Question

plz help me 11th pcbm Math problem (30 point I selected for this question)
plz help me ​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that f(0) = k

So,

Given function can be rewritten as

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{ {\bigg(\dfrac{3x + 2}{2 - 5x} \bigg) }^{\dfrac{1}{x} } \: when \: x \: \ne \: 0} \\ \\ &\sf{ \: \: \: \: \: \: \: \: \: k \: \: \: \: \: \: \: \: \: \: \: when \: x \: = \: 0} \end{cases}\end{gathered}\end{gathered}$

We know,

A function f(x) is said to be Continuous at x = a, iff

$\boxed{ \quad \sf{ \:\displaystyle\lim_{x \to a} \bf\: f(x) = f(a) \quad}}$

Now,

$\rm :\longmapsto\:f(0) = k$

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}{\bigg(\dfrac{3x + 2}{2 - 5x} \bigg) }^{\dfrac{1}{x} }$

If we substitute directly x = 0, we get indeterminant form.

$\rm \: = \:\displaystyle\lim_{x \to 0}{\bigg(1 + \dfrac{3x + 2}{2 - 5x} - 1\bigg) }^{\dfrac{1}{x} }$

$\rm \: = \:\displaystyle\lim_{x \to 0}{\bigg(1 + \dfrac{3x + 2 - (2 - 5x)}{2 - 5x} \bigg) }^{\dfrac{1}{x} }$

$\rm \: = \:\displaystyle\lim_{x \to 0}{\bigg(1 + \dfrac{3x + 2 - 2 + 5x}{2 - 5x} \bigg) }^{\dfrac{1}{x} }$

$\rm \: = \:\displaystyle\lim_{x \to 0}{\bigg(1 + \dfrac{8x}{2 - 5x} \bigg) }^{\dfrac{1}{x} }$

$\rm \: = \:\displaystyle\lim_{x \to 0}{\bigg(1 + \dfrac{8x}{2 - 5x} \bigg) }^{\dfrac{2 - 5x}{8x} \times \dfrac{8x}{x(2 - 5x)} }$

We know that

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \: {\bigg(1 + x\bigg) }^{ \dfrac{1}{x} } = e}}$

So, using this result, we get

$\rm \:  =  \:  \:{\bigg(e \bigg) }^{\displaystyle\lim_{x \to 0} \: \dfrac{8}{2 - 5x} }$

$\rm \:  =  \:  \:{\bigg(e \bigg) }^{ \dfrac{8}{2 - 5 \times 0} }$

$\rm \:  =  \:  \:{\bigg(e \bigg) }^{ \dfrac{8}{2 - 0} }$

$\rm \:  =  \:  \:{\bigg(e \bigg) }^{ \dfrac{8}{2 } }$

$\rm \:  =  \:  \: {e}^{4}$

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}{\bigg(\dfrac{3x + 2}{2 - 5x} \bigg) }^{\dfrac{1}{x} } = {e}^{4}$

Since, it is given that f(x) is continuous at x = 0.

$\bf\implies \:\quad \sf{ \:\displaystyle\lim_{x \to 0} \bf\: f(x) = f(0) \quad}$

$\bf\implies \:k = {e}^{4}$

$\bf\implies \:f(0) = {e}^{4}$

Additional Information :-

1. If f and g are two continuous functions at x = a, then

• f + g is also continuous at x = a

• f - g is also continuous at x = a

• f × g is also continuous at x = a

2.