If f ′′ ( x ) = − f ( x ) and g ( x ) = f ′ ( x ) and F ( x ) = f x 2 2 + g x 2 2 and given that F ( 5 ) = 5 , then F ( 10 ) is
If the function f defined on − 1 3 , 1 3 by f x = 1 x log e 1 + 3 x 1 − 24 , w h e n x ≠ 0 k , w h e n x = 0 is continuous, then k is equal to
Let f ( x ) = sin tan – 1 x + sin cot – 1 x 2 – 1 , | x | > 1 . If dy dx = 1 2 d dx sin – 1 ( f ( x ) ) and y ( 3 ) = π 6 then y ( – 3 ) is equal to:
Let f ( x ) = x | x | . The set of points where f ( x ) is twice differentiable is :
If f ( x ) = x e ( 1 / x ) – e ( – 1 / x ) e ( 1 / x ) + e ( – 1 / x ) ; x ≠ 0 0 ; x = 0 , then which of the following is true
Let f ( x ) = | | x − 1 | − 1 | then all the number of points where f ( x ) is not differentiable is (are)
f ( x ) = min { sin x , cos x } i n 0 , π 4 then f ‘ ( 0 ) =
Let S be the set of points where the functions f x = 2 − x − 3 , x ∈ R , is not differentiable. Then ∑ x ∈ S f f x is equal to .
If f x = sin ( a + 2 ) x + sin x x ; x < 0 b ; x = 0 x + 3 x 2 1 / 3 – x 1 / 3 x 4 / 3 ; x > 0 is continuous at x = 0 , then a + 2 b is equal to:
If f x + y 2 = f ( x ) + f ( y ) 2 for all real x and y and f 1 ( 0 ) exists and equals to – 1 and f ( 0 ) = 1 , then f 1 (x)=
Let f : 0 , π 2 R be a function defined by f ( x ) = max sin x , cosx , 3 4 , then number of points where f(x) is non differentiable is
The number of points at which g ( x ) = 1 1 + 2 f ( x ) is not differentiable, where f ( x ) = 1 1 + 1 x is
Let α ∈ R be such that the function f ( x ) = cos – 1 1 – x 2 sin – 1 1 – x { x } – { x } 3 , x ≠ 0 α , x = 0 is continuous at x = 0 , where { x } = x – [ x ] , [ x ] i s t h e g r e a t e s t i n t e g e r l e s s t h a n o r e q u a l t o x . T h e n :
If f 1 ( x ) = sin ( log x ) and y = f 2 x + 3 3 − 2 x , then d y d x at x = 1 is equal to (Base of all log = e )
N u m b e r o f p o i n t s o f d i s c o n t i n u i t y f o r f ( x )= s g n ( s i n x ), x ∈[0,4 π ] i s
Number of points of non-differentiability of the function g x = x 2 cos 2 4 x + x 2 cos 2 4 x + x 2 sin 2 4 x + x 2 cos 2 4 x + x 2 cos 2 4 x in ( − 50 , 50 ) where [ x ] and { x } denotes the greatest integer function and fractional part function of x respectively, is equal to
If the function g x = e p x + log e ( 1 + 4 x ) + q x 3 , x ≠ 0 is continuous at x = 0 , r , x = 0 then the value of 6 p q r is
If f ( x ) = a | sin x | + b e | x | + c | x | 3 and if f ( x ) is differentiable at x = 0 , then
The function f ( x ) = max { 1 − x , 1 + x , 2 } , x ∈ ( − ∞ , ∞ ) is
The function f ( x ) = ( x − 1 ) | x − 1 | + sin ( | x | ) is
Let f ( x ) = lim n ∞ a x ( x − 1 ) cot π x 4 n + p x 2 + 2 cot π x 4 n + 1 , if x ∈ ( 0 , 1 ) ∪ ( 1 , 2 ) 0 , if x = 1 . If f is differentiable at x = 1 , then the value of | a + p | is
If f ( x + y ) − f ( x ) − f ( y ) = | x | y + x y 2 ∀ x , y ∈ R and f ′ ( 0 ) = 0 then
If f ( x ) = − 3 x + 2 , x < 1 1 2 x 2 + 7 , x ≥ 1 , then which of the following is not true
Let f ( x ) = x 1 + 2 1 / x , if x ≠ 0 0 if x = 0 then f 1 0 − =
If f ( x ) = | x − 1 | − [ x ] (Where [ X ] is greatest integer less than or equal to x ) then
Let f ( x + y ) = f ( x ) f ( y ) and f ( x ) = 1 + ( sin 2 x ) g ( x ) where g ( x ) is continues. Then f 1 ( x ) equals
The number of points of discontinuity of f ( x ) = Lt n ∞ x 2 n − 1 x 2 n + 1 is
Let f ( x ) = [ 3 + 2 cos x ] , x ∈ − π 2 , π 2 where [.] denotes the greatest integer function. Then number of points of discontinuity of f ( x ) is
Let f and g be differentiable functions satisfying g ( a ) = b , g 1 ( a ) = 2 and f o g = I (identity function). Then f 1 ( b ) is equal to
Let g ( x ) = x 2 + x tan x − x tan 2 x a x + tan x − tan 3 x , x ≠ 0 0 , x = 0 If g 1 ( 0 ) exists and is equal to non-zero value b then b is equal to
The function f ( x ) = x tan 2 x sin 3 x ⋅ sin 5 x for x ≠ 0 is continuous at x = 0 then f ( 0 )
If f ( x ) = | x + 2 | tan − 1 ( x + 2 ) , x ≠ − 2 2 , x = – 2 then f ( x ) is
f ( x ) = sin − 1 2 x 1 + x 2 is differentiable on
If f ( x ) = 1 , x < 0 1 + sin x , 0 ≤ x < π 2 , Then f ′ ( 0 ) i s
If f ( x ) = a | sin x | + b e | x | + c | x | 3 andif f ( x ) is differentiableat x = 0 ,then
The function f ( x ) = 1 − 1 − x 2 is
If the derivative of the function f ( x ) = a x 2 + b , x < − 1 b x 2 + a x + 4 , x ≥ − 1 is everywhere continuous, then
If f ( x ) = ( x + 1 ) cot x is continuous at x = 0 , then f ( 0 ) is
The function f ( x ) = e 1 / x − 1 e 1 / x + 1 , x ≠ 0 is 0 , x = 0
The function f ( x ) = | x | + | x − 1 | is
Let f ( x ) = 2 x 3 − 5 . Then, number of points of discontinuity of f ( x ) in ( 1 , 2 )
The function f ( x ) = | x − 3 | , x ≥ 1 x 2 4 − 3 x 2 + 13 4 , x < 1
f ( x ) = | ⌈ x ] x | in − 1 ≤ x ≤ 2 is
The function f ( x ) = [ log ( 1 + a x ) − log ( 1 − b x ) ] x is not defined at x = 0 . The value, which should be assigned to f at x = 0 so that it is continuous at x = 0 , is
f ( x ) = ( 1 + p x ) − ( 1 − p x ) x , − 1 ≤ x < 0 2 x + 1 x − 2 , 0 ≤ x ≤ 1 is continuous in the interval [ − 1 , 1 ] then is equal to
If f ( x ) = x 3 Sgn, then
The function f ( x ) = x 2 / a , 0 ≤ x < 1 a , 1 ≤ x < 2 2 b 2 − 4 b / x 2 , 2 ≤ x < ∞ Is continuous for 0 ≤ x < ∞ , then the most suitable values of a and b are
Let f ′ ( x ) be continuous at x = 0 and f ′′ ( 0 ) = 4 . The value of lim x 0 2 f ( x ) − 3 f ( 2 x ) + f ( 4 x ) x 2
Value of A and B , so that the function f ( x ) defined by f ( x ) = x + A 2 sin x if 0 ≤ x < π 4 , 2 x + cot x + B if π 4 ≤ x < π 2 , becomes continuous are A cos 2 x − B sin x if π 2 ≤ x ≤ π
The values of x where the function f ( x ) = tan x log ( x − 2 ) x 2 − 4 x + 3 is discontinuous are given by
Let [ . ] denote the greatest integer function and f ( x ) = tan 2 x . Then,
Consider the following statements S and R . S : Both sin x and cos x are decreasing functions in π 2 , π R If a differentiable function decreases in an interval ( a , b ) , then its derivative also decreases in ( a , b ) . Which of the following is true?
Let f ( x ) = ∫ 0 x t sin 1 t d t . Then the number of points of discontinuity of the function f ( x ) in the open interval ( 0 , π ) is
Let f ( x ) = 1 | x | , | x | ≥ 1 a x 2 + b , | x | < 1 be continuous and differentiable every where then
The function f ( x ) = x 2 − 1 x 2 − 3 x + 2 + cos ( | x | ) is not differentiable at:
Let f ( x ) = a x 2 + 1 f o r x > 1 x + a f o r x ≤ 1 then f ( x ) is derivable at x = 1 , if
Let f ( x ) = x − 1 2 cos 1 x − 1 − x , x ≠ 1 − 1 , x = 1 . The set of points where f ( x ) is not differentiable is
The set of all points of discontinuity of the function f ( x ) = tan x log x 1 − cos 4 x contains
The domain of the derivative of the function f ( x ) = tan − 1 x if | x | ≤ 1 1 2 ( | x | − 1 ) if | x | > 1
Let f and g be differentiable functions such that f o g = I , the identity function, If g ′ ( a ) = 2 and g ( a ) = b , then f ( b ) =
Let f ( x ) = g ( x ) − g ( a ) x − a , x ≤ a g ′ ( a ) , x = a ,where g is a function derivable at x = a , then at x=a
If f ( x ) = ∫ x + sin x x + cos x d x , then lim x ∞ f ( x ) =
The left hand derivative of f ( x ) = [ x ] sin ( π x ) a t x = k , k an integer, is
If f ( x ) f ( y ) = f ( x + y ) for all x , y , suppose f ( 5 ) = 2 and f 1 ( 0 ) = 3 , t h e n f ( 5 ) i s e q u a l t o
Let f ( x + y ) = f ( x ) f ( y ) and f ( x ) = 1 + ( sin 2 x ) g ( x ) where g ( x ) is continuous. Then g ′ ( x ) is equal to
If f is differentiable function, the value of lim h 0 f x + h 2 − f x 2 2 h is equal to
The function defined by f x = x sin 1 x for x ≠ 0 0 for x = 0 at x = 0 is
f ( x ) = x 3 − log sin x x − 2 to be continuous at x = 0 then f ( 0 ) =
Let f : R R be defined by f ( x ) = α + sin [ x ] x if x > 0 2 if x = 0 Where [ x ] denotes the β + sin x − x x 3 x < 0 integral part of x . If f is continuous at x = 0 then β − α =
f ( x ) = 3 x 2 sin 2 x 2 if x < 0 x 2 + 2 x + c 1 − 3 x 2 if x ≥ 0 , x ≠ 1 3 then in order that f be continuous at x = 0 then 0 if x = 1 3 the value of c is
If f ( x ) be a continuous function for all real values of x and satisfies x 2 + x ( f ( x ) − 2 ) + 2 3 − 3 − 3 f ( x ) = 0 ∀ x ∈ R then the value of f ( 3 ) is
Let f ( x ) = e x − 1 2 n sin n x a log 1 + x a n for x ≠ 0 16 n for x = 0 and f is a continuous at x = 0 then the value of a is
If f ( x ) = x α cos 1 x , if x ≠ 0 0 , if x = 0 is continuous at x = 0 then
f ( x ) = min x , x 2 ∀ x ∈ R then f ( x ) is
Let f ( x ) = ( 256 + a x ) 1 8 − 2 ( 32 + b x ) 1 5 − 2 . If f is continuous at x = 0 then the value of a b is
If [.] denotes the greatest integer function then the number of points where f x = x + x + 1 3 + x + 2 3 is discontinuous for x ∈ 0 , 3 are
The function f ( x ) = cos − 1 ( cos x ) is
If f ( x ) = ( x − a ) g ( x ) and g ( x ) is continuous at x = a then f 1 ( a ) =
The function f ( x ) = [ x ] 2 − x 2 (where [ y ] is the tangent integer ≤ y ) is discontinuous at
If f ( x ) = 1 x 2 − 17 x + 66 then f 2 x − 2 is discontinuous at x is equal to
Let the function f ( x ) = sin 3 x 2 x ∀ x ≠ 0 0 ∀ x = 0
If a function f ( x ) is defined as f ( x ) = x x 2 , x ≠ 0 0 , x = 0 then
If f ( x ) = x e x then at x = 0
If f : R R is even function which is twice differentiable on R and f 11 ( π ) = 1 then f 11 − π is
If f ( x ) = − 3 x + 2 , x < 1 1 2 x 2 + 7 , x ≥ 1 then which of the following is not true
If f ( x ) is a continuous function ∀ x ∈ R and f ( x ) ∈ ( 1 , 30 ) and g ( x ) = f ( x ) a where [.] denotes the greatest integer function is continuous then the least positive integral value of a is
Number if points of non-differentiability of the function g ( x ) = x 2 cos 2 4 x + x 2 cos 2 4 x + x 2 sin 2 4 x + x 2 cos 2 4 x + x 2 cos 2 4 x in ( − 50 , 50 ) where [ x ] and { x } denotes the g.I.F and fractional point function of x respectively is equal to
Let f : R R be defined by f ( x ) = 2 x + 5 if x is irrational 2 x − 7 if x is rational then the number of points of continuity of f ( x ) for x ∈ R
If f ( x ) = x 2 + α for x ≥ 0 2 x 2 + 1 + β for x < 0 is continuous at x = 0 and f 1 2 = 2 then α 2 + β 2 =
If f ( x ) = ( cos x ) 1 sin x for x ≠ 0 k for x = 0 then value of k , so that f is differentiable at x = 0 is
If f : R R be a differentiable function such that f ( x + 2 y ) = f ( x ) + f ( 2 y ) + 4 x y ∀ x , y ∈ R then f 1 ( 1 ) − f 1 ( 0 ) =
If x = sint and y = sinpt then the value of 1 − x 2 d y d x 2 − x d y d x + p 2 y =
If sec x − y x + y = a then d y d x =
y = Tan − 1 1 − x 1 + x then d y d cos − 1 x =
For a real number ‘y’, let [y] denote the integral part of ‘y’. Then derivative of the function f ( x ) = T a n x − π π 1 + x 2 is
Let f be a differentiable function such that 8 f ( x ) + 6 f ( 1 / x ) − x = 5 ( x ≠ 0 ) and y = x 2 f ( x ) then d y d x at x = 1 is
