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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Finde the value of a and b, if the function f(x) defined by $\text{f(x)}\begin{cases}\text{x}^2+3\text{x}+\text{a}, &\text{x}\leq1\\\text{bx}+2, & \text{x}>1\end{cases}$is differentiable at x = 1.
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Answer

Given that f(x) is differentiable at x = 1, Therefore, f(x) is countinuous at x = 1.
$\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\text{f(1)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}(\text{x}^2+3\text{x}+\text{a})=\lim\limits_{\text{x}\rightarrow1}(\text{bx}+2)=1+3+\text{a}$
$\Rightarrow1+3+\text{a}=\text{b}+2$
$\Rightarrow\text{a}-\text{b}+2=0\dots(1)$
Again, f(x) is differentiable at x = 1. So,
(LHL at x = 1) = (RHL at x = 1)
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(z)}-\text{f}(1)}{\text{z}-1}=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(z)}-\text{f}(1)}{\text{z}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}^2+3\text{x}+\text{a})-(4+\text{a})}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{bx}+2)-(4+\text{a})}{\text{x}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2+3\text{x}-4}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{bx}-2-\text{a})}{\text{x}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}(\text{x}-4)=\lim\limits_{\text{x}\rightarrow1}\frac{\text{b}(\text{x}-1)}{\text{x}-1}$
$\Rightarrow5=\text{b}$
Hence, a = 3 and b = 5.

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