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iit jee Grade: 12
        


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

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
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 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
askIITians Faculty
3 years ago
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