JEE 2011 :: General Questions :: Mathematics

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lf z is any complex number satisfying$|z-3-2 i| \leq2$,  then the maximum value of |2z-6+5i| is 

Answer:

Explanation:

Given ,$|z-3-2 i| \leq2$....(i)

 To find  minimum of |2z-6+5i|

 or  $2|z-3+ \frac{5}{2}i|$

 Using triangle inequality

 i.e, $||z_{1}|-|z_{2}|| \leq |z_{1}+z_{2}|$

 $\therefore$    $|z-3+ \frac{5}{2} i|$

 $= |z-3-2i+2i+\frac{5}{2}i|$

 =$|(z-3-2i)+\frac{9}{2} i| \geq |z-3-2i| -\frac{9}{2}|$

 $\geq |2-\frac{9}{2} |\geq \frac{5}{2} \Rightarrow  |z-3+\frac{5}{2}i| \geq \frac{5}{2}$

   or    $ |2z-6+5i| \geq 5$

Consider the parabolu $y^{2}=8x$. Let $\triangle_{1}$ be the area of the triangle formed by the endpoints of its latus rectum and the point  $P(\frac{1}{2},2)$ on the parabola and $\triangle_{2}$ be the area of the triangle formed by drawing tangents at P and at the end points of the latusrectum. Then $\frac{\triangle_{1}}{\triangle_{2}}$ is 

Answer:

Explanation:

As, we know area of triangle formed by three points on the parabola is twice the area of the triangle formed by corresponding tangents i.e, area of $\triangle PQR$=2 area   of $\triangle T_{1}T_{2}T_{3}$

 $\triangle _{1}=2 \triangle _{2}$ or $\frac{ \triangle_{1}}{\triangle_{2}}=2$

Let $f(\theta)=\sin\left[ \tan^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right]$  where  

$-\frac{\pi}{4}< \theta < \frac{\pi}{4}$ .Then. the value of $\frac{d}{d(\tan \theta)} (f(\theta))$ is 

Answer:

Explanation:

 $f(\theta)= \sin\left(\tan^{-1}\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)-\frac{\pi}{4}< \theta <\frac{\pi}{4}$

 Let $\tan^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}d}=\phi$

 $\Rightarrow$   $\tan \phi = \frac{\sin \theta}{\sqrt{\cos 2 \theta}}$

 $\therefore$   $\sin \phi=\frac{\sin \theta}{\sqrt{ \sin^{2} \theta+\cos 2 \theta}}$

 =$\frac{\sin \theta}{\sqrt{1-\sin^{2} \theta}}= \frac{\sin \theta}{\cos \theta}=\tan \theta$

$\therefore$   $f(\theta)= \sin \phi=\tan \theta$

 $\Rightarrow$   $\frac{d f(\theta)}{d( \tan \theta)}=1$

Let  $f: [1,\infty) \rightarrow (2, \infty)$  be a differentiable function such that /(1) = 2. If $6\int_{1}^{x} f(t)dt=3x f(x)-x^{3}$  for all $x  \geq 1$  , then the value of f(2) is 

Answer:

Explanation:

 Given, $f(1)= \frac{1}{3}$ and  $6\int_{1}^{x} f(t) dt$

  = $3x f(x)-x^{3}$ for all x $\geq$ 1

 Using (Newton-Leibnitz formula),

On Differentiating both sides,

 $6 f(x).1-0-3f(x)+3x f'(x)-3x^{2}$

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

$\Rightarrow$   $f'(x)-\frac{1}{x} f(x)=x$

 $\Rightarrow$   $\frac{x f'(x)-f(x)}{x^{2}}=1 \Rightarrow  \frac{d}{dx}\left\{\frac{f(x)}{x}\right\}=1$

 On integreating  both sides,

 $\frac{f(x)}{x}=x+C$                $[ \because f(1)=\frac{1}{3}]$

 $\Rightarrow$  $\frac{1}{3}=1+C \Rightarrow C=-\frac{2}{3}$

 Now,  $f(x)=x^{2}-\frac{2}{3} x$

 $\Rightarrow$    $f(2)=4-\frac{4}{3}=\frac{8}{3}$

Note      Here, f(1) =2  does not satisfy given function

 $\therefore$    $f(1)=\frac{1}{3}$

 For that  , $f(x)=x^{2}-\frac{2}{3} x$

 and $f(2)=4-\frac{4}{3}= \frac{8}{3}$

The minimum value of the sum of real numbers $a^{-5}, a^{-4},3a^{-3},1,a^{8}$ and $a^{10}$ with a >0 is 

Answer:

Explanation:

Using AM $\geq$ GM

$\frac{a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^{8}+a^{10}}{8}$

 $\geq (a^{-5}.a^{-4}.a^{-3}.a^{-3}.a^{-3}.1.a^{8}.a^{10})^{1/8}$

$\Rightarrow$  $ a^{-5}+a^{-4}+3a^{-3}+1+a^{8}+a^{10}\geq 8.1$

 Hence, minimum value is 8

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