VITEEE 2015 :: Mathematics :: General questions

• Mathematics - General questions

The solution  of   $\frac{d^{2}x}{dy^{2}}-x=k$ , where k is a non-zero constant , vanishes  when y=0 and tends  of finite limit as y  tends to infinity , is 

Answer:

Explanation:

 We can  write given different equation as 

    (D²-1)x= k  .....(i) 

where , $D=\frac{d}{dy}$

 Its auxiliary equation  is m²-1=0 , so that  m=1,-1

 Hence, CF= C₁e^y  +C₂ e^-y

 Where C₁ , C₂ are arbitary constants

  Now, also   $PI=\frac{1}{D^{2}-1}k$

    $=k.\frac{1}{D^{2}-1}e^{0.y}$

$=K.\frac{1}{0^{2}-1}e^{0.y}=-K$

 So,  solution  of eq.(i) is 

$x=C_{1}e^{y}+C_{2}e^{-y}-k$  .....(ii)

 Given that x=0 , when y=0

 So, 0= $C_{1}+C_{2}-k$     (From (ii))

 $\Rightarrow$   $C_{1}+C_{2}=k$

Multiplying  both sides of eq.(ii) by e^-y . we get

$x.e^{-y}=C_{1}+C_{2}e^{-2y}-ke^{-y}$ .....(iv)

 Given that x → m when  y→ $\infty$ , m being a finite quanity

 So, eq (iv) becomes

 $x\times0=C_{1}+C_{2}\times0-(k\times0)$

$\Rightarrow$   C₁=0       .........(v)

 From  eqs. (iv) and (v) , we get

 C₁  =0  and C₂ =k

 Hence , eq.(ii) becomes

$x=ke^{-y}-k=k(e^{-y}-1)$

Which of the following inequality is true for x>0 ?

Answer:

Explanation:

Let f(x) = $ \log(1+x) -\frac{x}{1+x}$

 $\therefore$      $f'(x)=\frac{1}{1+x}-\frac{(1+x).1-x.1}{(1+x)^{2}}$

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

 Which is positive .         [$\because$ x>0]

 $\therefore$   f(x)  is monoatomic increasing , when x>0.

$\Rightarrow$  f(x)  > f (0)

 now, f(0) = log 1-0=0

 $\therefore$ f(x) >0

$\Rightarrow$          $\log(1+x)-\frac{x}{1+x} >0$

$\Rightarrow$         $\frac{x}{1+x}<  \log(1+x)$.....(i)

                  Also, for x >0,

 $x^{2}>0\Rightarrow x^{2}+x>x$

$\Rightarrow$   x(x+1) >x

$\Rightarrow$    $x > \frac{x}{x+1}$  ...........(ii)

 From eqs. (i) and (ii) , we get

 $\frac{x}{x+1}<\log (1+x)<x$

                         [ $\because$ log(1+x)<x for x >0]

Using  Rolle's theorm , the equation 

$a_{0}x^{n}+a_{1}x^{n-1}+.........+a_{n}=0$

 has atleast one root betwwen 0 and 1, if

Answer:

Explanation:

 Consider the function  f defined  by 

$f(x)=a_{0}\frac{x^{n+1}}{n+1}+a_{n}\frac{x^{n}}{n}+....+a_{n-1}\frac{x^{2}}{2}+a_{n} x$

 since, f(x) ia a polynomial , so it is continuous and differentiable  for all x. f(x)  is continuous in the closed interval [0,1] and differentiable in the open interval (0,1).

 Also, f(0)=0

 and

$f(1)=\frac{a_{0}}{n+1}+\frac{a_{1}}{n}+....+\frac{a_{n-1}}{2}+a_{n} =0$ [say]

i.e, f(0)= f(1)

Thus , all the three conditions of Rolle's theorm are satisfied . Hence , there is atleast one value of x in the open interval  (0,1)

 where f '(x)=0

 i.e,  $a_{0}x^{n}+a_{1}x^{n-1}+a_{n}=0$

 At t=0, the function   $f(t)=\frac{\sin t}{t}$ has 

Answer:

Explanation:

 Given $f(t)=\frac{\sin t}{t}$

 At t=0, we will check continuity of the  function 

LHL=f(0-h)

$=\lim_{h \rightarrow 0}\frac{\sin(0-h)}{(0-h)}=\lim_{h \rightarrow 0}\frac{-\sin h}{-h}=1$

 RHL=f(0+h)

$\lim_{h \rightarrow 0}\frac{\sin(0+h)}{(0+h)}=\lim_{h \rightarrow 0}\frac{\sin h}{h}=1$

and f(0)=1

LHL=RHL= f(0)

So, the function is continuous  at t=0

Now, we check the function is maximum or minimum

 $f '(t)=\frac{1}{t}\cos t-\frac{1}{t^{2}}\sin t$

and    $f ''(t)=\frac{-1}{t}\sin t-\frac{1}{t^{2}}\cos t-\frac{1}{t^{2}}\cos t+\frac{2}{t^{3}}\sin t$

 =$\frac{-\sin t}{t}-\frac{2\cos t}{t^{2}}+\frac{2\sin t}{t^{3}}$

 For maximum or minimum value  of  f(x) , put    f '(x)=0

$\Rightarrow\frac{\cos t}{t}-\frac{\sin t}{t^{2}}=0\Rightarrow\frac{\tan t}{t^{}}=1$

 Now  $\lim_{t \rightarrow 0}f ''(t)$

$=-\lim_{t \rightarrow 0}\left(\frac{\sin t}{t}\right)-2\lim_{t \rightarrow 0}\left(\frac{t\cos t-\sin t}{t^{3}}\right)$

                                                                    $[\frac{0}{0} form]$

 $= -1-2\lim_{t \rightarrow 0}\left(\frac{\cos t-t\sin t-\cos t}{3t^{2}}\right)$   [Using L' Hosptial rule ]

$= -1+\frac{2}{3}\lim_{t \rightarrow 0}\frac{\sin t}{t}$

$= -1+\frac{2}{3}\times1=\frac{-1}{3}<0$

 So , function f(t) is maximum at t=0

From the city population , the probabilty of  selecting a male or smoker is  $\frac{7}{10}$ , a male smoker is $\frac{2}{5}$ and a male, if a smoker is already selected , is $\frac{2}{3}$ . Then , the probability of 

Answer:

Explanation:

Suppose , A : a male is selected

 B: a smoker is selected 

 Given:

$P(A\cup B)=\frac{7}{10}, P(A\cap B)=\frac{2}{5}$ and    $P(\frac{A}{B})=\frac{2}{3}$

 The probability of selecting a smoker

$P(B)=\frac{P(A\cap B)}{P(\frac{A}{B})}$

=$\frac{2\times3}{5\times2}=\frac{3}{5}$

The  probability  of selecting  a non-smoker So, P(B)=1-P(B)

$=1-\frac{3}{5}=\frac{2}{5}$

 The probabilty of selecting a male 

$P(A)=P(A\cup B)+P(A\cap B)-P(B)$

$=\frac{7}{10}+\frac{2}{5}-\frac{3}{5}$

$=\frac{7+4-6}{10}=\frac{1}{2}$

 The probability of selecting  a smoker, if a male  is first  selected , is given by

$P(\frac{B}{A})=\frac{P(A\cap B)}{P(A)}$

=$\frac{2}{5}\times\frac{2}{1}=\frac{4}{5}$

• Mathematics - General questions