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.

Jitender Singh IIT Delhi
9 years ago
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 minima
Let
$f^{''}(0) = 2a < 0$
$a< 0$
$f^{''}(sin^{-1}\frac{-2a}{3}) = \frac{8}{9}a^{3}-2a > 0$
$a(a-\frac{3}{2})(a+\frac{3}{2})> 0$
$\Rightarrow \frac{-3}{2}< a< 0, a> \frac{3}{2}$
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 & Regards
Jitender Singh
IIT Delhi