Exercise 7.9 - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $7.9$
Question 1.
Find the asymptotes of the following curves:
(i) $f(x)=\frac{x^2}{x^2-1}$
(ii) $f(x)=\frac{x^2}{x+1}$
(iii) $f(x)=\frac{3 x}{\sqrt{x^2+2}}$
(iv) $f(x)=\frac{x^2-6 x-1}{x+3}$
(v) $f(x)=\frac{x^2+6 x-4}{3 x-6}$
Solution:
(i) $\lim _{x \rightarrow 1^{-}} \frac{x^2}{x^2-1}=-\infty$ and $\lim _{x \rightarrow 1^{+}} \frac{x^2}{x^2-1}=\infty$
So $x=-1$ and $x=1$ are vertical asymptotes.
as $\begin{aligned} \lim _{x \rightarrow \infty} \frac{x^2}{x^2-1} & =\lim _{x \rightarrow \infty} \frac{x^2 / x^2}{1-\frac{1}{x^2}} \\ & =\lim _{x \rightarrow \infty} \frac{1^1}{1-\frac{1}{x^2}}=1\end{aligned}$
$\mathrm{y}=1$ is a horizontal asymptote
So the asymptotes are $x=-1, x=+1, y=1$
(ii) Since the numerator is of higher degree than the denominator we have a slant asymptote to find that asymptote we have to divide the numerator by the denominator So the slant asymptote is $y=x-1$

Also $\lim _{x \rightarrow-1} \frac{x^2}{x+1}=\infty$
So $x=-1$ is a vertical asymptote

$
\begin{aligned}
& \text { (iii) } \lim _{x \rightarrow \infty^{+}} \frac{3 x}{\sqrt{x^2+2}}=3 \lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^2+2}} \\
& =3 \lim _{x \rightarrow \infty} \frac{x}{x \cdot \frac{1}{x} \sqrt{x^2+2}}=3 \lim _{x \rightarrow \infty} \frac{1}{\sqrt{1+\frac{2}{x^2}}} \\
& =\frac{3}{\lim _{x \rightarrow \infty} \sqrt{1+\frac{2}{x^2}}}=\frac{3}{\sqrt{\lim _{x \rightarrow \infty}\left(1+\frac{2}{x^2}\right)}} \\
& =\frac{3}{1}=3 \\
& \text { RHL : } \lim _{x \rightarrow \infty^{+}} \frac{3 x}{\sqrt{x^2+2}}=3 \lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^2+2}}=3 \\
& \text { LHL: } \lim _{x \rightarrow \infty^{-}}\left(\frac{-3 x}{\sqrt{x^2+2}}\right)=-3 \lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^2+2}}=-3 \\
&
\end{aligned}
$
$\therefore \mathrm{y}=3$ and $\mathrm{y}=-3$ are the horizontal asymptotes and there is no slant asymptote

(iv) Since the numerator is of highest degree than the denominator. We have a slant asymptote to find it we have to divide numerator by the denominator.

So the equation of asymptotes is $y=x-9$ and $x=-3$
(v) Since the numerator is of highest degree than the denominator. We have a slant asymptote to find it we have to divide the numerator by the denominator.

Question 2.
Sketch the graphs of the following functions:
(i) $y=-\frac{1}{3}\left(x^3-3 x+2\right)$
(ii) $y=x \sqrt{4-x}$
(iii) $y=\frac{x^2+1}{x^2-4}$
(iv) $y=\frac{1}{1+e^{-x}}$
(v) $y=\frac{x^3}{24}-\log x$
Solution:
(i) $y=-\frac{1}{3}\left(x^3-3 x+2\right)$
Factorizing we get
$
y=-\frac{1}{3}(x-1)^2(x+2)=f(x)
$
- The domain and the range of the given function $f(x)$ are the entire real line.
- Putting $y=0$ we get $x=1,1,-2$. Hence the $x$ intercepts are $(1,0)$ and $(-2,0)$ and by putting $x=0$ we get $y=$ $-\frac{2}{3}$

$\therefore$ The $y$ intercept is $\left(0,-\frac{2}{3}\right)$
- $f^{\prime}(x)=-\frac{\left(3 x^2-3\right)}{3}=-\left(x^2-1\right)=\left(1-x^2\right)$ $f^{\prime}(x)=0 \Rightarrow 1-x^2=0 \Rightarrow x=\pm 1$
the critical points of the curve occur at $x=\pm 1$
- $f^{\prime \prime}(x)=-2 x$
$f^{\prime \prime}(1)=-2<0 \quad \therefore f(x)$ is maximum at $x=1$
and local maximum is $f(1)=0$
$f^{\prime \prime}(-1)=2>0 \Rightarrow f(x)$ is minimum at
$x=-1$ and the local minimum is
$f(-1)=-\frac{4}{3}$
- $f^{\prime \prime}(x)=-2 x<0 \forall x>c$

$\therefore$ The function is concave upward in the negative real line.
- Since $f^{\prime}(x)=0$ at $x=0$ and $f^{\prime}(x)$ changes its sign when passing through $x=0, x=0$ is a point of inflection is
$
\left(0,-\frac{2}{3}\right)
$
- The curre has no asymptotes.

(ii)
$
y=x \sqrt{4-x}=f(x)
$
$y=x \sqrt{4-x}=f(x)$
- When $x>4$ the curve doesnot exists and it exists for $x \leq 4$

The domain is $[-\infty, 4]$ and the range is $\left(-\infty, \frac{16}{3 \sqrt{3}}\right]$.
- The curve passes through the origin. The curve intersect $x$ axis at $(4,0)$.
- $f^{\prime}(x)=\frac{-x}{2 \sqrt{4-x}}+\sqrt{4-x}=\frac{8-3 x}{2 \sqrt{4-x}}$ $\therefore f^{\prime}(x)=0 \Rightarrow 8-3 x=0 \Rightarrow x=\frac{8}{3}$
$\therefore$ Critical point of the curve occur at $x=\frac{8}{3}$
- $f^{\prime \prime}(x)=\frac{3 x-16}{4(4-x)^{3 / 2}}$ $f^{\prime \prime}\left(\frac{8}{3}\right)=-\frac{3 \sqrt{3}}{4}<0 \Rightarrow f(x)$ is maximum at $x=\frac{8}{3}$ and the local maximum is $f\left(\frac{8}{3}\right)=\frac{16}{3 \sqrt{3}}$ and the local minimum is 0 at $x=4$ (from the graph)
- $f^{\prime \prime}(x)=\frac{3 x-16}{4(4-x)^{3 / 2}}<0 \forall x<4$
The curve concave downward in the negative real line
- No point of inflection exists.
- as $x \rightarrow \infty, y=\pm \infty$ and so the curve does not have any asymptotes

(iii)
$
y=\frac{x^2+1}{x^2-4}
$

- The domain of the given function $f(x)$ is $(-\infty,-2) \cup$
$
\begin{aligned}
& (-2,2) \cup(2, \infty) \\
& \text { (i.e.) } x<-2 \text { or }-2<x<2 \text { or } x \\
& >2
\end{aligned}
$
Range of $f(x)$ is
$
\begin{aligned}
& \left(-\infty, \frac{-1}{4}\right] \cup(1, \infty) \\
& \text { (i.e.,) } f(x) \leq \frac{-1}{4} \text { or } f(x)>1
\end{aligned}
$
- Putting $y=0 . x$ is unreal hence there is no ' $x$ ' intercept. By putting $x=0$ we get

$
y=\frac{-1}{4}
$
$y$ intercept is $\left(0, \frac{-1}{4}\right)$.
$f^{\prime}(x)=\frac{-10 x}{\left(x^2-4\right)^2}$
$
\therefore f^{\prime}(x)=0 \Rightarrow x=0 \text {. }
$
$\therefore$ The critical point is at $x=0$
$f^{\prime \prime}(x)=\frac{10\left(x^2-4\right)\left(3 x^2+4\right)}{\left(x^2-4\right)^4}$
$f^{\prime \prime}(0)=\frac{-5}{8}<0 . \therefore f(x)$ is maximum at $x=0$
and the local maximum is $f(0)=-\frac{1}{4}$
- No points of reflection
- When $\mathrm{x}=\pm 2, \mathrm{y}=\infty$, Vertical asymptotes are $\mathrm{x}=2$ and $\mathrm{x}=-2$ and horizontal asymptote is $\mathrm{y}=1$

(iv) $y=\frac{1}{1+e^{-x}}=f(x)$
- The domain of the function $f(x)$ is the entire real line
(i.e.,) $(-\infty, \infty)$ (i.e.,) $-\infty<x<\infty$ and the range is $(0,1)$ (i.e.,) $0<f(x)<1$

- No ' $x$ ' intercept for $f(x)$ and when $x=0, y=\frac{1}{2}$
$\therefore$ The $y$ intercept is $\left(0, \frac{1}{2}\right)$
- $f^{\prime}(x)=\frac{e^{-x}}{\left(1+e^{-x}\right)^2}$
$f(x)=0 \Rightarrow e^{-x}=0$ which is not possible hence there is no extremum.
- No vertical asymptote for the curve and the horizontal asymptotes are $y=1$ and $y=0$

(v)
$
y=\frac{x^3}{24}-\log x=f(x)
$
- The curve exists only for positive values of $(x>0)$
(i.e.,) domain is $(0, \infty)$ and the range is $\left(\frac{1}{3}-\log e^2>\infty\right)$
- No intersection points are possible
$f^{\prime}(x)=\frac{x^2}{8}-\frac{1}{x}$
(i.e.,) domain is $(0, \infty)$ and the range is $\left(\frac{1}{3}-\log e^2>\infty\right)$
- No intersection points are possible
$f^{\prime}(x)=\frac{x^2}{8}-\frac{1}{x}$
$
f^{\prime}(x)=0 \Rightarrow x^3-8=0 \Rightarrow x^3=8 \Rightarrow x=2
$
$\therefore$ Critical point occur at $x=2$
- $f^{\prime \prime}(x)=\frac{x}{4}+\frac{1}{x^2}$
$f^{\prime \prime}(2)=\frac{3}{4}>0$
$\therefore f(x)$ is minimum at $x=2$ and
the local minimum is $f(2)=\frac{1}{3}-\log e^2$

- No point of inflection.
- No horizontal asymptote is possible.
But the vertical asymptote is $x=0$ ( $y$ axis).