JEE 2011 :: General Questions :: Mathematics

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Let U₁ and U₂ be two urns such that U₁ contains 3 white and 2 red balls and U₂ contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U₁ and put into U₂. However, if tail appears then 2 balls are drawn at random from U₁ and put into U₂. Now, 1 ball is drawn at random from U₂

Given that the drawn ball from U₂ is white. the probability that head appeared on the coin is

Answer:

Explanation:

P( head appeared /white from $V_{2}$)

 =$P(H).\frac{\left\{\frac{^{3}C_{1}}{^{5}C_{1}}\times\frac{^{2}C_{1}}{^{2}C_{1}}+\frac{^{2}C_{1}}{^{5}C_{1}}\times\frac{^{1}C_{1}}{^{2}C_{1}}\right\}}{\frac{23}{30}}$

 =$\frac{1}{2}\frac{\left\{\frac{3}{5}\times1+\frac{2}{5}\times\frac{1}{2}\right\}}{\frac{23}{30}}=\frac{12}{23}$

Let U₁ and U₂ be two urns such that U₁ contains 3 white and 2 red balls and U₂ contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U₁ and put into U₂. However, if tail appears then 2 balls are drawn at random from U₁ and put into U₂. Now, 1 ball is drawn at random from U₂

The probability of the drawn ball from U₂ being  white is

Answer:

Explanation:

Now , probability of the drawn ball from $U_{2}$ being  white is 

 $\Rightarrow$ P( white /U₂)

 =  $P(H).\left\{\frac{^{3}C_{1}}{^{5}C_{1}}\times\frac{^{2}C_{1}}{^{2}C_{1}}+\frac{^{2}C_{1}}{^{5}C_{1}}\times\frac{^{1}C_{1}}{^{2}C_{1} }\right\}$ $+ P(T)\left\{\frac{^{3}C_{1}}{^{5}C_{1}}\times\frac{^{3}C_{2}}{^{3}C_{2}}+\frac{^{2}C_{2}}{^{5}C_{2}}\times\frac{^{1}C_{1}}{^{3}C_{2}}$

$+\frac{^{3}C_{1}.^{2}C_{1}}{^{5}C_{2}}\times\frac{^{2}C_{1}}{^{3}C_{2}}\right\}$

Now ,P(white/U₂)=$\frac{1}{2}\left\{\frac{3}{5}\times1+\frac{2}{5}\times\frac{1}{2}\right\}$

$+\frac{1}{2}\left\{\frac{3}{10}\times1+\frac{1}{10}\times\frac{1}{3}+\frac{6}{10}\times\frac{2}{3}\right\}=\frac{23}{30}$

Let $ f : R \rightarrow R $ be a function such that f(x + y)= f(x) + f(y),$ \forall x,y \epsilon R $. If f(x) is  differentiable at x = 0, then

Answer:

Explanation:

 $f(x+y)=f(x)+f(y)$ as f(x) is differentiable  at x=0

 $\Rightarrow$    $f'(0)=k$ ....(i)

 Now, $f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h \rightarrow 0}\frac{f(x)+f(h)-f(x)}{h}$

 =$\lim_{h \rightarrow 0}\frac{f(h)}{h}$    $[ \frac{0}{0}$ from]

 Given ,$f(x+y)=f(x)+f(y), \forall x,y$

$\therefore$ $f(0)=f(0)+f(0)$

when $x=y=0 \Rightarrow f(0)=0$

 Using L' Hosptial 's rule 

 $\lim_{h\rightarrow 0}\frac{f'(h)}{1}=f'(0)=k$ .....(ii)

 $\Rightarrow$  $f'(x)=k, $on integrating  both sides

$f(x)=kx+C, $ as f(0)=0 $\Rightarrow$ C=0

 So, f(x)=kx

$\therefore$  f(x) is continuous for all $x \epsilon R$ and 

 $f'(x)=k$, i.e constant  for all $x \epsilon R$

 hence , both (b) and (c) are correct.

The vector(s) which is/are coplanar with vectors   $\widehat{i}+\widehat{j}+2\widehat{k}$ and $\widehat{i}+2\widehat{j}+\widehat{k}$ , are  perpendicular to the vector  $\widehat{i}+\widehat{j}+\widehat{k}$ is/are 

Answer:

Explanation:

Let $a= \widehat{i}+\widehat{j}+2\widehat{k},b=\widehat{i}+2\widehat{j}+\widehat{k}$ and

$c= \widehat{i}+\widehat{j}+\widehat{k}$

$\therefore$ Avector coplanar to a and b and perpendicular to c

 Now  $\lambda ( a \times b) \times c$

 $\Rightarrow$         $\lambda {(a.c)b-(b.c)a}$

 $\Rightarrow$  $\lambda {(1+1+4)(\widehat{i}+2 \widehat{j}+\widehat{k})}$

                                           ${-(1+2+1)(\widehat{i}+\widehat{j}+2 \widehat{k}})$

$\Rightarrow$  $\lambda (6 \widehat{i}+12 \widehat{j}+6 \widehat{k}-6\widehat{i}-6\widehat{j}-12\widehat{k})$

 $\Rightarrow$  $\lambda(6 \widehat{j}-6 \widehat{k}) \Rightarrow  6 \lambda (\widehat{j}-\widehat{k})$

 for  $\lambda=\frac{1}{6} \Rightarrow$ Option (a) is correct

for $\lambda=-\frac{1}{6} \Rightarrow$ Option (d) is correct

Let the eccentricity of the hyperbola  $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$ =1 be reciprocal to that of the ellipse x² + 4y² = 4. If the hyperbola passes through a focus of the ellipse, then

Answer:

Explanation:

 Here , equation of ellipse

$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$

 $\Rightarrow$   $e^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{1}{4}=\frac{3}{4}$

 $\therefore$   $e= \frac{\sqrt{3}}{2}$ and focus $(\pm ae,0)$

  $= ( \pm \sqrt{3},0)$

 For hyperbola    $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

 $e_{1}^{2}=1+\frac{b^{2}}{a^{2}}$  where, $e_{1}^{2}=\frac{1}{e^{2}}=\frac{4}{3}$

$\Rightarrow$  $1+ \frac{b^{2}}{a^{2}}=\frac{4}{3} \Rightarrow \frac{b^{2}}{a^{2}}=\frac{1}{3}$....(i)

and hyperbola passes through $(\pm \sqrt{3},0)$

Now, $\frac{3}{a^{2}}=1 \Rightarrow  a^{2}=3$  ....(ii)

 from eqs. (i) and (ii) , we get

 $b^{2}=1$  ......(iiii)

 $\therefore$ Equation of hyperbola is 

$\frac{x^{2}}{3}-\frac{y^{2}}{1}=1$

 Focus is $(\pm ae,0)$

Now, $\left(\pm \sqrt{3}.\frac{2}{\sqrt{3}},0\right) \Rightarrow  (\pm2,0)$

 hence , both options (b) and (d)  are correct

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