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At $x=\Large\frac{5}{6}$$,f(x)=2\sin 3x+3\cos 3x$ is:

\begin{array}{1 1}(A)\;maximum & (B)\;minimum\\(C)\;zero & (D)\;neither\;maximum\;nor\;minimum\end{array}

1 Answer

  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $f'(x)$ and put $f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$f(x)=2\sin 3x+3 \cos 3x$
Differentiating w.r.t $x$ we get,
$f'(x)=2(3 \cos3x)+3(-3 \sin 3x)$
$ = 6 \cos 3x-9 \sin 3x$
Differentiating again w.r.t $x$ we get,
$ f''(x)=6(-3 \sin 3x)-9(3 \cos 3x)$
$ = -18 \sin 3x-27 \cos 3x$
$ \Rightarrow -3 (6 \sin 3x+9 \cos x) < 0$
Since $f''(x)<0$, the function has maximum value.
Hence A is the correct option.
answered Aug 12, 2013 by thanvigandhi_1