Competition Level Test: Continuity And Differentiability- 1
30 Questions MCQ Test Mathematics For JEE | Competition Level Test: Continuity And Differentiability- 1
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A function f(x) is defined as below
x ≠ 0 and f(0) = a, f(x) is continuous at x = 0 if a equals
x ≠ 0 and f(0) = a, f(x) is continuous at x = 0 if a equals 
, -1 ≤ x < 0 is continuous in the interval [–1, 1], then `p' is equal to:
, -1 ≤ x < 0 is continuous in the interval [–1, 1], then `p' is equal to:lim f(x) = -½
Apply L hospital
lim(x → 0) p/2(1+px)½ + p/2(1-px)½ = -½
⇒ 2p = -1
p = -1/2
Let f(x) = 
when – 2 ≤ x ≤ 2. Then (where [ * ] represents greatest integer function)
when – 2 ≤ x ≤ 2. Then (where [ * ] represents greatest integer function)At (x = 0) f(0) = |(0+½) .[0]| = 0
f(1) = |(1+½) .[1]| = 3/2
f(2) = |(2+½) .[2]| = 5
f(-1) = |(-1+½) .[-1]| = ½
LHL at x = 2
lim x-->2 f(x) = |(x+½) .[2]|
=|(5/2) * 1| = 5/2
f(2) = limx-->2 f(x) ( continuous at x=2)
Let f(x) = sgn (x) and g(x) = x (x2 – 5x + 6). The function f(g(x)) is discontinuous at
If y = 
where t = 
, then the number of points of discontinuities of y = f(x), x ∈ R is
The equation 2 tan x + 5x – 2 = 0 has
Let f(x) = 2tan x + 5x – 2
Now, f(0) = – 2 < 0
and f(π/4 ) = 2 + 5π/4 – 2
= 5π/5 > 0,
the line intersects the x-axis at x = ⅖
since 0 < ⅖ < π/4
Thus, f(x) = 0 has atleast real solution root in ( 0, π/4)
If f(x) = 
, then indicate the correct alternative(s)
The function f (x) = 1 + | sin x l is
If f(x) = 
be a real valued function then
The function f(x) = sin-1 (cos x) is
Let f(x) be defined in [–2, 2] by f(x) = 
then f(x)
If f(x) is differentiable everywhere, then
Even if f(x) is differentiable everywhere, |f(x)| need not be differentiable everywhere. An example being where the function changes its sign from negative to positive. f(x)=x at x=0B.
∣f∣= {f if f>0 −f if f<0}
So,∣f∣2 = {f2 if f>0 −f2 if f<0}
Differentiating ∣f∣2,
{2ff′ if f>0 −2ff′ if f<0}
At f=0, LHD=RHD=∣f(0)∣2=0, so function is differentiable.
Let f(x + y) = f(x) f(y) all x and y. Suppose that f(3) = 3 and f'(0) = 11 then f'(3) is given by
f(x + y)=f(x) f(y) wrt x we get:
f′(x+y) (1 + y′) = f(x) f′(y) (y′) + f′(x) f(y)
Now on substituting x = 0 and y = 3, we get :
f′(3) (1 + 0)=f(0) f′(3) 0+ f(3) f′(0)
f′(3) = 3 × 11 = 33
If f : R → R be a differentiable function, such that f(x + 2y) = f(x) + f(2y) + 4xy 
x, y ∈ R, then
Let f(x) = x – x2 and g(x) = 
. Then in the interval [0, ¥)
Let [x] denote the integral part of x ∈ R and g(x) = x – [x]. Let f(x) be any continuous function with f(0) = f(1) then the function h(x) = f(g(x))
If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is
If f(x) = 
, then f(x) is
The functions defined by f(x) = max {x2, (x – 1)2, 2x (1 – x)}, 0 £ x £ 1
Let f(x) = x3 – x2 + x + 1 and g(x) = 
then
Let f²(x) be continuous at x = 0 and f²(0) = 4 then value of 
is
Let f : R → R be a function such that f 
, f(0) = 0 and f¢(0) = 3, then
Suppose that f is a differentiable function with the property that f(x + y) = f(x) + f(y) + xy and 
f(h) = 3 then
If a differentiable function f satisfies f 
x, y Î R, find f(x)
Let f : R → R be a function defined by f(x) = Min {x + 1, |x| + 1}. Then which of the following is true ?
f(x) = Min {x + 1,∣x∣ + 1}
x = 0 ⇒ ∣x∣ = x
f(x) = min {x + 1,x + 1}
= x + 1
x < 0,∣x∣ = −x
f(x) = min {x + 1,−x + 1}
−x + 1> x + 1
x < 0
⇒x < 0, f(x) = x + 1
x > 0,f (x) = x + 1
⇒ f(x) = x + 1 for all x
∴f(x) is differential everywhere (continuous and constant slope)
The function f : R /{0} → R given by vf(x) = 
can be made continuous at x = 0 by defining f(0) as
Function f(x) = (|x – 1| + |x – 2| + cos x) where x Î [0, 4] is not continuous at number of points
Let f(x + y) = f(x) f(y) for all x, y, where f(0) ≠ 0. If f'(0) = 2, then f(x) is equal to
A function f : R → R satisfies the equation f(x + y) = f(x) . f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f'(0) = 2 then f'(x) =
Let f(x) = [cos x + sin x], 0 < x < 2p where [x] denotes the greatest integer less than or equal to x. the number of points of discontinuity of f(x) is
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