Maharashtra State BoardHSC Science (Electronics) 11th
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Discuss the continuity of the following function at the point(s) or on the interval indicated against them: f(x) =2x2-2x+5,for 0≤x≤2=1-3x-x21-x,for 2<x<4=x2-25x-5,for 4≤x≤7andx≠5=7,for x=5 - Mathematics and Statistics

Sum

Discuss the continuity of the following function at the point(s) or on the interval indicated against them:

f(x) `{:(= 2x^2 - 2x + 5",", "for"  0 ≤ x ≤ 2),(= (1 - 3x - x^2)/(1 - x) "," , "for"  2 < x < 4),(= (x^2 - 25)/(x - 5)",", "for"  4 ≤ x ≤ 7 and x ≠ 5),(= 7",", "for"  x = 5):}`

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Solution

The domain of f is [0, 7]

For 0 ≤ x ≤ 2, f(x) = 2x2 – 2x + 5, being a polynomial function is continuous.

For 2 < x < 4, f(x) = `(1 - 3x - x^2)/(1 - x)`, being a rational function is continuous except at the point where its denominator 1 – x = 0, i.e., at the point x = 1 which does not belong to (2, 4).

For 4 ≤ x ≤ 7, x ≠ 5, f(x) = `(x^2 - 25)/(x - 5)`, being a rational function is continuous except at the point where its denominator x – 5 = 0, i.e., at the point x = 5.

Continuity at x = 5

f(5) = 7     ...(Given)   ...(1)

`lim_(x -> 5) "f"(x) =  lim_(x -> 5) (x^2 - 25)/(x - 5)`

= `lim_(x -> 5) ((x - 5)(x + 5))/(x - 5)`

= `lim_(x -> 5) (x + 5)`  ...[∵ x → 5, x ≠ 5 ∴ x  – 5 ≠ 0]

= 5 + 5

= 10    ...(2)

From (1) and (2),

`lim_(x -> 5) "f"(x) ≠  "f"(5)`

∴ f is not continuous at x = 5.

Hence, f is continuous on its domain [0, 7] except at the point x = 5, where it is discontinuous.

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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 8 Continuity
Miscellaneous Exercise 8 | Q II. (2) | Page 177
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