

                                                  $[1^{\infty}$ from ]

$\Rightarrow$    $e^{\lim_{x \rightarrow 0} ( x \log [1+b^{2}))}.\frac{1}{x}$

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

 Given, 

    $\lim_{x \rightarrow 0}{(1+x \log (1+b^{2}))}^{1/x}=2b \sin^{2} \theta$

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

$\therefore$    $\sin^{2} \theta = \frac{1+b^{2}}{2b}$

By AM $\geq$ GM,

$\frac{b+\frac{1}{b}}{2}\geq \left( b.\frac{1}{b}\right)^{1/2} \Rightarrow \frac{b^{2}+1}{2b}\geq 1$...(iii)

 From Eqs.(ii) and (iii) , we get  $\sin^{2} \theta=1$

$\Rightarrow$   $\theta = \pm \frac{\pi}{2} $ as $\theta \epsilon (- \pi, \pi )$

Let $f:[-1,2] \rightarrow [0, \infty]$ be a continuous function such that f(x) = f(1 - x) for all, $x\epsilon [-1,2]$. Let $R_{1}=\int_{-1}^{2} x f(x) dx$ and $R_{2}$  be the area of the region bounded by y = f(x),x = - 1, x = 2and the X-axis. Then

Answer:

Explanation:

Here  $R_{1}= \int_{-1}^{2} xf(x) dx$

 Using, $\int_{a}^{b} f(x) dx=\int_{a}^{b}  f(a+b-x) dx$

 $R_{1}=\int_{-1}^{2} (1-x)f(1-x) dx, $ 

                                           $[ \because   f(x)=f(1-x)]$

 $\therefore$      $R_{1}=\int_{-1}^{2} (1-x)f(x) dx$.........(ii)

 Given , $R_{2}$ is area bounded by

 f(x),x-1 and x=2

 $\therefore$   $R_{2}==\int_{-1}^{2} f(x) dx$ ...(iii)

 Adding Eqs.(i) and (ii) , we get

  $2R= =\int_{-1}^{2} f(x) dx$......(iv)

$\therefore$   From Eqs.(iii) and (iv) , we get

  $2R_{1}=R_{2}$

Let f(x)= x² and g(x)= $\sin x $ for all x $\epsilon$ R. Then, the set of all x satisfying (fogogof)(x) = (gogof)(x), where (fog)(x) = f(g(x))is

Answer:

Explanation:

 $f(x)=x^{2},g(x)= \sin x$

 $(g o f)(x)= \sin x^{2}$

 $g o(gof)(x)=\sin(\sin x^{2})$

 $(f ogogof)(x)=(\sin(\sin x^{2}))^{2}$.......(i)

 Again  , $(gof)(x)=\sin x^{2}$

 $(gogof)(x)=\sin (\sin x^{2})$.....(ii)

 Given,  $(fogogof)(x)=(gogof)(x)$

$\Rightarrow$  $(\sin (\sin x^{2}))^{2}= \sin (\sin x^{2})$

 $\Rightarrow$  $\sin (\sin x^{2}) ( \sin (\sin x^{2})-1)=0$

 $\Rightarrow$   $\sin (\sin x^{2})=0 $ or  $\sin (\sin x^{2})=1$

 $\Rightarrow$  $\sin x^{2}=0$   or $\sin x^{2}=\frac{\pi}{2}$

 $\therefore$  $x^{2}=n \pi$

  (i.e, not possible as $ -1 \leq \sin \theta \leq1)$

 $ x=\pm \sqrt{n \pi}$

Let(x, y)be any point on the parabola $y^{2}=4x$. Let P be the point that divides the line segment from (0, 0)to (x, y) in the ratio 1 : 3. Then, the locus of P is

Answer:

Explanation:

By section formula

 $h=\frac{x+0}{4}, k=\frac{y+0}{4}$

 $\therefore$  x-=4h, and y=4k

 Substituting in y²=4x

 $(4k)^{2}=4(4h) \Rightarrow  k^{2}=h$

or    $y^{2}=x$ is required locus

Let P(6, 3) be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If the normal at the point P intersects the X-axis at (9, 0), then the eccentricity of the hyperbola is

Answer:

Explanation:

Equation of normal to hyperbola at $(x_{1},y_{1})$ is

 $\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}+b^{2}$

 $\therefore$  at (6,3)   , $\frac{a^{2}x}{6}+\frac{b^{2}y}{3}=a^{2}+b^{2}$

   It passes throught (9.0)

Now, $\frac{a^{2}.9}{6}=a^{2}+b^{2}$

 $\Rightarrow$  $\frac{3a^{2}}{2}-a^{2}=b^{2}\Rightarrow \frac{a^{2}}{b^{2}}=2$

 $\therefore$  $e^{2}=1+\frac{b^{2}}{a^{2}}=1+\frac{1}{2} \Rightarrow  e=\sqrt{\frac{3}{2}}$

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