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Continuity and Differentiability Ncert Solutions Maths class 12th Maths Class 12th Continuity and Differentiability

10. $f (x) =$ ${\begin{matrix} x^{3} - 3, if x \leq 2 & x^{2} + 1, if x > 2 \end{matrix}$

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alok kumar singh | Contributor-Level 10

4 months ago

10. Given f (x) = ${\begin{matrix} x^{3} - 3 if x | ? | 2 & x^{2} + 1 if x > 2 \end{matrix}$

For x = c < 2,

f (c) = c3 3

$\underset{x \to c}{l i m}$ f (x) = $\underset{x \to c}{l i m}$ x3 3 = c3 3.

So f is continuous at x $| < |$ 2.

For x = c > 2

f (c) = x2 + 1 = c2 + 1

$\underset{x \to 2}{l i m}$ f (x) = $\underset{x \to 2}{l i m}$ x2 + 1 = c2 + 1 = f (c)

So, f is continuous at x $| > |$ 2.

For x = c = 2, f (2) = 23 3 = 8 3 = 5.

L.H.L. $\underset{x \to 2^{-}}{l i m}$ f (x) = $\underset{x \to 2^{-}}{l i m}$ x3 3 = 23 3 = 5.

R.H.L. $\underset{x \to 2^{+}}{l i m}$ f (x) = $\underset{x \to 2^{+}}{l i m}$ x2 + 1 = 22 + 1 = 5

∴ R.H.L. = L.H.L. = f (2).

So, f is continuous at x = 2

Hence f has no point of discontinuity.

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$= \frac{f (x) \cdot (n_{1} + n_{2})}{(x - 3) (x - 5)} (x - \frac{(5 n_{1} + 3 n_{2})}{n_{1} + n_{2}})$

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10. $f (x) =$ ${\begin{matrix} x^{3} - 3, if x \leq 2 & x^{2} + 1, if x > 2 \end{matrix}$