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Question: If \({x_i} > 0,\)for \(1 \leqslant i \leqslant n\)and \({x_1} + {x_2} + ..... + {x_n} = \pi \) then ...

If xi>0,{x_i} > 0,for 1⩽i⩽n1 \leqslant i \leqslant nand x1+x2+.....+xn=π{x_1} + {x_2} + ..... + {x_n} = \pi then the greatest value of the sumsin⁡xi+sin⁡x2+....+sin⁡xn=......\sin {x_i} + \sin {x_2} + .... + \sin {x_n} = ......
A. n
B. π\pi
C. n sin(πn)\left( {\dfrac{\pi }{n}} \right)
D. 0

Explanation

Solution

HINT- We can solve this question by applying Jensen's inequality for concave downward function. Johan Jensen is the first person who described mathematical rule and the mathematical rule is also known generally as Jensen's Inequality. First, we identify the nature of the f(x)=sin⁡xf(x) = \sin xcurve and then Applying Jensen's inequality in the interval (0, π).

Complete step-by-step solution -
We need to calculate,
greatest value of the sum
sin⁡xi+sin⁡x2+....+sin⁡xn\sin {x_i} + \sin {x_2} + .... + \sin {x_n}
And as given in question,
xi>0,1⩽i⩽n{x_i} > 0,1 \leqslant i \leqslant n and x1+x2+x3+....+xn=π{x_1} + {x_2} + {x_3} + .... + {x_n} = \pi
∴xi⩽π\therefore {x_i} \leqslant \pi for all 1⩽i⩽n1 \leqslant i \leqslant n
f(x)=sin⁡xf(x) = \sin x
f′(x)=cos⁡xf'(x) = \cos x
f′′(x)=−sin⁡xf''(x) = - \sin x
In the interval (0, π) and f′′(x)<0f''(x) < 0
So, the graph of f(x)=sin⁡xf(x) = \sin xis concave downward graph
For this question we are Applying Jensen's inequality for concave downward function we have
f(x1+x2+x3+....xnn)⩾f\left( {\dfrac{{{x_1} + {x_2} + {x_3} + ....{x_n}}}{n}} \right) \geqslant f(x1)+f(x2)+f(x3)+......+f(xn)n\dfrac{{f({x_1}) + f({x_2}) + f({x_3}) + ...... + f({x_n})}}{n}
Since f(x)=sin⁡rf(x) = \sin r
sin⁡(x1+x2+x3+....xnn)⩾\sin \left( {\dfrac{{{x_1} + {x_2} + {x_3} + ....{x_n}}}{n}} \right) \geqslant sin⁡x1+sin⁡x2+sin⁡x3+......sin⁡xnn\dfrac{{\sin {x_1} + \sin {x_2} + \sin {x_3} + ......\sin {x_n}}}{n}
sin⁡(πn)⩾\sin \left( {\dfrac{\pi }{n}} \right) \geqslant sin⁡x1+sin⁡x2+sin⁡x3+......sin⁡xnn\dfrac{{\sin {x_1} + \sin {x_2} + \sin {x_3} + ......\sin {x_n}}}{n}
since x1+x2+x3+......+xn=π{x_1} + {x_2} + {x_3} + ...... + {x_n} = \pi
sin⁡x1+sin⁡x2+sin⁡x3+.......sin⁡xn⩽nsin⁡(πn)\sin {x_1} + \sin {x_2} + \sin {x_3} + .......\sin {x_n} \leqslant n\sin \left( {\dfrac{\pi }{n}} \right)
Therefore, the greatest value of the sumsin⁡xi+sin⁡x2+....+sin⁡xn\sin {x_i} + \sin {x_2} + .... + \sin {x_n} is n sin(πn)\left( {\dfrac{\pi }{n}} \right).
Thus, option (C) is the correct answer.

Note- We need to remember, In the interval (0, π), the graph of f(x)=sin⁡xf(x) = \sin x is a concave downward graph. While In the interval (π, 2 π), the graph of f(x)=sin⁡xf(x) = \sin x is a concave upward graph. And also, in this question, students should be careful while considering all the cases which add up to n as missing out any change gives an overall change in the result. and also avoid basic calculations mistakes during calculations.