Applications of Derivatives- EXERCISE 6.2

Chapter 6 Applications of Derivatives

EXERCISE 6.2

Question 1: how that the function given by f(x) = 3x + 17 is strictly increasing on R.

Solution:

Let x₁ and x₂ be any two numbers in R.

Then,

$\displaystyle {{x}_{1}}<{{x}_{2}}$

$\displaystyle \Rightarrow 3{{x}_{1}}+17<3{{x}_{2}}+17$

$\displaystyle \Rightarrow f({{x}_{1}})<f({{x}_{2}})$

Thus, f is strictly increasing on R.

Question 2: Show that the function given by f(x) = e²^x is strictly increasing on R.

Solution:

Let x₁ and x₂ be any two numbers in R.

Then,

x₁< x₂= 2x₁< 2x₂ ⟹ e²^x¹ < e²^x² = f(x₁) < f(x₂)

Thus, f is strictly increasing on R.

Question 3: Show that the function given by f(x) = sin x is

Strictly increasing in $\displaystyle \left( {0,\frac{\pi }{2}} \right)$

(b) Strictly decreasing in $\displaystyle \left( {\frac{\pi }{2},\pi } \right)$

(c) Neither increasing nor decreasing in $\displaystyle \left( {0,\pi } \right)$

Solution:

It is given that f(x) = sin x

Hence, $\displaystyle {{f}^{'}}\left( x \right)=\cos x$

(a) Here, $\displaystyle x\in \left( {0,\frac{\pi }{2}} \right)$

⟹ cos x > 0

⟹ $\displaystyle {{f}^{'}}\left( x \right)>0$

Thus, f is strictly increasing in $\displaystyle \left( {0,\frac{\pi }{2}} \right)$ .

Here, $\displaystyle x\in \left( {\frac{\pi }{2},\pi } \right)$

⟹ cos x < 0

⟹ $\displaystyle {{f}^{'}}(x)<0$

Thus, f is strictly decreasing in $\displaystyle \left( {\frac{\pi }{2},\pi } \right)$ .

Here, x $\displaystyle \in$ (0,π)

The results obtained in (a) and (b) are sufficient to state that f is neither increasing nor

decreasing in (0, π).

Question 4: Find the intervals in which the function f given by f(x) = 2x² – 3x is

(a) Strictly increasing (b) Strictly decreasing

Solution:

The given function is f(x) = 2x² – 3x

Hence, $\displaystyle {{f}^{'}}\left( x \right)=4x-3$

Therefore, $\displaystyle {{f}^{'}}\left( x \right)=0$

⟹ x = $\displaystyle \frac{3}{4}$

In $\displaystyle \left( {-\infty ,\frac{3}{4}} \right),{{f}^{'}}\left( x \right)=4x-3<0$

Hence, f is strictly decreasing in $\displaystyle \left( {-\infty ,\frac{3}{4}} \right)$

In $\displaystyle \left( {\frac{3}{4},\infty } \right),{{f}^{'}}\left( x \right)=4x-3>0$

Hence, f is strictly increasing in $\displaystyle \left( {\frac{3}{4},\infty } \right)$ .

Question 5: Find the intervals in which the function f given f(x) = 2x³ – 3x² – 36x + 7 is

(a) Strictly increasing (b) Strictly decreasing

Solution:

The given function is f(x) = 2x³ – 3x² – 36x + 7

Hence, f’(x) = 6x² – 6x – 36

= 6(x² – x – 6)

=6(x + 2)(x – 3)

Therefore, f’(x) = 0

⟹ x = -2, 3

In $\displaystyle \left( {-\infty ,-2} \right)$ and $\displaystyle \left( {3,\infty } \right)$ , $\displaystyle f'\left( x \right)>0$

In (-2, 3), f’(x) < 0

Hence, f is strictly increasing in $\displaystyle \left( {-\infty ,-2} \right),\left( {3,\infty } \right)$ and strictly decreasing in (-2, 3).

Question 6: Find the intervals in which the following functions are strictly increasing or decreasing.

(a) x² + 2x – 5 (b) 10 – 6x – 2x²

(c) –2x³ – 9x² – 12x + 1 (d) 6 – 9x – 9x²

(e) (x + 1)³ (x – 3)³

Solution: (a) f(x) = x² + 2x – 5

Hence, f’(x) = 2x + 2

Therefore,

⟹ f’(x) = 0

⟹ x = –1

x = – 1 divides the number line into intervals $\displaystyle \left( {-\infty ,1} \right)$ and $\displaystyle \left( {-1,\infty } \right)$

In $\displaystyle \left( {-\infty ,1} \right)$ , f’(x) = 2x + 2 < 0

Thus, f is strictly decreasing in $\displaystyle \left( {-\infty ,1} \right)$

In $\displaystyle \left( {-1,\infty } \right)$ , f’(x) = 2x + 2 > 0

Thus, f is strictly increasing in $\displaystyle \left( {-1,\infty } \right)$

(b) f(x) = 10 – 6x – 2x²

Hence, $\displaystyle {{f}^{'}}\left( x \right)=-6-4x$

Therefore, f’(x) = 0

⟹x = $\displaystyle -\frac{3}{2}$

x = $\displaystyle -\frac{3}{2}$ , divides the number line into two intervals $\displaystyle \left( {-\infty ,-\frac{3}{2}} \right)$ and $\displaystyle \left( {-\frac{3}{2},\infty } \right)$

In $\displaystyle \left( {-\infty ,-\frac{3}{2}} \right)$ , f’(x) = – 6 – 4x < 0

Hence, f is strictly increasing for $\displaystyle \left( {x<-\frac{3}{2}} \right)$

In $\displaystyle \left( {-\frac{3}{2},\infty } \right)$ , f’(x) = – 6 – 4x > 0

Hence, f is strictly increasing for $\displaystyle \left( {x>-\frac{3}{2}} \right)$

(c) f(x) = –2x³ – 9x² – 12x + 1

Hence, f’(x) = – 6x² – 18x – 12

= – 6(x² + 3x + 2)

= – 6(x + 1)(x + 2)

Therefore, f’(x) = 0

⟹ x = – 1, 2

Therefore,

⟹f’(x) = 0

⟹ x = – 1 , 2

x = – 1 and x = –2 divide the number line into intervals $\displaystyle \left( {-\infty ,-2} \right),\left( {-2,-1} \right)$ and $\displaystyle \left( {-1,\infty } \right)$ .

In $\displaystyle \left( {-\infty ,-2} \right)$ and $\displaystyle \left( {-1,\infty } \right)$ , f’(x) = – 6(x + 1)(x + 2) < 0

Hence, f is strictly decreasing for x < – 2 and x > – 1

In (-2,-1), f’(x) = – 6(x + 1)(x + 2) > 0

Hence f is strictly increasing in -2 < x < -1

(d) 6 – 9x – 9x²

Hence, f’(x) = – 9 – 2x

Therefore,

⟹f’(x) = 0

⟹x = $\displaystyle -\frac{9}{2}$

In $\displaystyle \left( {-\frac{9}{2},\infty } \right)$ , f’(x) < 0

Hence, f is strictly decreasing for $\displaystyle \left( {x>-\frac{9}{2}} \right)$

In $\displaystyle \left( {-\infty ,-\frac{9}{2}} \right)$ . f’(x) > 0

Hence, f is strictly decreasing in $\displaystyle \left( {x>-\frac{9}{2}} \right)$

(e) (x + 1)³ (x – 3)³

Hence,

f’(x) = 3(x + 1)² (x – 3)³ + 3(x – 3)² (x + 1)³

= 3(x + 1)² (x – 3)² {x – 3 + x + 1}

= 3(x + 1)² (x – 3)² (2x – 2)

= 6 (x + 1)² (x – 3)² (x – 1)

Therefore, f’(x) = 0

⟹x = –1,3,1

x = –1,3,1 divides the number line into four intervals $\displaystyle \left( {-\infty ,-1} \right)$ , (-1,1),(1,3) and $\displaystyle \left( {3,\infty } \right)$

In $\displaystyle \left( {-\infty ,-1} \right)$ and (-1,1), f’(x) = 6(x+1)²(x-3)²(x-1) < 0

Hence, f is strictly decreasing in $\displaystyle \left( {-\infty ,-1} \right)$ and (-1,1)

In (1,3) and $\displaystyle \left( {3,\infty } \right)$ , f’(x) = 6(x+1)²(x-3)²(x-1) > 0

Hence, f is strictly increasing in (1,3) and $\displaystyle \left( {3,\infty } \right)$

Question 7: Show that $\displaystyle y=\log \left( {1+x} \right)-\frac{{2x}}{{2+x}},x>-1$ , is an increasing function of x throughout its domain.

Solution:

It is given that $\displaystyle y=\log \left( {1+x} \right)-\frac{{2x}}{{2+x}}$

Therefore,

$\displaystyle \frac{{dy}}{{dx}}=\frac{1}{{1+x}}-\frac{{\left( {2+x} \right)\left( 2 \right)-2x\left( 1 \right)}}{{{{{\left( {2+x} \right)}}^{2}}}}$

= $\displaystyle \frac{1}{{1+x}}-\frac{4}{{{{{\left( {2+x} \right)}}^{2}}}}$

= $\displaystyle \frac{{{{x}^{2}}}}{{\left( {1+x} \right){{{\left( {2+x} \right)}}^{2}}}}$

Now, $\displaystyle \frac{{dy}}{{dx}}=0$

Hence,

⟹ $\displaystyle \frac{{{{x}^{2}}}}{{{{{\left( {2+x} \right)}}^{2}}}}$ = 0

⟹ x² = 0

⟹x = 0

Since, x > – 1, x = 0 divides domain $\displaystyle \left( {-1,\infty } \right)$ in two intervals – 1 < x < 0 and x > 0

When, – 1 < x < 0

Then,

x < 0 ⟹ x² > 0

x > – 1 ⟹ (2 + x) > 0

⟹(2 + x)² > 0

Hence, $\displaystyle y=\frac{{{{x}^{2}}}}{{{{{\left( {2+x} \right)}}^{2}}}}>0$

When, x > 0

Then,

x > 0 ⟹ x² > 0

⟹(2 + x)² > 0

Hence, $\displaystyle y=\frac{{{{x}^{2}}}}{{{{{\left( {2+x} \right)}}^{2}}}}>0$

Thus, f is increasing throughout the domain.

Question 8: Find the values of x for which y = [x (x – 2)]² is an increasing function.

Solution:

We have, y = [x (x – 2)]²

= [x² – 2x]²

Therefore,

$\displaystyle \frac{{dy}}{{dx}}=\frac{d}{{dx}}{{\left[ {{{x}^{2}}-2x} \right]}^{2}}$

= 2(x² – 2x)(2x – 2)

= 4x(x – 2)(x – 1)

Now, $\displaystyle \frac{{dy}}{{dx}}$ = 0

Hence, ⟹4x(x – 2)(x – 1)

x = 0, x = 2, x = 1

x = 0, x = 1 and x = 2 divide the number line intervals $\displaystyle \left( {-\infty ,0} \right)$ , (0,1),(1,2) and $\displaystyle \left( {2,\infty } \right)$

In $\displaystyle \left( {-\infty ,0} \right)$ and (1,2), $\displaystyle \frac{{dy}}{{dx}}<0$

Hence, y is strictly decreasing in intervals $\displaystyle \left( {-\infty ,0} \right)$ and (1,2)

In (0,1) and $\displaystyle \left( {2,\infty } \right)$ , $\displaystyle \frac{{dy}}{{dx}}>0$

Hence, y is strictly increasing in intervals (0,1) and $\displaystyle \left( {2,\infty } \right)$

Question 9: Prove that $\displaystyle y=\frac{{4\sin \theta }}{{\left( {2+\cos \theta } \right)}}-\theta$ is an increment function of $\displaystyle \theta$ in $\displaystyle \left[ {0,\frac{\pi }{2}} \right]$ .

Solution:

We have, $\displaystyle y=\frac{{4\sin \theta }}{{\left( {2+\cos \theta } \right)}}-\theta$

Therefore,

$\displaystyle \frac{{dy}}{{d\theta }}=\frac{{\left( {2+\cos \theta } \right)\left( {4\cos \theta } \right)-4\sin \theta \left( {-\sin \theta } \right)}}{{{{{\left( {2+\cos \theta } \right)}}^{2}}}}-1$

= $\displaystyle \frac{{8\cos \theta +4{{{\cos }}^{2}}\theta +4{{{\sin }}^{2}}\theta }}{{{{{\left( {2+\cos \theta } \right)}}^{2}}}}-1$

= $\displaystyle \frac{{8\cos \theta +4}}{{{{{\left( {2+\cos \theta } \right)}}^{2}}}}-1$

Now, $\displaystyle \frac{{dy}}{{d\theta }}=0$

Hence,

$\displaystyle \frac{{8\cos \theta +4}}{{{{{\left( {2+\cos \theta } \right)}}^{2}}}}=1$

⟹8cos𝜃 + 4 = 4 + cos²𝜃 + 4 cos𝜃

⟹ cos²𝜃 – 4 cos𝜃 = 0

⟹ cos𝜃 (cos𝜃 – 4) = 0

⟹ cos𝜃 = 0 or cos𝜃 = 4

Since, $\displaystyle \cos \theta \ne 4$

Therefore, cos𝜃 = 0

⟹𝜃 = $\displaystyle \frac{\pi }{2}$

Now,

$\displaystyle \frac{{dy}}{{d\theta }}=\frac{{8\cos \theta +4-\left( {4+{{{\cos }}^{2}}\theta +4\cos \theta } \right)}}{{{{{\left( {2+\cos \theta } \right)}}^{2}}}}$

= $\displaystyle \frac{{4\cos \theta -{{{\cos }}^{2}}\theta }}{{{{{\left( {2+\cos \theta } \right)}}^{2}}}}$

= $\displaystyle \frac{{\cos \left( {4-\cos \theta } \right)}}{{{{{\left( {2+\cos \theta } \right)}}^{2}}}}$

In interval $\displaystyle \left[ {0,\frac{\pi }{2}} \right]$ , we have cos𝜃 > 0

so, 4 > cos𝜃

⟹ 4 – cos𝜃 > 0

Hence, cos𝜃(4-cos𝜃) > 0 and (2 + cos𝜃)²> 0

Therefore,

$\displaystyle \frac{{\cos \left( {4-\cos \theta } \right)}}{{{{{\left( {2+\cos \theta } \right)}}^{2}}}}>0$

Hence, $\displaystyle \frac{{dy}}{{d\theta }}>0$

So, y is strictly increasing in $\displaystyle \left( {0,\frac{\pi }{2}} \right)$ and the given function is continuous at x = 0 and x = $\displaystyle {\frac{\pi }{2}}$

Thus, y is increasing in interval $\displaystyle \left( {0,\frac{\pi }{2}} \right)$ .

Question 10: Prove that the logarithmic function is strictly increasing on $\displaystyle \left( {0,\infty } \right)$ .

Solution:

The given function is f(x) = log x

Therefore, $\displaystyle f'\left( x \right)=\frac{1}{x}$

For, $\displaystyle x>0,f'\left( x \right)=\frac{1}{x}>0$

Thus, the logarithmic function is strictly increasing in interval $\displaystyle \left( {0,\infty } \right)$ .

Question 11: Prove that the function f given by f(x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (-1,1)

Solution:

The given function is f(x) = x² – x + 1

Therefore, f’(x) = 2x – 1

Now, f’(x) = 0

⟹x = $\displaystyle \frac{1}{2}$

$\displaystyle x=\frac{1}{2}$ divides the interval v into $\displaystyle \left( {-1,\frac{1}{2}} \right)$ and $\displaystyle \left( {\frac{1}{2},1} \right)$