मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

For what values of a and b is the function f(x) =x2-4x-2,for x<2=ax2-bx+3,for 2≤x<3=2x-a+b,for x≥3} continuous for every x on R?

Advertisements
Advertisements

प्रश्न

For what values of a and b is the function

f(x) `{:(= (x^2 - 4)/(x - 2)",", "for"  x < 2),(= "a"x^2 - "b"x + 3",", "for"  2 ≤ x < 3),(= 2x - "a" + "b"",", "for"  x ≥ 3):}}` continuous for every x on R?

बेरीज
Advertisements

उत्तर

f(x) is continuous for every x on R.

∴ f(x) is continuous at x = 2 and x = 3.

f(x) is continuous at x = 2.

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

∴ `lim_(x -> 2) (x^2 - 4)/(x - 2) =  lim_(x -> 2) ("a"x^2 - "b"x + 3)`

∴ `lim_(x -> 2) ((x - 2)(x + 2))/(x - 2) =  lim_(x -> 2) ("a"x^2 - "b"x + 3)`

∴ `lim_(x -> 2)(x + 2) =  lim_(x -> 2) ("a"x^2 - "b"x + 3)  ...[(because x -> 2  therefore x ≠ 2),(therefore x - 2 ≠ 0)]`

∴ 2 + 2 = a(2)2 – b(2) + 3

∴ 4a – 2b + 3 = 4

∴ 4a – 2b = 1   ...(i)

Also f(x) is continuous at x = 3

∴ `lim_(x -> 3^-) "f"(x) = lim_(x -> 3^+) "f"(x)`

∴ `lim_(x -> 3)("a"x^2 - "b"x + 3) =  lim_(x -> 3) (2x - "a" + "b")`

∴ a(3)2 – b(3) + 3 = 2(3) – a + b

∴ 9a – 3b + 3 = 6 – a + b

∴ 10a – 4b = 3    ...(ii)

Multiply (i) by 2 and (ii) by 1, we get

8a – 4b = 2    ...(iii)

10a – 4b = 3   ...(iv)

Subtracting (iv) from (iii)

– 2a = – 1

∴ a = `1/2`

Substituting a = `1/2` in (i), we get

`4(1/2) - 2"b"` = 1

∴ 2 – 2b = 1

∴ 1 = 2b

∴ b = `1/2`

∴ a = `1/2` and b = `1/2`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Continuity - EXERCISE 8.1 [पृष्ठ १७४]

APPEARS IN

संबंधित प्रश्‍न

Examine whether the function is continuous at the points indicated against them :

f(x) `{:(= x/(tan3x) + 2",",   "for"  x < 0),(= 7/3",",  "for"  x ≥ 0):}}  "at"  x = 0`


Find all the point of discontinuities of f(x) = [x] on the interval (− 3, 2).


Test the continuity of the following function at the point or interval indicated against them :

f(x) `{:(= 4x + 1",",  "for"  x ≤  8/3),(= (59 - 9x)/3 ",",  "for"  x > 8/3):}}  "at"  x = 8/3`


Show that following function have continuous extension to the point where f(x) is not defined. Also find the extension :

f(x) = `(x^2 - 1)/(x^3 + 1)` for x ≠ – 1


Discuss the continuity of the following function at the point indicated against them :

f(x) = `{:(=( sqrt(3) - tanx)/(pi - 3x)",", x ≠ pi/3),(= 3/4",", x = pi/3):}}  "at"  x = pi/3`


Discuss the continuity of the following function at the point indicated against them :

f(x)  `{:(=(4^x - 2^(x + 1) + 1)/(1 - cos 2x)",",  "for"  x ≠ 0),(= (log 2)^2/2",",  "for"  x = 0):}}` at x = 0.


The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it become continuous :

f(x) `{:(= log_((1 + 3x)) (1 + 5x)",", "for"  x > 0),(=(32^x - 1)/(8^x - 1)",",  "for"  x < 0):}}` at x = 0


If f(x) = `(cos^2 x - sin^2 x - 1)/(sqrt(3x^2 + 1) - 1)` for x ≠ 0, is continuous at x = 0 then find f(0)


If f(x) = `(4^(x - π) + 4^(π - x) - 2)/(x - π)^2` for x ≠ π, is continuous at x = π, then find f(π).


If f(x) `{:(= (24^x - 8^x - 3^x + 1)/(12^x - 4^x - 3^x + 1)",",  "for"  x ≠ 0), (= "k"",",  "for"  x = 0):}}` is continuous at x = 0, find k


Show that there is a root for the equation x3 − 3x = 0 between 1 and 2.


Let f(x) = ax + b (where a and b are unknown)

= x2 + 5 for x ∈ R

Find the values of a and b, so that f(x) is continuous at x = 1


Suppose f(x) `{:(= "p"x + 3",", "for"  "a" ≤ x ≤ "b"),(= 5x^2 − "q"",", "for"  "b" < x ≤ "c"):}`

Find the condition on p, q, so that f(x) is continuous on [a, c], by filling in the blanks.

f(b) = ______

`lim_(x -> "b"^+) "f"(x)` = _______

∴ pb + 3 = _______ − q

∴ p = `"_____"/"b"` is the required condition


Select the correct answer from the given alternatives:

If f(x) `{:(= "a"x^2 + "b"x + 1",", "for"  |x −1| ≥ 3), (= 4x + 5",", "for"  -2 < x < 4):}` is continuous everywhere then,


Select the correct answer from the given alternatives:

f(x) `{:(= ((16^x - 1)(9^x - 1))/((27^x - 1)(32^x - 1))",", "for"  x ≠ 0),(= "k"",", "for"  x = 0):}` is continuous at x = 0, then ‘k’ =


Select the correct answer from the given alternatives:

If f(x) = [x] for x ∈ (–1, 2) then f is discontinuous at


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

f(x) `{:(= (|x + 1|)/(2x^2 + x - 1)",", "for"  x ≠ -1),(= 0",", "for"  x = -1):}}` at x = – 1


Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:

f(x) `{:(= x^2 + 5x + 1"," , "for"  0 ≤ x ≤ 3),(= x^3 + x + 5"," , "for"  3 < x ≤ 6):}`


Find f(a), if f is continuous at x = a where,

f(x) = `(1 - cos[7(x - pi)])/(5(x - pi)^2`, for x ≠ π at a = π


Solve using intermediate value theorem:

Show that 5x − 6x = 0 has a root in [1, 2]


If f(x) = `{:{(tan^-1|x|; "when"  x ≠ 0), (pi/4;  "when"  x = 0):}`, then ______ 


If f(x) is continuous at x = 3, where

f(x) = ax + 1, for x ≤ 3

= bx + 3, for x > 3 then.


If f(x) = `{{:((sin5x)/(x^2 + 2x)",", x ≠ 0),(k + 1/2",", x = 0):}` is continuous at x = 0, then the value of k is ______.


If the function f(x) defined by

f(x) = `{{:(x sin  1/x",", "for"  x = 0),(k",", "for"  x = 0):}`

is continuous at x = 0, then k is equal to ______.


The function f(x) = x – |x – x2| is ______.


If f(x) = `{{:((x - 4)/(|x - 4|) + a",",  "for"  x < 4),(a + b",",  "for"  x = 4  "is continuous at"  x = 4","),((x - 4)/(|x - 4|) + b",",  "for"  x > 4):}`

then ______.


For x > 0, `lim_(x rightarrow 0) ((sin x)^(1//x) + (1/x)^sinx)` is ______.


If f(x) = `{{:((3 sin πx)/(5x),",", x ≠ 0),(2k,",", x = 0):}`

is continuous at x = 0, then the value of k is ______.


Which condition defines continuity of a real-valued function \[f\] at \[x=c\]?


Which one-sided-limit condition is required for continuity at \[x=c\]?


Which is a necessary condition for continuity at \[x=c\]?


For a piecewise function, at which points should continuity be checked especially?


For \[f(x)=x^2\], what verifies continuity at \[x=0\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×