JEE 2011 :: General Questions :: Mathematics

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• General Questions - Mathematics

Let $\omega = e^{i \pi/3}$ and a,b,c,x,y,z be non  zero complex numbers such that a+b+c=x, $a+b \omega+c\omega^{2}=y, a+b \omega^{2}+c \omega =z$

then the value of $\frac{|x|^{2}+|y|^{2}+|z|^{2}}{|a|^{2}+|b|^{2}+|c|^{2}}$ is 

Answer:

Explanation:

 Here, $\omega= e^{12 \pi/3}$,  then only integer solution exists

 Then $\frac{|x|^{2}+|y|^{2}+|z|^{2}}{|a|^{2}+|b|^{2}+|c|^{2}}$=3

Let $a =-\widehat{i}-\widehat{k},b=-\widehat{i}+\widehat{j}$ and $c= \widehat{i}+2 \widehat{j}+3 \widehat{k}$  be the three vectors. If r is a  vector such that  r x b=c x b and r.a=0 , then the value of r.b is 

Answer:

Explanation:

 r x b= c x b

 $\Rightarrow$  (r-c) x b=0  $\Rightarrow$ $r-c+\lambda b$

  or  $r=c+\lambda b$..........(i)

 Given, r.a=0 , taking dot product with a for eq.(i)

 Now, $r.a=a.c+\lambda a.b$

$\therefore$  $\lambda=\frac{\overrightarrow{a}.\overrightarrow{c}}{a.b}$ $[\because \overrightarrow{r}.\overrightarrow{a}=0]$......(ii)

  From Eq.(i) and (ii) , we get

 $r=c-\frac{a.c}{a.b}b$

 Taking dot with b, we get

 $r.b=c.b-\frac{a.c}{a.b} (b.b)$

 where, $\begin{bmatrix}a=-\widehat{i}-\widehat{k}  \\b=-\widehat{i}+\widehat{j}\\c=\widehat{i}+2\widehat{j}+3\widehat{k} \end{bmatrix}$

 =$(-1+2)- \frac{(-1-3)}{(1)}(1+1)=1+8=9$

Let  $y'(x)+y(x)g'(x)=g(x)g'(x)y(0)=0, x \epsilon R,$ where f'(x)  denoted $\frac{d f(x)}{dx}$ and g(x) is a given non-constant differentiable function on R with g(0)=g(2)=0,  Then the value of y(2) is .....

Answer:

Explanation:

$\frac{dy}{dx}+y.g'(x)=g(x)g'(x)$

 IF= $e^{\int g'(x)dx}= e^{g(x)}$

 $\therefore$ solution is 

              $y(e^{g(x)})=\int g(x).g'(x).e^{g(x)} dx+C$

Put   $g(x)=t, g'(x)dx=dt$

 $y(e^{g(x)})=\int t.e^{t} dt+C$

  =$t.e^{t}-\int 1.e^{t} dt+C=t.e^{t}-e^{t}+C$

 $y e^{g(x)}=(g(x)-1)e^{g(x)}+C$........(i)$

 Given , y(0)=0,g(0)=g(2)=0

$\therefore$  Eq.(i) becomes 

        $y(0).e^{g(0)}=(g(0)-1).e^{g(0)}+C$

 $\Rightarrow$   $0= (-1).1+C \Rightarrow C=1$

 $\therefore$  $y(x).e^{g(x)}=(g(x)-1)e^{g(x)}+1$

$\Rightarrow$ $y(2).e^{g(2)}=(g(2)-1)e^{g(2)}+1$

 where g(2)=0

 $\Rightarrow$ $y(2).1= (-1).1+1 \Rightarrow y(2)=0$

Let $f: (0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-bx}$ where b is a constant such that 0<b<1, then 

Answer:

Explanation:

Here , $f(x)=\frac{b-x}{1-bx}$

 where, 0<b<1,0<x<1

For function to be invertible it should
be one-one onto

 $\therefore$ Check range

 Let    f(x)=y $\Rightarrow y=\frac{b-x}{1-bx}$

 $\Rightarrow$  $y-bxy=b-x \Rightarrow x(1-by)=b-y$ 

$\Rightarrow$   $x= \frac{b-y}{1-by}$

 where , 0<x<1

 $\therefore$ $0 < \frac{b-y}{1-by} <1$

  $\frac{b-y}{1-by} >0$ and $\frac{b-y}{1-by} <1$

 $\Rightarrow$    y<b  or $y > \frac{1}{b}$ .......(i)

  $\frac{(b-1)(y+1)}{1+by}  <-1 < y < \frac{1}{b}$.......(ii)

 From eqs.(i) and (ii) , we get

  $Y \in \left(-1, \frac{1}{b}\right)\subset$ Codomain

 Thus , f(x)  is not invertible 

Let E and F be two independent events. The probability that exactly one of them occurs is 11/25 and the probability of none of them occurring is 2/25.lf P(T) denotes the probability of occurrence of the event T, then

Answer:

Explanation:

  $P( E \cup F)-P(E \cap F)=\frac{11}{25}$.....(i)

 (i.e, only E or only F)

 Neither of them occurs =$\frac{2}{25}$

 $\Rightarrow$    $P(\overline{E} \cap \overline{F})=\frac{2}{25}$......(ii)

 From . Eq.(i), we get

 $P(E)+P(F)-2P(E \cap F)=\frac{11}{25}$ ...(iiI)

 From . Eq.(ii), we get

 $(1-P(E))(1-P(F))=\frac{2}{25}$

$\Rightarrow$  $1-P(E)-P(F)+P(E).P(F)=\frac{2}{25}$.....(iv)

 From eqs.(iii)  and (iv) , we get

 $P(E)+P(F)=\frac{7}{5}$  $P(E).P(F)=\frac{12}{25}$

 $\therefore$  $P(E) .\left\{ \frac{7}{5}-P(E)\right\}=\frac{12}{25}$

$\Rightarrow$   $(P(E))^{2}-\frac{7}{5} P(E)+\frac{12}{25}=0$

 $\Rightarrow$  $\left(P(E)-\frac{3}{5}\right)\left(P(E)-\frac{4}{5}\right)=0$

 $\therefore$    $P(E)= \frac{3}{4}$ or $\frac{4}{5}$  $\Rightarrow  P(F)= \frac{4}{5}$ or $\frac{3}{5}$

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