Solve the Following Question.(2 Marks)

Question 12 Marks

Solve using intermediate value theorem:
Show that $5^x-6 x=0$ has a root in $[1,2]$

Answer

Let $f(x)=5^x-6 x$
$5^x$ and $6 x$ are continuous functins for all $x \in R$.
$\therefore 5^{ x }-6 x$ is also continuous for all $x \in R$.
i.e. $f(x)$ is continuous for all $x \in R$.
A root of $f(x)$ exists if $f(x)=0$ for at least one value of $x$.
$ f(1)=5^1-6(1)$
$=-1<0$
$f(2)=(5)^2-6(2)$
$=13>0$
$\therefore f(1)<0 \text { and } f(2)>0 $
By intermediate value theorem, there has to be a point ' $c$ ' between $1$ and $2$ such that $f(c)=0$.
$\therefore$ There is a root of the given equation in $[1,2]$.

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Question 22 Marks

Find $f(a)$, if $f$ is continuous at $x=a$ where, $f(x)=\frac{1-\cos [7(x-\pi)]}{5(x-\pi)^2}$, for $x \neq \pi$ at $a=\pi$

Answer

$f$ is continuous at $x=\pi$
$\therefore f (\pi)=\lim _{x \rightarrow \pi} f (x)=\lim _{x \rightarrow \pi} \frac{1-\cos [7(x-\pi)]}{5(x-\pi)^2}$
Put $x-\pi=h$, as $x \rightarrow \pi, h \rightarrow 0$
$\therefore f (\pi)=\lim _{ h \rightarrow 0} \frac{1-\cos 7 h }{5 h ^2}$
$=\lim _{ h \rightarrow 0} \frac{2 \sin ^2\left(\frac{7 h }{2}\right)}{5 h ^2}$
$=\frac{2}{5} \lim _{ h \rightarrow 0} \frac{\sin ^2\left(\frac{7 h }{2}\right)}{\left(\frac{7 h }{2}\right)^2} \times\left(\frac{7}{2}\right)^2$
$=\frac{2}{5}\left|\lim _{ h \rightarrow 0} \frac{\sin \left(\frac{7 h }{2}\right)}{\left(\frac{7 h }{2}\right)}\right|^2 \times \frac{49}{4}$
$=\frac{2}{5} \times(1)^2 \times \frac{49}{4} \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$\left.\therefore f (\pi)=\frac{49}{10}\right]$

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Question 32 Marks

Discuss the continuity of f on its domain, where
f(x) = |x + 1|, for -3 ≤ x ≤ 2
= |x – 5|, for 2 < x ≤ 7

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Question 42 Marks

$ \text {If } \mathrm{f}(\mathrm{x})=\frac{\sin 2 x}{5 x}-\mathrm{a}, \text { for } \mathrm{x}>0$
$ =4 \text { for } \mathrm{x}=0 $
$ =\mathrm{x}^2+\mathrm{b}-3, \text { for } \mathrm{x}<0$
is continuous at $x=0$, find $a$ and $b$.

Answer

$f(x)$ is continuous at $x=0$ (given)
$ \therefore \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)=\mathrm{f}(0)$
$\therefore \lim _{x \rightarrow 0^{+}}\left(\frac{\sin 2 x}{5 x}-\mathrm{a}\right)=4$
$\therefore \quad \lim _{x \rightarrow 0^{+}} \frac{\sin 2 x}{5 x}-\lim _{x \rightarrow 0^{+}} \mathrm{a}=4$
$\therefore \quad \frac{1}{5} \lim _{x \rightarrow 0^{+}} \frac{\sin 2 x}{2 x} \times(2)-\lim _{x \rightarrow 0^{+}} \mathrm{a}=4$
$\therefore \frac{1}{5}(1)(2)-\mathrm{a}=4 \quad[\because x \rightarrow 0,2 x \rightarrow 0$
$\therefore \frac{2}{5}-\mathrm{a}=4 \quad\left[ \text { and } \lim _{\theta \rightarrow 0^{+}} \frac{\sin \theta}{\theta}=1 \right]$
$\therefore \quad \frac{2}{5}-4=\mathrm{a}$
$\therefore \quad \mathrm{a}=-\frac{18}{5}$
$\therefore \lim _{x \rightarrow 0^{-}}\left(x^2+\mathrm{b}-3\right)=4 $

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Question 52 Marks

If $f(x)=\frac{5^x+5^{-x}-2}{x^2}$, for $x \neq 0$
$=k_{\text {, }}$ for $x=0$
is continuous at $x=0$, then find $k$.

Answer

$f(x)$ is continuous at $x=0$
..(given)
$
\begin{aligned}
\therefore \quad f(0) & =\lim _{x \rightarrow 0} f(x) \\
& =\lim _{x \rightarrow 0} \frac{5^x+5^{-x}-2}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{5^x+\frac{1}{5^x}-2}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\left(5^x\right)^2+1-2\left(5^x\right)}{5^x \cdot x^2} \\
& =\lim _{x \rightarrow 0} \frac{\left(5^x-1\right)^2}{5^x \cdot x^2} \\
& =\lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}\right)^2 \cdot \frac{1}{5^x} \\
& =\lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}\right)^2 \times \lim _{x \rightarrow 0} \frac{1}{5^x} \\
& =\left(\log _5 5\right)^2 \times \frac{1}{5^0} \quad \ldots\left[\because \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a\right]\\
\therefore K=(log 5)^5
\end{aligned}
$

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Question 62 Marks

Which of the following functions has a removable discontinuity?
$\begin{array}{ll}
f(x)=3 x+2, & \text { for }-4 \leq x \leq-2 \\
=2 x-3, & \text { for }-2\end{array}$

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Question 72 Marks

Identify discontinuities for the following functions as either a jump or a removable discontinuity :
$f(x)=4+\sin x$, for $x<\pi=3-\cos x$ for $x>\pi$.

Answer

$f(x)=4+\sin x, x<\pi=3-\cos x, x>\pi$
$\sin x$ and $\cos x$ are continuous for all $x \in R$.
4 and 3 are constant functions.
$\therefore 4+\sin x$ and $3-\cos x$ are continuous for all $x \in R$.
$\therefore f ( x )$ is continuous for both the given intervals.
Let us test the continuity at $x=\pi$.
$ \lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow x^{-}}(4+\sin x)$
$=4+\sin \pi$
$=4+0$
$=4$
$\lim _{x \rightarrow x ^{+}} f (x)=\lim _{x \rightarrow x ^{+}}(3-\cos x)$
$=3-\cos \pi$
$=3-(-1)$
$=4$
$\therefore \quad \lim _{x \rightarrow s ^{-}} f (x)=\lim _{x \rightarrow z ^{+}} f (x)$
$\therefore \quad \lim _{x \rightarrow \pi} f(x)=4$
But $f(\pi)$ is not defined.
$\therefore f ( x )$ has a removable discontinuity at $x =\pi$.

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Question 82 Marks

Identify discontinuities for the following functions as either a jump or a removable discontinuity :
$f(x)=x^2-3 x-2$, for $x<-3=3+8 x$, for $x>-3$.

Answer

$f(x)=x^2-3 x-2, x<-3=3+8 x, x>-3$
$f(x)$ is a polynomial function for both the intervals.
$\therefore f ( x )$ is continuous for both the given intervals.
Let us test the continuity at $x =-3$.
$\lim _{x \rightarrow-3^{-}} f (x)=\lim _{x \rightarrow-3^{-}}\left(x^2-3 x-2\right)$
$=(-3)^2-3(-3)-2$
$=9+9-2$
$=16$
$\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{+}}(3+8 x)$
$=3+8(-3)$
$=-21$
$\therefore \quad \lim _{x \rightarrow-3^{-}} f (x) \neq \lim _{x \rightarrow-3^{+}} f (x)$
$\therefore \quad \lim _{x \rightarrow-3} f (x) \text { does not exist. }$
$\therefore f ( x )$ is discontinuous at $x =-3$.
$\therefore f ( x )$ has a jump discontinuity at $x =-3$

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Question 92 Marks

Identify discontinuities for the following functions as either a jump or a removable discontinuity :
$f(x)=x^2+3 x-2$, for $x \leq 4$
$=5 x+3$, for $x>4$.

Answer

$ f(x)=x^2+3 x-2, x \leq 4$
$=5 x+3, x>4 $
$f(x)$ is a polynomial function for both the intervals.
$\therefore f ( x )$ is continuous for both the given intervals.
Let us test the continuity at $x =4$.
$ \lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{-}}\left(x^2+3 x-2\right)$
$=(4)^2+3(4)-2$
$=26$
$\lim _{x \rightarrow 4^{+}} f (x)=\lim _{x \rightarrow 4^{+}}(5 x+3)$
$=5(4)+3$
$=23$
$\therefore \quad \lim _{x \rightarrow 4^{-}} f (x) \neq \lim _{x \rightarrow 4^{+}} f (x)$
$\therefore \quad \lim _{x \rightarrow 4} f (x) \text { does not exist. }$
$\therefore f ( x )$ is discontinuous at $x =4$.
$\therefore f ( x )$ has a jump discontinuity at $x =4$.

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Question 102 Marks

Identify discontinuities for the following functions as either a jump or a removable discontinuity : $f(x)=\frac{x^2-10 x+21}{x-7}$

Answer

Given, $f(x)=\frac{x^2-10 x+21}{x-7}$
It is a rational function and is discontinuous if $x-7=0$,
i.e., $x=7$
$\therefore f ( x )$ is continuous for all $x \in R$, except at $x =7$.
$\therefore f (7)$ is not defined.
Now, $\lim _{x \rightarrow 7} f (x) =\lim _{x \rightarrow 7} \frac{x^2-10 x+21}{x-7}$
$ =\lim _{x \rightarrow 7} \frac{(x-7)(x-3)}{x-7}$
$=\lim _{x \rightarrow 7}(x-3) \ldots\left[\begin{array}{l} \because x \rightarrow 7, x \neq 7, \\ \therefore x-7 \neq 0 \end{array}\right]$
$=7-3$
$ =4$
Thus, $\lim _{x \rightarrow 7} f (x)$ exist but $f (7)$ is not defined.
$\therefore f ( x )$ has a removable discontinuity.

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Question 112 Marks

Examine whether the function is continuous at the points indicated against them.:

f(x) =$\frac{x}{\text{tan}3x}$+2, for x < 0
=$\frac{7}{3}$,for x ≥ 0, at x = 0.

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Question 122 Marks

Examine whether the function is continuous at the points indicated against them.:

f(x) = $\frac{x^2 +18x-19}{x-1}$for x ≠ 1
= 20, for x = 1, at x = 1.

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Question 132 Marks

Examine whether the function is continuous at the points indicated against them.:

f(x) = x3 – 2x + 1, if x ≤ 2
= 3x – 2, if x > 2, at x = 2.

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Maths Solve the Following Question.(2 Marks)

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