A) \[\frac{{{\pi }^{5}}}{32}-\frac{{{\pi }^{4}}}{64}+\frac{{{\pi }^{3}}}{32}+1\]
B) \[\frac{{{\pi }^{5}}}{16}-\frac{{{\pi }^{4}}}{32}+\frac{{{\pi }^{3}}}{24}-1\]
C) \[\frac{{{\pi }^{5}}}{32}-\frac{{{\pi }^{4}}}{32}+\frac{{{\pi }^{3}}}{16}\]
D) \[\frac{{{\pi }^{5}}}{32}-\frac{{{\pi }^{4}}}{32}+\frac{{{\pi }^{3}}}{24}+1\]
Correct Answer: D
Solution :
[d] \[f(x)={{x}^{2}}+\cos x\] \[\Rightarrow f'(x)=2x-sinx\] \[\Rightarrow f'\left( \frac{\pi }{2} \right)=\pi -1\] So, equation of normal at \[x=\frac{\pi }{2}\]is \[\left( y-\frac{{{\pi }^{2}}}{4} \right)=\frac{1}{1-\pi }\left( x-\frac{\pi }{2} \right)\] It meets x-axis at \[x=\frac{(\pi -1){{\pi }^{2}}}{4}+\frac{\pi }{2}\] Also, \[f'(x)=2x-\sin \,x>0\] for \[x>0\] And \[f'(x)=2x-\sin x<0\] for \[x<0\] So, \[f(x)\] increases for \[x\in (0,\infty )\]and decreases for \[x\in (-\infty ,0).\] Graph of the function is as shown in the following figure.You need to login to perform this action.
You will be redirected in
3 sec