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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
, 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(1px)^{½ }= ½
⇒ 2p = 1
p = 1/2
Let f(x) = 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 (x^{2} – 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 xaxis 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} = {f^{2} if f>0 −f^{2} 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 – x^{2} 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 {x^{2}, (x – 1)^{2}, 2x (1 – x)}, 0 £ x £ 1
Let f(x) = x^{3} – x^{2} + 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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