JEE 2017 :: Main 2017 :: Mathematics 2017

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• Main 2017 - Mathematics 2017

If 20 m of wire is available for fencing off a flower -bed in the form of a circular sector, then the maximum area ( in sq m) of the flower-bed is 

Answer:

Explanation:

   Total length = 2r+rθ =20

       $\Rightarrow \theta =\frac{20-2r}{r}$

 Now, area of flower-bed,

          $A= \frac{1}{2}r^{2}\theta \Rightarrow A =\frac{1}{2}r^{2}(\frac{20-2r)}{r}$

  $\Rightarrow A=10r-r^{2}$

  $\therefore  \frac{\text{d}A}{\text{d}r}=10-2r$

For maxima or minima , put  $\therefore  \frac{\text{d}A}{\text{d}r}=0$

$\therefore 10-2r=0\Rightarrow r=5$

  $\therefore  A_{max}= \frac{1}{2}(5)^{2}[\frac{20-2(5)}{5}]$

               $        = \frac{1}{2}\times25\times2=25 sq.m$

For three events  A. B and C. If P ( exactly one of A or B occurs)= P (exactly one of B or C occurs)= $\frac{1}{4}$  and P (all the three events occurs simultaneously )= $\frac{1}{16}$ , then the probability that atleast one of the events occurs, is

Answer:

Explanation:

We have P (exactly one of A or B occurs)

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

=$=P(A)+P(B)-2P(A\cap B)$

   According to the question,

     $P(A)+P(B)-2P(A\cap B)=\frac{1}{4}$   .........(i)

     $P(B)+P(C)-2P(B\cap C)=\frac{1}{4}$   .........(ii)

and     $P(C)+P(A)-2P(C\cap A)=\frac{1}{4}$     ........(iii)

 On adding Eqs(i),(ii) and (iii) , we get

    $2[P(A)+P(B)+P(C)-P(A\cap B)-P (B\cap C)-P(C\cap A)]=\frac{3}{4}$

  $\Rightarrow P(A)+P(B)+P(C)-P(A\cap B)-P (B\cap C)-P(C\cap A)=\frac{3}{8}$

  P (atleast one event occurs)

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

                                                  $+P(A\cap B\cap C)=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}$

                               $[ \because P(A\cap B\cap C=\frac{1}{16})]$

• Main 2017 - Physics 2017
• Main 2017 - Chemistry 2017
• Main 2017 - Mathematics 2017