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  • Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 37 maths textbook solution

Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 37 maths textbook solution

Answers (1)

Answer:

 The correct option is (e)

Hint:

Use this definition of continuity to solve this problem. A function f(x) is said to be continuous at x=a

if

\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)

Given:

f(x)=\left\{\begin{array}{cl} (1+a x)^{1 / x} & , x<0 \\ b & , x=0 \\ \frac{(x+c)^{1 / 3}-1}{(x+1)^{1 / 2-1}} & , x>0 \end{array}\right.

may be continuous at x = 0

Solution:

Now at x = 1

f(x)=\left\{\begin{array}{cl} (1+a x)^{1 / x} & , x<0 \\ b & , x=0 \\ \frac{(x+c)^{1 / 3}-1}{(x+1)^{1 / 2-1}} & , x>0 \end{array}\right.

To find the value of a, b and c we will use definition of continuous function

So we know if  f(x) is continuous at x = 0 then we have,

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)-(1)

Using equation (1) We have,

\lim _{x \rightarrow 0^{-}} f(x)=f(0) \Rightarrow \lim _{h \rightarrow 0} f(0-h)=f(0)

\begin{aligned} &\left.\Rightarrow \lim _{h \rightarrow 0}\{1+a(-h)\}^{\frac{1}{h}}=f(0) \quad [\because f(x)=(1+a x)^{1 / x}, x<0\right] \\ &\Rightarrow \lim _{h \rightarrow 0}(1-a h)^{-\frac{1}{h}}=f(0)=b \quad[\because \boldsymbol{f}(\boldsymbol{x})=\boldsymbol{b}, \boldsymbol{x}=\mathbf{0}] \end{aligned}

Taking log both sides, then

\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0} \log (1-a b)^{-1 / n}=\log b \\ &\Rightarrow \lim _{h \rightarrow 0}-\frac{1}{n} \log (1-a h)=\log b \quad\left[\because \log a^{m}=\boldsymbol{m l o g} a\right] \end{aligned}

\Rightarrow \lim _{h \rightarrow 0} \frac{a \log (1-a h)}{-a h}=\log b \quad \text { [multiply and divide a on L.H.S] }

\begin{aligned} &\Rightarrow a \lim _{h \rightarrow 0} \frac{\log \{1+(-a b)\}}{(-a b)}=\log b\\ &\Rightarrow a \cdot 1=\log b &\left[\therefore \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]\\ &\Rightarrow a=\log b \qquad - (2) \end{aligned}

Again, using equation (1) we get,

\begin{aligned} &\lim _{x \rightarrow 0+} f(x)=\{(0) \\ &\Rightarrow \lim _{h \rightarrow 0} f(0+h)=f(0) \end{aligned}

\begin{aligned} &\left.\Rightarrow \lim _{h \rightarrow 0} f(n)=f^{\prime} (0) \right) \\ &\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(h+c)^{1 / 3}-1}{(h+1)^{1 / 2}-1}\right)=f(0) \quad\left[\because \boldsymbol{f}(x)=\frac{(x+\boldsymbol{c})^{\mathbf{1} / 3}-4}{(\boldsymbol{x}+\mathbf{1})^{\mathbf{1} / 2}-\mathbf{1}}, x>\mathbf{0}\right] \end{aligned}

\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(h+c)^{1 / 3}-1}{(n+1)^{1 / 2}-1} \times \frac{(n+1)^{1 / 2}+1}{(n+1)^{1 / 2}+1}\right)=\mathrm{b} \quad[\because f(x)=b, x=0]

\Rightarrow \lim _{h \rightarrow 0} \frac{\left((h+r)^{4} 3-1\right) \times\left((n+1)^{42}+1\right)}{\left((h+1)^{1 / 2}-1\right)\left((h+1)^{1 / 2}+1\right)}=b

\Rightarrow \lim _{h \rightarrow 0} \frac{\left((h+c)^{1 / 3}-1\right)\left((n+1)^{4 / 2}+1\right)}{\left((h+1)^{1 / 2}\right)^{2}-(1)^{2}}=\mathrm{b} \quad\left[\because(\boldsymbol{a}+\boldsymbol{b})(\boldsymbol{a}-\boldsymbol{b})=\boldsymbol{a}^{2}-\boldsymbol{b}^{2}\right]

\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0} \frac{\left((h+c)^{1 / 3}-1\right)\left((h+1)^{42}+1\right)}{h+1-1}=b \\ &\Rightarrow \lim _{h \rightarrow 0} \frac{\left.\left((h+c)^{1 / 3}-1\right)(n+1)^{4 / 2}+1\right)}{h}=b \end{aligned}

\Rightarrow \lim _{h \rightarrow 0} \frac{\left((h+c)^{1 / 3}-1\right)}{h} \times \lim _{h \rightarrow 0}\left((n+1)^{1 / 2}+1\right)=b \quad\left[\begin{array}{ll} \because \lim _{x \rightarrow a} f(x) \cdot g(x)= & \lim _{x \rightarrow a} f(x) \\ x \lim _{x \rightarrow a} g(x) \end{array}\right]

\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(h+c)^{1 / 3}-1^{4 / 3}}{h}\right) \times(0+1)^{1 / 2}+1=b \\ &\Rightarrow \lim _{h \rightarrow 0} \frac{(h+c)^{1 / 3}-1^{1 / 3}}{(h+c)-c} \times 1^{1 / 2}+1=b \\ &\Rightarrow g \times 2=b \end{aligned}

\begin{aligned} &\text { where } g \therefore=\lim _{h \rightarrow 0} \frac{(h+c)^{1 / 3}-1^{1 / 3}}{(h+c)-c}-(a) \\ &\Rightarrow g \cdots=\frac{6}{2} \quad-(3) \end{aligned}

To find the value of g, we will use

\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-\mathrm{L}}

But this formula is applicable on g only if c=L

So we take c = L then we have,

g=\lim _{h \rightarrow 0} \frac{(h+1)^{1 / 3}-1^{1 / 3}}{(h+1)-1}                    \left[\because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right]

\begin{aligned} &=\frac{1}{3} \cdot 1^{1 / 3-1} \\ &=\frac{1}{3} 1^{-2 / 3}=\frac{1}{3} \cdot 1=\frac{1}{3} \\ &\therefore g=\frac{1}{3} \end{aligned}

Equation (3)

\begin{aligned} &\Rightarrow \frac{1}{3}=\frac{b}{2} \\ &\Rightarrow b=\frac{2}{3} \end{aligned}

Then equation 2 becomes,

\begin{aligned} &\Rightarrow a=\log \frac{2}{3} \\ &\therefore a=\frac{\log 2}{3}, b=\frac{2}{3}, c=1 \end{aligned}

Posted by

Gurleen Kaur

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