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Continuity and Differentiability

Short Answer Type

451.

$Let f (x) = x, [\frac{x}{2}] for - 10 < x < 10, where [t] denotes the greatest integer function . Then the number of points of discontinuity of f is equal to____$

452.
$Let f : R \to R be defined as f (x) = \{x^{5} \sin (\frac{1}{x}) + 5 x^{2}, x < 0 0, x = 0 x^{5} \cos (\frac{1}{x}) + {λx}^{2}, x > 0\} The value of λ for which f'' (0) exists, is$

Ans : 5

$f' (x) = \{5 x^{4} \sin (\frac{1}{x}) - x^{3} \cos (\frac{1}{x}) + 10 x, x < 0 0, x = 0 5 x^{4} \cos (\frac{1}{x}) + x^{3} \sin (\frac{1}{x}) + 2 λx, x > 0\} f'' (x) = (20 x^{3} \sin (\frac{1}{x}) - 5 x^{2} \cos (\frac{1}{x}) - 3 x^{2} \cos (\frac{1}{x}) - xsin (\frac{1}{x}) + 10, x < 0 0, x = 0 20 x^{3} \cos (\frac{1}{x}) + 5 x^{2} \sin (\frac{1}{x}) + 3 x^{2} \sin (\frac{1}{x}) - xcos (\frac{1}{x}) + 2 λ, x > 0) f'' (0^{+}) = f'' (0^{-}) \Rightarrow 2 λ = 10 \Rightarrow λ = 5$

Previous Year Papers

Test Series

Test Yourself

Short Answer Type

452. Let f : R → R be defined asfx = x5sin1x + 5x2, x < 00, x = 0x5cos1x + λx2, x > 0The value of λ for which f''0 exists, is

452.
$Let f : R \to R be defined as f (x) = \{x^{5} \sin (\frac{1}{x}) + 5 x^{2}, x < 0 0, x = 0 x^{5} \cos (\frac{1}{x}) + {λx}^{2}, x > 0\} The value of λ for which f'' (0) exists, is$