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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY2 Marks
Question
For what value of $\lambda$ is the function defined by
$\text{f(x)}= \begin{cases}\lambda(\text{x}^{2} - 2\text{x}), \text{if}\ \text{x} \leq0\\ \text{4x} + 1,\ \ \ \ \ \ \ \text{if}\ \text{x} > 0\end{cases}$
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Answer

$\text{f(x)}= \begin{cases}\lambda(\text{x}^{2} - 2\text{x}), \text{if}\ \text{x} \leq0\\ \text{4x} + 1,\ \ \ \ \ \ \ \text{if}\ \text{x} > 0\end{cases}$ At x = 0$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\lambda(\text{x}^{2} - \text{2x}) = \lambda(0 - 0) = 0$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}(4\text{x} + 1) = 4(0) + 1 = 0 + 1 = 1$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} \neq ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x }) $ At x = 1$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}}(4\text{x} + 1) = 4 + 1 = 5$
Also f(1) = 4 + 1 = 5 $\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(1)}$ $\therefore$ f is continuous at x = 1 whatever value of $\lambda$ be.

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