Continuity and Differentiability questions
732 published questions with worked solutions.
- If f(x) = x^2 - (a + 2)x + a x - 2 , for x 2, = 2, for x = 2 is continuous at x = 2, then a is
- If f(x) = [4] x - [4] a [3] x - [3] a , for x a, = 3 4 , for x = a is continuous at x = a, then a is
- If f(x) = x ( 1 x ), for x 0, = k, for x = 0 is continuous at x = 0, then k is
- If f(x) = (1 + cx)^ 1/x , for x 0, = e^5, for x = 0 is continious at x = 0, then c is
- If f(x) = cases x^2 - 3x - 2 x - 3 , & 0 x 4 [2mm] x^2 + 1 x - 2 , & 4 < x 6 cases then f is
- If f(x) = |x| x , for x 0, = 0, for x = 0, then f is
- If f is continuous at x = 1 and x = 2, where f(x) = cases x + a, & x < 1 3x + b, & 1 x < 2 x^2, & x 2 cases then a and b are
- If f(x) = x + x x - x in ( - 2 , 2 ) then
- If function f(x) = ( ^2 x)^ ^2 x , x 0, = k, x = 0 is continuous at x = 0, find k.
- Find the value of k, if f(x) = x^6 - 1 64 x^3 - 1 8 , x 1 2 , = k, x = 1 2 is continuous at x = 1 2
- Find the value of k, if the function f(x) is continuous and defined by f(x) = kx x , for x 0, = 4 + x, for x = 0 is
- If f(x) = 2x - 3, x < 1; = ax + b, 1 x 2; = x^2 - 1, x 2 is continues for all x, find a and b
- Find the value of k if f(x) = (e^ kx - 1) kx x^2 , for x 0 and f(0) = 4 is continuous at x = 0
- A function f is defined as follows f(x) = cases 4x 5x + a, & x > 0 x + 4 - b, & x < 0 1, & x = 0 cases is continuous at x = 0,...
- A function defined by f(x) = 3 - x - 3x , x 3 , = k, x = 3 is continuous at x = 3 then the value of k is
- f continuous on [- , ] where f(x) = cases -2 x, & - x < - 2 x + , & - 2 < x < 2 x, & 2 x cases then value of and is
- The function f is defined in [-2, 2] by f(x) = cases x + a, & x < 0 x, & 0 x < 1 b - x, & x 1 cases and is continuous. Then a and...
- If f(x) = x^3 + x^2 - 16x + 20 (x - 2)^2 , for x 2, = a, for x = 2 and f is continuous at x = 2, then a =
- If f(x) = 1 x - 2 + 6x 8 - x^3 , for x 2, = a, for x = 2 and f is continuous at x = 2, then =
- If f(x) = x^ 1/3 + x^ 1/9 - 2 x - 1 , for x 1 = , for x = 1 and f is continuous at x = 1, then =
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