In the following figure, the graphs of two functions y = f(x) and y = sin x are givne. They intersect at origin, `A(a,f(a)),B(pi,0) and C(2pi,0).A_(i)(i=1,2,3)` is the area bounded by the curves as shown in the figure, respectively, for `x in (0,a), x in (a, pi), x in (pi,2pi)`. If `A_(1)=1+(a-1)cosa - sin` a, then The function f(x) is
A
`x^(2) sinx`
B
`xsinx`
C
`2x sinx`
D
`x^(3)sinx`
Text Solution
Verified by Experts
The correct Answer is:
B
`A_(1)=int_(0)^(a)(sinx-f(x))dx=1+(a-1)cosa-sina` Differentiating w.r.t. 'a' both sides. `therefore" "sina-f(a)=cosa-(a-1)sin a -cosa` `therefore" "f(a)=asina` `therefore" "f(x)=x sin x` `"Solving y = sin x wity y = x sin x, we have x sin x = sin x"` `therefore" "x=1 or x=pi,2pi` `therefore" "A_(2)=int_(1)^(pi)(x sin x-sinx)dx` `=(-x cosx)_(1)^(pi)-int_(1)^(pi)-cosxdx+(cosx)_(1)^(pi)` `=(pi+cos1)+(0-sin1)+(-1-cos1)` `=(pi-sin1-1)"sq. units"`
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