VITEEE 2016 :: Mathematics :: General questions

• Mathematics - General questions

Convert the hexadecimal numeral ABCD into Binary numeral

Answer:

Explanation:

Replacing each hexadecimal digit by the corresponding  4-digit binary numeral, we have 

 (ABCD)₁₆ = (1010101111001101)₂

What is the area of loop of the curve r =a sin 3 $\theta$ ?

Answer:

Explanation:

If curve r = a sin3$\theta$

 To trace the curve , we consider the following table:

 Thus there is a loop between $\theta$ =0 &  $\theta=\frac{\pi}{3}$

 as r  varies from r=0 to r=0.

 Hence, the area of the loop lying in the

positive quadrant $=\frac{1}{2}\int_{0}^{\frac{\pi}{3}} r^{2}d\theta$

 $=\frac{1}{2}\int_{0}^{\frac{\pi}{3}}\sin^{2}\phi .\frac{1}{3}d\phi$

[on putting ,  $=3\theta=\phi\Rightarrow d\theta=\frac{1}{3}d\phi]$

$=\frac{a^{2}}{6}\int_{0}^{\frac{\pi}{2}} \sin^{2}\phi d\phi$

$=\frac{a^{2}}{6}\int_{0}^{\frac{\pi}{2}} \frac{1-\cos2\phi}{2}d\phi [\because \cos2\theta=1-2\sin^{2}\theta]$

$=\frac{a^{2}}{12}\left[\phi+\frac{\sin 2 \phi}{2}\right]^{\frac{\pi}{2}}_{0}$

$=\frac{a^{2}}{12}\left[\frac{\pi}{2}+\sin\pi\right]=\frac{a^{2}\pi}{24}$

If  $e^{x}= y+\sqrt{1+y^{2}}$  , then the value of y is 

Answer:

Explanation:

 Given   $e^{x}=y+\sqrt{1+y^{2}}$

 $\Rightarrow e^{x}-y=\sqrt{1+y^{2}}$

Squaring both side, we have

$e^{2x}+y^{2}-2e^{x}y=1+y^{2}$

 $\Rightarrow 2e^{x}y=e^{2x}-1$

$\Rightarrow y=\frac{e^{2x}-1}{2e^{x}}\Rightarrow y=\frac{1}{2}[e^{x}-e^{-x}]$

 The equation of one of the common tangent to the parabola y² =8x   and x²+y² -12x+4=0  is 

Answer:

Explanation:

Any tangent to parabola y² =8x is  $y= mx+\frac{2}{m}$   .............(i)

 It touches the circle  $x^{2}+y^{2}-12x+4=0$

 if the length of perpendicular from the centre (6,0) is equal to radis  $\sqrt{32}$

$\therefore$     $\frac{6m+\frac{2}{m}}{\sqrt{m^{2}+1}}=\pm\sqrt{32}$

 $\Rightarrow \left(3m+\frac{1}{m}\right)^{2}=8(m^{2}+1)$

 $\Rightarrow(3m^{2}+1)^{2}=8(m^{4}+m^{2})$

 $\Rightarrow m^{4}-2m^{2}+1=0\Rightarrow m=\pm1$

 Hence, the required tangents are y=x+2 and y=-x-2

If g(x)  is a polynomial satisfying g(x) g(y) =g(x)+g(y) +g(xy)-2

 For all real x and y and  g(2)=5 then  $Lt_{x \rightarrow 3}g(x)$  is 

Answer:

Explanation:

 g(x).g(y) =g(x)+g(y)+g(xy)-2  ...(i)

 Put x=1, y=2 then

 g(1). g(2)=g(1)+g(2)+g(2)-2

 5g(1)- g(1) +5+5-2

 4g(1)= 8

$\therefore$   g(1)=2

  Put y = $\frac{1}{x}$ in equation  (i) , we get

 g(x) , g($\frac{1}{x}$ )= g(x) + g( $\frac{1}{x}$)+g(1)-2

g(x) . g($\frac{1}{x}$) = g(x) + g($\frac{1}{x}$)+2-2

                                                      [$\because$ g(1)=2]

 This valid only for the polynomial 

  $\therefore$ g(x) = 1$\pm$xⁿ .................(2)

 Now g(2)=5      (Given)

 $\therefore$   1$\pm$ 2ⁿ= 5     [Using equation(2)]

 $\pm$  2ⁿ =4, $\Rightarrow$ 2ⁿ =4,-4

 Since , the  value of 2ⁿ cannot be -Ve

 So , 2ⁿ =4 . $\Rightarrow$  n=2

 Now, put  n=2 in equation (2) , we get

 g(x) =1  $\pm$ x²

 $\therefore Lt_{x \rightarrow 3}g(x)=Lt_{x \rightarrow 3} (1\pm x^{2})= 1\pm (3)^{2}$

      =  $ 1\pm 9=10,-8$

• Mathematics - General questions