A) \[-\,1\]
B) \[0\]
C) \[1\]
D) \[2\]
Correct Answer: A
Solution :
[a]| We have, |
| \[y=\frac{{{x}^{2}}}{8}+x\cos x+\cos 2x\] |
| \[\Rightarrow \]\[y=\frac{1}{8}[{{x}^{2}}+8x\cos x+8\cos 2x]\] |
| \[\Rightarrow \]\[y=\frac{1}{8}[{{x}^{2}}+8x\cos x+4{{\cos }^{2}}x+8\,(2{{\cos }^{2}}x-1)\]\[-\,4{{\cos }^{2}}x]\] |
| \[\Rightarrow \]\[y=\frac{1}{8}[{{(x+2\cos x)}^{2}}+12{{\cos }^{2}}x-8]\] |
| \[\Rightarrow \]\[y=\frac{1}{8}{{(x+2\cos x)}^{2}}+\frac{1}{8}(12\,{{\cos }^{2}}x-8)\] |
| \[\therefore \]Minimum value of y is \[\frac{1}{8}(0-8)=-\,1\] |
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