Let f (n, x) = ∫ ncos (nx)dx, with f (n, 0) = 0. If the expression ∑ _ x = 1 ^ 89 f(1, x) simplifies to (sin asin b)/(sin c), then the value of (b)/(ac) is (where a > b )

Correct answer: C. (89)/(45) f(n, x) = (nsin (nx))/(n) + C As f(n, 0) = C = 0 ⇒ f(n, x) = (nx) Thus, ∑ _ x = 1 ^ 89 f(1, x) = 1 + 2 + . . . . . + 89 = ((1 + 89)/(2)) sin (89 × 1)/(2) ((1)/(2)) =( (45) ((89)/(2)))/( ((1)/(2)))

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Let f (n, x) = ∫ ncos (nx)dx, with f (n, 0) = 0. If the expression...

Let

f

\left(n, x\right) = ∫ ncos

\left(nx\right)dx,

with

f \left(n, 0\right) = 0.

If the expression

\mathop{∑}_{x = 1}^{89}f\left(1, x\right)

simplifies to

\dfrac{sin asin b}{sin c},

then the value of

\dfrac{b}{ac}

is (where

a > b

)

Options

A $45$
B $89$
C $\dfrac{89}{45}$
D $\dfrac{45}{89}$

Correct answer

C. $\dfrac{89}{45}$

Step-by-step solution

f\left(n, x\right) = \dfrac{nsin \left(nx\right)}{n} + C

f\left(n, 0\right) = C = 0 ⇒ f\left(n, x\right) = \sin \left(nx\right)

Thus,

\mathop{∑}_{x = 1}^{89}f\left(1, x\right) = \sin 1 + \sin 2 + . . . . . + \sin 89 = \dfrac{\sin \left(\dfrac{1 + 89}{2}\right)\text{ sin }\dfrac{89 \times 1}{2}}

{\sin \left(\dfrac{1}{2}\right)}=\dfrac{\sin \left(45\right)\sin \left(\dfrac{89}{2}\right)}{\sin \left(\dfrac{1}{2}\right)}

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