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Let f(x) = sin (cube) x + a sin (square) x,                      [for –pi/2<x<pi/2]. Find the interval in which a should lie in order that f(x) has exactly one minimum and one maximum.

7 years ago

Jitender Singh
IIT Delhi
158 Points
										Ans:$f(x)=sin^{3}x+asin^{2}x$Lets 1stfind the crical points:$f^{'}(x) = 0$$f^{'}(x)=3sin^{2}xcosx+2asinxcosx=0$$sinx.cosx(3sinx+2a) = 0$$sin2x = 0, sinx = \frac{-2a}{3}$$x= 0, sin^{-1}\frac{-2a}{3}$$f^{''}(x) = -3sin^{3}x + 6sinx.cos^{2}x+2acos2x$$f^{''}(0) = -3.0 + 6.0.1+2a.1 = 2a$$f^{''}(sin^{-1}\frac{-2a}{3}) = -3(\frac{-2a}{3})^{3} + 6.\frac{-2a}{3}.(1-\frac{4a^{2}}{9})+2a(1-\frac{8a^{2}}{9})$$f^{''}(sin^{-1}\frac{-2a}{3}) = \frac{8}{9}a^{3}-2a$Both of these values should have opposite sign for exactly one maxima & one minimaLet$f^{''}(0) = 2a < 0$$a< 0$$f^{''}(sin^{-1}\frac{-2a}{3}) = \frac{8}{9}a^{3}-2a > 0$$a(a-\frac{3}{2})(a+\frac{3}{2})> 0$$\Rightarrow \frac{-3}{2}< a< 0, a> \frac{3}{2}$So the combined solution is$\frac{-3}{2}< a< 0$Lets$f^{''}(0)=2a> 0$$a> 0$$f^{''}(sin^{-1}\frac{-2a}{3})=a(a-\frac{3}{2})(a+\frac{3}{2})< 0$$\Rightarrow 0< a< \frac{3}{2}, a< \frac{-3}{2}$So the combined solution is$0< a< \frac{3}{2}$So ‘a’ should lie in$(\frac{-3}{2},0)\bigcup (0, \frac{3}{2})$Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
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