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in Complex number and Quadratic equations by (25 points)
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Roots of a quadratic equation \( x^{2}+5 x+3=0 \) are \( 4 \cos ^{2} \alpha+a \) and \( 4 \sin ^{2} \alpha+a \). Another quadratic equation is given as \( x ^{2}+ px + q =0 \) where \( p , q \in N \) and \( p , q \in[1,10] \). If roots of second quadratic equation are real then the probability that they are \( 4 \cos ^{4} \alpha+b \) and \( 4 \sin ^{4} \alpha+b \) is 

a) \( \frac{1}{16} \) 

b) \( \frac{3}{32} \) 

c) \( \frac{1}{32} \) 

d) \( \frac{5}{100} \)

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1 Answer

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by (40.5k points)

Roots of quadratic equation x2 + 5x + 3 = 0

are 4 cos2 α  + a and 4 sinα  + a

\(\therefore\) sum of roots = \(\cfrac{-b}a\) = -5

= (4 cos2 α + a) + (4 sinα +a) = - 5

4 (sinα + cos2 α) + 2a = - 5

= 2a + 4 = -5

= 2a = -5 - 4 = -9

= a = \(\cfrac{-9}2\)....(1)

product of roots = \(\cfrac{c}a\) = 3

= (4cos2 α + a) (4 sinα + a) = 3

= 16 sin2 α cos2 α + 4a (sin2 α + cos2 α) + a2 = 3

= 4(2 sin α cos α)2 + 4 x \(\cfrac{-9}2\) + \(\left(\cfrac{-9}2\right)^2\) = 3 (From (1))

4 sin2 2 α - 18 + \(\cfrac{81}4\) = 3

= 4 sin2 2 α = 21 - \(\cfrac{81}4\) = \(\cfrac{84-81}4\)

\(\cfrac34\)

= sin2 2 α = \(\cfrac{3}{16}\)

= sin 2 α = \(\cfrac{\sqrt3}{4}\)....(2)

Second quadratic equation is x2 + px + q = 0,

p, q ε N and p,q ε (1,10)

Total possible value for p and q = 8 x 8 = 64

sum of roots = (4 cos4 α + b) + (4 sinα + b) = -p

= -p = 4(sin4 α + cos4 α) + 2b

= -p = 4 ((sin2 α)2 + (cos2 α)+ 2sincos2 α - 2 sin2 α cos2 α) + 2b

= - p = 4 ((sin2 α + cos2 α)- 2 sincos2 α) + 2b

 = - p = 4 (1 - \(\cfrac12\)(2sin α cos α)2) + 2b

= - p = 4 (1 - \(\cfrac12\)sin2 2 α) + 2b 

= - p = 4 (1 - \(\cfrac12\) x \(\cfrac{3}{16}\)) + 2b  (From (2))

= - p = 4 - \(\cfrac{3}{8}\) + 2b

= 2b = \(\cfrac{3}{8}\) - 4 - p

= 2b = \(\cfrac{3-32-8p}8\)

= b = \(\cfrac{-29-8p}{16}\)....(3)

product of roots = (4cos4 α + b)(4sin4 α + b) = q

= q = 16 cos4 α sin4 α + 4b(sin4 α + cos4 α) + b2

= q = (2 sin α co α)4 + 4b ((sin2 α + cos2 α)2 - 2sinα cos2 α) +b2

= (sin 2α)4 + 4b (1 - \(\cfrac12\)(sin 2α)2) + b2

\(\left(\cfrac{\sqrt3}4\right)^4\) + 4b  (1 - \(\cfrac12\)  x \(\cfrac{3}{16}\)) + b(From (3))

\(\cfrac{9}{256}\) + \(\cfrac{29\,b}8\) + b2

\(\cfrac{9}{256}\) + \(\cfrac{29}8\) x - \(\cfrac{(29+8p)}{16}\) + \(\left(\cfrac{-(29+8p)}{16}\right)^2\)

\(\cfrac{9}{256}\) - \(\cfrac{841+232p)}{128}\) + \(\cfrac{841+464p+64p^2}{256}\)

\(\cfrac{9-1682-464p+841+464p+64p^2}{256}\)

\(\cfrac{64p^2-832}{256}\) = \(\cfrac{64(p^2-13)}{256}\)

= q = \(\cfrac{p^2-13}{4}\)....(4)

\(\because\) p and q are natural number such that p,q ε [1,10]

For possible values of p and q are

(p,q) = (5,3), (7,9)

\(\therefore\) Favorable outcomes = 2

\(\therefore\) Required probability = \(\cfrac{2}{64}\) = \(\cfrac1{32}\)

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