Maharashtra State BoardHSC Science (General) 11th
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Select the correct answer from the given alternative: If f(x) =x2+sinx+1for x≤0=x2-2x+1for x≤0 then - Mathematics and Statistics

MCQ

Select the correct answer from the given alternative:

If f(x) `{:( = x^2 + sin x + 1, "for"  x ≤ 0),(= x^2 - 2x + 1, "for"  x ≤ 0):}` then

Options

  • f is continuous at x = 0, but not differentiable at x = 0

  • f is neither continuous nor differentiable at x = 0

  • f is not continuous at x = 0, but differentiable at x = 0

  • f is both continuous and differentiable at x = 0

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Solution

f is continuous at x = 0, but not differentiable at x = 0

Explanation;

f(x) `{:( = x^2 + sin x + 1,","  x ≤ 0),(= x^2 - 2x + 1,","  x ≤ 0):}` 

`lim_(x -> 0^-) "f"(x) = lim_(x -> 0^-) (x^2 + sinx + 1)` = 0 + 0 + 1 = 1

`lim_(x -> 0^+) "f"(x) = lim_(x -> 0^+) (x^2 - 2x + 1)` = 0 – 0 + 1 = 1

∴ f is continuous at x = 0

Lf'(0) = `lim_("h" -> 0^-) ("f"(0 + "h") - "f"(0))/"h"`

= `lim_("h" -> 0^-) ("h"^2 + sin "h" + 1 - (0 + 0 + 1))/"h"`

= `lim_("h" -> 0^-) ("h" + sin"h"/"h")` = 0 + 1 = 1

Rf'(0) = `lim_("h" -> 0^+) ("f"(0 + "h") - "f"(0))/"h"`

= `lim_("h" -> 0^+) ("h"^2 - 2"h" + 1 - 1)/"h"`

= `lim_("h" -> 0) ("h" - 2)`

= – 2

∵ Rf'(0) ≠ Lf'(0)

∴ f is not differentiable at x = 0.

  Is there an error in this question or solution?
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 9 Differentiation
Miscellaneous Exercise 9 | Q I. (7) | Page 195
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