Solve the following Question.(1 Marks)

Question 11 Mark

Let $f(x) = ax + b$ (where a and b are unknown)
$= x^2 + 5$ for $x \in R$
Find the values of a and b, so that $f(x)$ is continuous at $x = 1$

Answer

$f(x)=x^2+5, x \in R$
$\therefore f(1)=1+5=6$
If $f(x)=a x+b$ is continuous at $x=1$, then
$f(1)=\lim _{x \rightarrow 1}(a x+b)=a+b$
$\therefore 6=a+b \text { where, } a, b \in R$
$\therefore$ There are infinitely many values of a and b .

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Question 21 Mark

Show that there is a root for the equation $x^3 – 3x = 0$ between $1$ and $2$.

Answer

Let $f(x) = x^3 – 3x$
$f(x)$ is a polynomial function and hence it is continuous for all $x ∈ R$.
A root of $f(x)$ exists, if $f(x) = 0$ for at least one value of $x$.
$f(1) = (1)^3 – 3(1) = -2 < 0$
$f(2) = (2)^3 – 3(2) = 2 > 0$
$\therefore f(1) < 0$ and $f(2) > 0$
$\therefore $ By intermediate value theorem,
there has to be point $‘c’$ between $1$ and $2$ such that $f(c) = 0$.
There is a root of the given equation between $1$ and $2$.

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Question 31 Mark

Show that there is a root for the equation $2x^3 – x – 16 = 0$ between $2$ and $3$.

Answer

Let $f(x) = 2x^3 – x – 16$
f(x) is a polynomial function and hence it is continuous for all $x ∈ R$.
A root of f(x) exists, if f(x) = 0 for at least one value of $x$.
$f(2) = 2(2)^3 – 2 – 16 = -2 < 0$
$f(3) = 2(3)^3 – 3 – 16 = 35 > 0$
$\therefore f(2) < 0$ and $f(3) > 0$
$\therefore $ By intermediate value theorem,
there has to be point $‘c’$ between $2$ and $3$ such that $f(c) = 0$.
$\therefore $ There is a root of the given equation between $2$ and $3$.

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Question 41 Mark

Examine the continuity of :
$f(x)=\frac{x^2-9}{x-3}$, for $x \neq 3$
$=8$ for $x=3$, at $x=3$.

Answer

$f(3)=8$ (given)
$\lim _{x \rightarrow 3} f (x)= \lim _{x \rightarrow 3} \frac{x^2-9}{x-3}$
$= \lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{x-3}$
$= \lim _{x \rightarrow 3}(x+3)$
$\quad \ldots[\because x \rightarrow 3, x \neq 3 \therefore x-3 \neq 0]$
$= 3+3=6$
$\lim _{x \rightarrow 3} f (x) \neq f (3) $
$\therefore f ( x )$ is discontinuous at $x =3$.

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Question 51 Mark

Examine the continuity of :
$f(x)=\sin x$, for $x \leq \frac{\pi}{4}$ $=\cos x$, for $x>\frac{\pi}{4}$, at $x=\frac{\pi}{4}$

Answer

$ f (x)=\sin x ; \quad x \leq \frac{\pi}{4}$
$=\cos x ; \quad x>\frac{\pi}{4}$
$\lim _{x \rightarrow \frac{x^{-}}{4}} f(x)=\lim _{x \rightarrow \frac{x^{-}}{4}}(\sin x)$
$=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
$\lim _{x \rightarrow \frac{x^{+}}{4}} f (x)=\lim _{x \rightarrow \frac{x^{+}}{4}}(\cos x)$
$=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
Also, $f \left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}$
$ =\frac{1}{\sqrt{2}} $
$\therefore \quad \lim _{x \rightarrow \frac{\pi^{-}}{4}} f (x)=\lim _{x \rightarrow \frac{\pi^*}{4}} f (x)= f \left(\frac{\pi}{4}\right)$
$\therefore f (x)$ is continuous at $x=\frac{\pi}{4}$.

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Question 61 Mark

Examine the continuity of :
$f(x) = x^3 + 2x^2 – x – 2$ at $x = -2$

Answer

Given, $f(x)=x^3+2 x^2-x-2$
$f(x)$ is a polynomial function and hence it is continuous for all $x \in R$. $\therefore f ( x )$ is continuous at $x =-2$.

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