VITEEE 2014 :: Mathematics :: General questions

• Mathematics - General questions

 The negation of   $(\sim p\wedge q)\vee(p\wedge \sim q)$ is 

Answer:

Explanation:

Let S: $(\sim p\wedge q)\vee(p\wedge \sim q)$

$\Rightarrow$   $\sim S=[(\sim p\wedge q)\vee(p\wedge \sim q)]$

$\Rightarrow$     $\sim S=\sim (\sim p \wedge q)\wedge \sim(p \wedge \sim q)]$

$\Rightarrow$     $\sim S=( p \vee \sim q)\wedge \sim(\sim p \vee  q)$

If $x+iy=\frac{3}{2+\cos \theta+i \sin \theta}$, then $x^{2}+y^{2}$ is equal to 

Answer:

Explanation:

$x+iy$

   $=\frac{3}{2+\cos \theta+i \sin \theta}=\frac{3(2+\cos \theta-i \sin \theta)}{(2+\cos \theta)^{2}+\sin^{2} \theta}$

  $=\frac{6+3\cos \theta-3 i \sin \theta}{4+\cos^{2} \theta+4 \cos \theta+\sin^{2} \theta}$

$\frac{6+3 \cos \theta-3i \sin \theta}{5+4 \cos \theta}$

   $=\left(\frac{6+3\cos \theta}{5+4 \cos \theta}\right)+\left(\frac{-3 \sin \theta}{5+4 \cos \theta}\right)$

 On equating  real and imaginary  parts, we get

$x= \frac{3(2+\cos \theta)}{5+4\cos \theta}$

and $y=\frac{-3 \sin \theta}{5+4 \cos \theta}$

$\therefore$   $x^{2}+y^{2}= \frac{9[4+\cos^{2} \theta+4 \cos \theta+\sin^{2}\theta]}{(5+4 \cos \theta)^{2}}$

$=\frac{9}{5+4 \cos \theta}=4\left(\frac{6+3 \cos \theta}{5+4 \cos^{2} \theta}\right)-3$

   =4x-3

The statement $(p \Rightarrow q)\Leftrightarrow(\sim p\wedge q)$  is 

Answer:

Explanation:

 Hence, given statement is neither tautology nor contradiction

The triangle formed by the tangent to the curve $f(x)=x^{2}+bx-b$ at the point (1,1) and the coordinate axes lie in the first quadrant . if its area is 2 , then the value of b is 

Answer:

Explanation:

 Given curve is y=$f(x)=x^{2}+bx-b$ 

$\Rightarrow$  f'(x)=2x+b

 The equation of tangent at point (1,1) is 

$y-1=\left(\frac{dy}{dx}\right)_{(1,1)}(x-1)$

$\Rightarrow$    $y-1=(b+2)(x-1)$

$\Rightarrow$  $2+b)x-y=1+b$

$\Rightarrow$  $\frac{x}{\left(\frac{1+b}{2+b}\right)}-\frac{x}{(1+b)}=1$

So, $OA=\frac{1+b}{2+b}$

 and OB=-(1+b)

 Now, area  of $\triangle AOB= \frac{1}{2} \times \frac{(1+b)[-(1+b)]}{(2+b)}=2$

$\Rightarrow$     $4(2+b)+(1+b)^{2}=0$

$\Rightarrow$  $b^{2}+6b+9=0$

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

The minimum value of $\frac{x}{\log x}$ is

Answer:

Explanation:

 Let  f(x)= $\frac{x}{\log x}$

 $\Rightarrow$    $f'(x)=\frac{\log x-1}{(\log x)^{2}}$

 For maxima  and minima , put f'(x)=0

 $\log x-1=0$

 $\Rightarrow$   $x=e$

 Now, f"(x)

    =  $\frac{(\log x)^{2}.\frac{1}{x}-(\log x-1).\frac{2 \log x}{x}}{(\log x)^{4}}$

  $\Rightarrow$     $f"(e)=\frac{\frac{1}{e}-0}{1}=\frac{1}{e} >0$

 $\therefore$   f(x)  is minimum  at x=e

Hence, minimum value of f(x) at x=e is 

    $f(e) =\frac{e}{\log e}=e$

• Mathematics - General questions