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The function is continuous at
If the function is continuous at x = 0 then a =
is continuous at x = 0 then
The function is discontinuous at the points
f (x) is discontinuous when x^{2}  3x + 2 = 0
⇒ x^{2}  3x + 2 = 0 ⇒ x = 1, 2
The values of a and b if f is continuous at x = 0, where
is continuous at then k =
so that f(x) is continuous at then
By LHospital rule
where [.] denotes greatest integer function and the function is continuous then
is continuous everywhere. Then the equation whose roots are a and b is
where [x] is the greatest integer function. The function f (x) is
The function is continuous at exactly two points then the possible values of ' a ' are
f (x) is continuous when x^{2}  ax + 3 =2  x
⇒ x^{2}  a 1 x + 1 = 0. This must have two distinct roots ⇒ Δ > 0 ⇒ (a 1)^{2}  4 > 0
If the function is continuous for every x ∈ R then
x^{2} + kx + 1>0 and x^{2}  k must not have any real root ;
∴ k^{2}  4 < 0 &k < 0
⇒ k ∈ [2, 2] and k < 0 ⇒ k ∈ [2, 0)
x is not differentiable at x = 0
x is continuous at x = 0
The function f (x) = cos^{1} (cos x) is
f (x) is continuous at x = π,  π
then which is correct
Let f (x) = x  1 + x + 1
Since g(x) = x is a continuous function and so f is continuous function. In particular f is continuous at a = 1 and x = 4) f is clearly not differentiable at x = 4) Since g(x) = x is not differentiable at x = 0. Now
The set of all points where the function is differentiable is
is not differentiable only at x = 0
If then derivative of f(x) at x = 0 is
If f : R → R be a differentiable function, such that f (x + 2y) = f (x) + f (2y) + 4xy for all x, y ∈ R then
f (x + 2y) = f (x) + f (2 y) + 4xy for x, y ∈ R putting x = y = 0, we get f (0) = 0
Let f be a differentiable function satisfying the condition for all
, then f ' (x) is equal to
replacing x and y both by 1, we get
The function is not differentiable at
By verification f ' (2 ) ≠ f ' (2 +)
∴ f(x) is not differentiable at x = 2
then set of all points where f is differentiable is
The function is clearly differentiable except possible at x = 2, 3
Let h(x) = min {x, x^{2}} for Then which of the following is correct
From the graph it is clear that h is continuous. Also h is differentiable except possible at x = 0 & 1
so h is not differentiable at 1
similarly h' (0 +)= 0 but h ' (0  ) = 1
If f (x + y) = 2f (x) f (y) for all x, y ∈ R where f ' (0) = 3 and f (4) = 2, then f ' (4) is equal to
If and f ' (0) = 1, f (0) = 1, then f (2) =
Take f (x) = ax+ b
Let f (x) be differentiable function such that and y. If
Let f : R → R be a function defined by f (x) = min {x + 1, x + 1}, Then which of the following is true?
209 videos218 docs139 tests

209 videos218 docs139 tests
