Question

Let $\sf{\sum\limits_{m=1}^\infty}y^{2m-1}=1}$ and y = 2 cosθ. If least value of cosθ is of the form $\sf{\dfrac{\sqrt{a}-b}{c} }$ (where a is a prime number), then the value of (a + c - b) is? Class 11 ↑ DON'T SPAM ✖

Answer 1

Basic question of series

Solution:- According to question

y+y^3+y^5+y^7+.........+y^(infinite)=1

take y common from LHS

y{1+y^2+y^4+y^6+.......+y^(infinite)}=1

y[1+{y^2+y^4+y^6+.......+y^(infinite)}]=1

we know that sum of infinite terms of GP=a/(1-r)

y(1+y^2/1-y^2)=1

y(1-y^2+y^2)/(1-y^2)=1

y/(1-y^2)=1

y=1-y^2

y^2+y-1=0

y={-1+-(1+4)^1/2}/2 (using quadratic rule)

y=(√5-1)/2 or (-√5-1)/2

putting y=2cos∅ we get

2cos∅=(√5 -1)/2

cos∅=(√5-1)/4

also it is given that cos∅=(√a -b)/c

so

(√a-b)/c=(√5-1)/4

comparing both side we get

a=5, b=1 and c=4

now,

a+c-b=5+4-1=8

This will be in the case of maximum value

In case of minimum value

2cos∅=(-√5-1)/2

cos∅=(√5+1)/-4

comparing this value with (√a -b)/c

a=5, b=-1 and c=-4

a+b-c=5-(-4)-(-1)=2

Answer 2

GIVEN :–

$\\ \: \: \: {\huge{.}} \sf \: \: \: \sf{\sum\limits_{m=1}^\infty}y^{2m-1}=1$

$\\ \: \: \: {\huge{.}} \sf \: \: \: y=2 \cos( \theta)$

$\\ \: \: \: {\huge{.}} \sf \: \: \: Least \: \: value \: \: of \: \: \cos \theta \:\:is = {\dfrac{\sqrt{a}-b}{c} }$

TO FIND :–

• Value of (a + c - b) = ?

SOLUTION :–

$\\ \implies \sf{\sum\limits_{m=1}^\infty}(y)^{2m-1}=1$

• We should write this as –

$\\ \implies \sf y + {y}^{3} + {y}^{5} + ........ \infty =1$

• It's a G.P. series And we know that sum of infinite G.P. is –

$\\ \dashrightarrow \large{ \boxed{ \sf sum = \dfrac{a}{(1 - r)}}}$

• Here –

$\\ \: \: \: {\huge{.}} \sf \: \: \: a=y$

$\\ \: \: \: {\huge{.}} \sf \: \: \: r= {y}^{2}$

• So that –

$\\ \implies \sf \dfrac{y}{1 - {y}^{2} } =1$

$\\ \implies \sf y = 1 - {y}^{2}$

$\\ \implies \sf {y}^{2} + y - 1 = 0$

• We can find the value of 'y' by using quadratic formula –

$\\ \implies \sf y = \dfrac{ - (1) \pm \sqrt{ {(1)}^{2} - 4(1)( - 1)} }{2(1)}$

$\\ \implies \sf y = \dfrac{ - 1 \pm \sqrt{1+ 4} }{2}$

$\\ \implies \sf y = \dfrac{ - 1 \pm \sqrt{5} }{2}$

• Hence , Least value of 'cosθ' is –

$\\ \implies \sf 2 \cos \theta = \dfrac{ - 1 - \sqrt{5} }{2}$

• We should write this as –

$\\ \implies \sf \cos \theta = \dfrac{ \sqrt{5} + 1 }{ - 4}$

• Now compare –

$\\ \: \: \: {\huge{.}} \sf \: \: \: a=5$

$\\ \: \: \: {\huge{.}} \sf \: \: \: b=-1$

$\\ \: \: \: {\huge{.}} \sf \: \: \: c=-4$

• So that –

$\\ \sf \implies a + c - b=5 + ( - 4) - ( - 1)$

$\\ \sf \implies a + c - b=5 - 4 + 1$

$\\\implies \large{ \boxed{ \sf a + c - b=2}}$