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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
Question
For what value of λ is the function defined by $f(x)=\left\{\begin{array}{ll} {\lambda\left(x^{2}-2 x\right),} & {\text { if } x \leq 0} \\ {4 x+1,} & {\text { if } x>0} \end{array}\right.$ continuous at x = 0? What about continuity at x = 1?
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Answer

It is given that $f(x)=\left\{\begin{array}{c} {\lambda\left(x^{2}-2 x\right), \text { if } x \leq 0} \\ {4 x+1, \text { if } x>0} \end{array}\right.$
Let f be continuous at x = 0, then, we get
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {\lambda \left( {{x^2} - 2x} \right)} \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {\lambda \left( {{0^2} - 2 \times 0} \right)} \right) = 0$
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} (4x + 1)$ = 4 $\times$ 0 + 1 = 1
$\therefore \mathop {\lim }\limits_{x \to {0^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {0^ + }} f(x)$
Therefore, there is no value of $\lambda$ for which f is continuous at x = 0
f(1) = 4x + 1 = 4 × 1 + 1 = 5
$\mathop {\lim }\limits_{x \to 1} (4x + 1)$ = 4 × 1 + 1 = 5
Then $\mathop {\lim }\limits_{{\mathbf{x}} \to 1} {\text{f}}({\text{x}}) = {\text{f}}(1)$
Therefore, for any values of $\lambda$, f is continuous at x = 1

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