# Applications of Derivatives Set 12

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Example - 26

Find the minimum value of

$f(x) = |\sin x + \cos x + \tan x + \cot x + \sec x + {\rm{cosec}}\;x\;|$

Solution:         \begin{align}f(x) = \left| {\;\sin x + \cos x + \frac{1}{{\sin x\cos x}} + \frac{{\sin x + \cos x}}{{\sin x\cos x}}\;} \right|\end{align}

Using $$\sin x + \cos x = t,$$ and therefore $$\sin x\cos x = \frac{{{t^2} - 1}}{2}$$, we have

\begin{align}&f(t) = \left| {\;t + \frac{{2(1 + t)}}{{{t^2} - 1}}\;} \right| = \left| {\;t + \frac{2}{{t - 1}}\;} \right|\\\\&\,\,\,\,\,\,\,\,\,\,\, = \left| {\;(t - 1) + \frac{2}{{t - 1}}\; + 1\;} \right|\\\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \le 2\sqrt 2 + 1\end{align}

The maximum value is $$2\sqrt 2 + 1$$

Example - 27

Let $$f(x) = \left( {{x^2} + bx + c} \right){e^x}$$

(a) Does $$f > 0\;{\rm{imply}}\;f' > 0$$             (b) Does $$f' > 0\;{\rm{imply}}\;f > 0$$

Solution:         We have \begin{align}f'(x) = \left( {{x^2} + (b + 2)x + b + c} \right){e^x}\end{align}. Thus,

$f > 0 \Leftrightarrow {b^2} - 4c < 0\;\;{\rm{and}}\;\;f' > 0\;\;\; \Leftrightarrow \;\;\;{b^2} - 4c + 4 < 0$

(a) Let$$f > 0$$. Thus,$${b^2} - 4c < 0$$, but nothing can be said about the sign of $${b^2} - 4c + 4$$, i.e., it may be positive $$f > 0$$ Thus,  does not necessarily imply  $$f' > 0$$.

(b) Let  $$f > 0.$$ Thus,$${b^2} - 4c + 4 < 0$$, which necessarily implies that $${b^2} - 4c < 0,\;{\rm{i}}{\rm{.e}}.,\;f > 0$$.

Example - 28

Find the minimum value of $$|f(x)|$$ given by

$f(x) = \frac{{{{\left( {x + \frac{1}{x}} \right)}^6} - \left( {{x^6} + \frac{1}{{{x^6}}}} \right) - 2}}{{{{\left( {x + \frac{1}{x}} \right)}^3} + \left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}$

Solution:         Substitute \begin{align}{\left( {x + \frac{1}{x}} \right)^3} = a\;\;{\rm{and}}\;\;{x^3} + \frac{1}{{{x^3}}} = b\end{align}. Thus,

\begin{align}&f(x) = \left| {\;\frac{{{a^2} - {b^2}}}{{a + b}}\;} \right| = \;|a - b|\; = \left| {\;{{\left( {x + \frac{1}{x}} \right)}^3} - {x^3} - \frac{1}{{{x^3}}}\;} \right|\\\\&\,\,\,\,\,\,\,\,\,\,\, = 3\left| {\;x + \frac{1}{x}\;} \right| \ge 6\end{align}

The minimum value of $$|f(x)|$$is 6.

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