Competition Level Test: Continuity And Differentiability- 2
30 Questions MCQ Test Mathematics (Maths) Class 12 | Competition Level Test: Continuity And Differentiability- 2
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The function 
is continuous at
is continuous at 
If the function 
is continuous at x = 0 then a =
is continuous at x = 0 then a = 
is continuous at x = 0 then
is continuous at x = 0 then
The function 
is discontinuous at the points
f (x) is discontinuous when x2 - 3|x| + 2 = 0
⇒ |x|2 - 3|x| + 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 L-Hospital 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 x2 - ax + 3 =2 - x
⇒ x2 - 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
x2 + kx + 1>0 and x2 - k must not have any real root ;
∴ k2 - 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, x2} 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?
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