Let f(n , x)= displaystyle ∫ ncos(n x)dx, with f(n , 0)=0. If the expression displaystyle ∑ x = 189 f (1 , x) simplifies to (sin a sin â¡ b/sin â¡ c), then the value of (b/a c) is (where ab )

Question

Let f(n , x)= displaystyle ∫ ncos(n x)dx, with f(n , 0)=0. If the expression displaystyle ∑ x = 189 f (1 , x) simplifies to (sin a sin â¡ b/sin â¡ c), then the value of (b/a c) is (where a>b ) (A) 45 (B) 89 (C) (89/45) (D) (45/89)

Tardigrade Admin · Accepted Answer

f(n , x)=(n s i n (n x)/n)+C As f(n , 0)=C=0 ⇒ f(n , x)=sin(n x) Thus, displaystyle ∑ x = 189 f (1 , x) = sin 1 + sin â¡ 2 + . . . . . + sin â¡ 89 =(s i n ((1 + 89/2)) s i n (89 × 1/2)/s i n ((1)2)) =(s i n (45) s i n ((89/2))/s i n ((1)2))

Q. Let $f (n, x) = \int n cos (n x) d x,$ with $f (n, 0) = 0.$ If the expression $x = 1 \sum 89 f (1, x)$ simplifies to $\frac{s ina s in b}{s in c},$ then the value of $\frac{b}{a c}$ is (where $a > b$ )

$45$

$89$

$\frac{89}{45}$

$\frac{45}{89}$

Q. Let f(n,x)=∫ncos(nx)dx, with f(n,0)=0. If the expression x=1∑89​f(1,x) simplifies to sin⁡csinasin⁡b​, then the value of acb​ is (where a>b )

45

89

4589​

8945​

f(n,x)=nnsin(nx)​+C As f(n,0)=C=0 ⇒f(n,x)=sin(nx) Thus, x=1∑89​f(1,x)=sin1+sin⁡2+.....+sin⁡89 =sin(21​)sin(21+89​)sin289×1​​ =sin(21​)sin(45)sin(289​)​

Q. Let $f (n, x) = \int n cos (n x) d x,$ with $f (n, 0) = 0.$ If the expression $x = 1 \sum 89 f (1, x)$ simplifies to $\frac{s ina s in b}{s in c},$ then the value of $\frac{b}{a c}$ is (where $a > b$ )

$45$

$89$

$\frac{89}{45}$

$\frac{45}{89}$