# Applications of Derivatives Set 3

Go back to  'SOLVED EXAMPLES'

Example - 6

Find the values of a for which the function\begin{align}f\left( x \right) = \sin x - a\sin 2x - \frac{1}{3}\sin 3x + 2ax\,\,{\text{increases}}\,\,{\text{on}}\,\,\mathbb{R}\end{align}

Solution: We want $$f'\left( x \right) \ge \,0\,\forall x \in \mathbb{R}$$

\begin{align}&f'\left( x \right) = \cos x - 2a\cos 2x - \cos 3x + 2a\\\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \cos x - 2a\left( {2{{\cos }^2}x - 1} \right) - \left( {4{{\cos }^3}x - 3\cos x} \right) + 2a\\\\&\,\,\,\,\,\,\,\,\,\,\,\;\, = 4a + 4\cos x - 4a{\cos ^2}x - 4{\cos ^3}x\\\\&\,\,\,\,\,\,\,\,\,\,\,\;\, = 4a{\sin ^2}x + 4\cos x{\sin ^2}x\\\\&\,\,\,\,\,\,\,\,\,\,\;\,\, = 4{\sin ^2}x\left( {a + \cos x} \right)\end{align}.

This is always non-negative  $$a \ge 1$$  if (since the minimum value of cos x is –1).

Therefore, the required values of $$a$$ are:

$$a \in \left[ {1,\infty } \right)$$

Example - 7

Determine the points of maxima and minima of the function \begin{align} f(x)=\frac{1}{8}\ln x-bx+{{x}^{2}},x>0\end{align}  where  $$b \ge 0$$ is a constant.

Solution: We need to evaluate the sign of $$f'\left( x \right)$$in different intervals of x:

\begin{align}&f'\left( x \right) = \frac{1}{{8x}} - b + 2x\\\\&\qquad \;\;= \frac{{16{x^2} - 8bx + 1}}{{8x}} = \frac{{g\left( x \right)}}{{8x}}\end{align}

The denominator of $$f'\left( x \right)$$ is always positive since $$x > 0$$. Hence, the sign of $$f'\left( x \right)$$ will depend on the sign of the numerator g(x) in various intervals, which will in turn depend on $$b$$.

\begin{align}&\qquad\;\, g\left( x \right)=16{{x}^{2}}-8bx+1\qquad\qquad ........ (i) \\\\ & \text{D of g}\left( \text{x} \right)=64{{b}^{2}}-64 \\\\ & \qquad\qquad\;=64\left( {{b}^{2}}-1 \right) \\ \end{align}

\begin{align}\boxed{0\le b<1} & \Rightarrow\quad D<0 \\\\ & \Rightarrow \quad g\left( x \right)>0\,\,\ \ \ \ \ \forall \,\,x\in \,\,\mathbb{R} \\\\ & \Rightarrow \quad f'\left( x \right)>0\,\,\ \ \ \ \forall \,\,x\in \,\,\mathbb{R} \\\\ & \Rightarrow \quad f(x)\text{ is strictly increasing}\,\,\forall x\in \mathbb{R} \\ \end{align}

\begin{align}\boxed{b=1}\quad& \Rightarrow\quad \,g\left( x \right)=16{{x}^{2}}-8x+1\qquad\qquad\qquad\quad\left( from(i) \right) \\\\ & \qquad\qquad\;\;\;={{\left( 4x-1 \right)}^{2}} \\\\ & \Rightarrow\quad f'\left( x \right)=\frac{{{\left( 4x-1 \right)}^{2}}}{8x}\ge 0 \\\\ & \Rightarrow\quad f'\left( x \right)=0\,\,\text{for}\,\,x=\frac{1}{4} \\\\ & \Rightarrow \quad{{\left. f''\left( x \right) \right|}_{x=\frac{1}{4}}}=0\,\,\text{and}\,{{\left. f'''\left( x \right) \right|}_{x=\frac{1}{4}}}=16\ne 0\,\,\,\left( Verify \right) \\\\ & \Rightarrow \quad x=\frac{1}{4}\,\text{is point of inflexion} \\\\ & \Rightarrow\quad f\left( x \right)\,\,\,\text{has no local maxima or minima} \\ \end{align}

\begin{align}\boxed{b>1}\quad & \Rightarrow \quad D>0 \\\\ & \Rightarrow\quad \,g\left( x \right)\,\left( \text{and}\,f'\left( x \right) \right)\text{has roots}\frac{b\pm \sqrt{{{b}^{2}}-1}}{4} \\\\ & \Rightarrow \quad f'\left( x \right)>0\,if\,x<\frac{b-\sqrt{{{b}^{2}}-1}}{4}\,\,\text{or}\,\,x>\frac{b+\sqrt{{{b}^{2}}-}}{4} \\\\ & and\,\,f'\left( x \right)<0\,\,\;\text{if}\,\;\frac{b-\sqrt{{{b}^{2}}-1}}{4}<\,\,x<\frac{b+\sqrt{{{b}^{2}}-1}}{4} \\ \end{align}

\begin{align} & \Rightarrow\quad \,f'\left( x \right)\text{ changes sign from +ve to } \\\\ & \qquad\quad x=\frac{b-\sqrt{{{b}^{2}}-1}}{4}\,\,\,\text{and from -ve to +ve in the neighbourhood of} \\\\ & \qquad\quad x=\frac{b-\sqrt{{{b}^{2}}-1}}{4} \\\\ & \Rightarrow\quad x=\frac{b-\sqrt{{{b}^{2}}-1}}{4}\text{is a local maximum for }f(x)\text{ and}\,\,\,x=\frac{b+\sqrt{{{b}^{2}}-1}}{4}\,\,\,\text{is a local minimum for }f(x) \\ \end{align}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school