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All questions of Chapter 5 - Continuity and Differentiability for JEE Exam
Therefore, f(x) continuous everywhere.
If the function f (x) = x2– 8x + 12 satisfies the condition of Rolle’s Theorem on (2, 6), find the value of c such that f ‘(c) = 0- a)6
- b)4
- c)8
- d)2
Correct answer is option 'B'. Can you explain this answer?
If the function f (x) = x2– 8x + 12 satisfies the condition of Rolle’s Theorem on (2, 6), find the value of c such that f ‘(c) = 0
a)
6
b)
4
c)
8
d)
2
|
Dr Manju Sen answered |
f (x) = x2 - 8x + 12
Function satisfies the condition of Rolle's theorem for (2,6).
We need to find c for which f’(c) = 0
f’(x) = 2x – 8
f’(c) = 2c – 8 = 0
c = 4
Function satisfies the condition of Rolle's theorem for (2,6).
We need to find c for which f’(c) = 0
f’(x) = 2x – 8
f’(c) = 2c – 8 = 0
c = 4
When Rolle’s Theorem is verified for f(x) on [a, b] then there exists c such that- a)c ε [a, b] such that f'(c) = 0
- b)c ε (a, b) such that f'(c) = 0
- c)c ε (a, b] such that f'(c) = 0
- d)c ε [a, b) such that f'(c) = 0
Correct answer is option 'B'. Can you explain this answer?
When Rolle’s Theorem is verified for f(x) on [a, b] then there exists c such that
a)
c ε [a, b] such that f'(c) = 0
b)
c ε (a, b) such that f'(c) = 0
c)
c ε (a, b] such that f'(c) = 0
d)
c ε [a, b) such that f'(c) = 0
|
Arun Yadav answered |
Answer is
B) c ∈ (a, b) such that f'(c) = 0.
Statement for Rolle’s Theorem :
Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then if f(a)=f(b), then there exists at least one point c in the open interval (a,b) for which f′(c)=0.
B) c ∈ (a, b) such that f'(c) = 0.
Statement for Rolle’s Theorem :
Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then if f(a)=f(b), then there exists at least one point c in the open interval (a,b) for which f′(c)=0.
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
a)
b)
c)
d)
|
Lavanya Menon answered |
On differentiating both sides with respect to x
Differentiate sin2(θ2 + 1) with respect to θ2- a)sin(2θ2 + 1)
- b)cos(2θ2 + 2)
- c)sin(2θ2 + 2)
- d)cos(2θ2 + 1)
Correct answer is option 'C'. Can you explain this answer?
Differentiate sin2(θ2 + 1) with respect to θ2
a)
sin(2θ2 + 1)
b)
cos(2θ2 + 2)
c)
sin(2θ2 + 2)
d)
cos(2θ2 + 1)
|
Gunjan Lakhani answered |
y = sin2(θ2+1)
v = θ2
dy/d(v) = dydθ/dvdθ
dy/dthη = sin2(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ2+1)cos(θ2+1)
= sin2(θ2+1).
v = θ2
dy/d(v) = dydθ/dvdθ
dy/dthη = sin2(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ2+1)cos(θ2+1)
= sin2(θ2+1).
- a)3 cos2 x. cos 4x
- b)cos 3x. cos3x
- c)-9 cos 3x. cos2x sinx
- d)3 sin2x.sin 4x
Correct answer is option 'D'. Can you explain this answer?
a)
3 cos2 x. cos 4x
b)
cos 3x. cos3x
c)
-9 cos 3x. cos2x sinx
d)
3 sin2x.sin 4x
|
Deepak Kapoor answered |
Given: y = sin 3x . sin3
= 3 sin2x.sin 4x
= 3 sin2x.sin 4x
Whta is the derivatve of y = log5 (x)- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Whta is the derivatve of y = log5 (x)
a)
b)
c)
d)
|
Tanuja Kapoor answered |
y = log5 x = ln x/ln 5 → change of base
= ln x/ln 5
dy/dx = 1/ln5⋅1/x → 1/ln5 is a constant, so we don't change it
= 1/(x ln 5)
= ln x/ln 5
dy/dx = 1/ln5⋅1/x → 1/ln5 is a constant, so we don't change it
= 1/(x ln 5)
If xy = 2, then dy/dx is- a)-x2/2
- b)y2
- c)-2/x2
- d)-y/x
Correct answer is option 'D'. Can you explain this answer?
If xy = 2, then dy/dx is
a)
-x2/2
b)
y2
c)
-2/x2
d)
-y/x
|
Hansa Sharma answered |
xy = 2
x dy/dx + y = 0
x dy/dx = -y
dy/dx = -y/x
x dy/dx + y = 0
x dy/dx = -y
dy/dx = -y/x
A real function f is said to be continuous if it is continuous at every point in …… .- a)[-∞,∞]
- b)The range of f
- c)The domain of f
- d)Any interval of real numbers
Correct answer is option 'A'. Can you explain this answer?
A real function f is said to be continuous if it is continuous at every point in …… .
a)
[-∞,∞]
b)
The range of f
c)
The domain of f
d)
Any interval of real numbers
|
Saranya Choudhury answered |
Its domain. This means that for any point x in the domain of f, as x approaches a certain value a, the value of f(x) approaches f(a). In other words, there are no sudden jumps or gaps in the graph of f.
More formally, a function f is continuous at a point a if:
1. f(a) is defined (i.e. a is in the domain of f).
2. The limit of f(x) as x approaches a exists (i.e. the left and right-hand limits are equal).
3. The limit of f(x) as x approaches a is equal to f(a).
If a function is continuous at every point in its domain, it is called a continuous function. Continuous functions have many useful properties and are often used in mathematical models and real-world applications.
More formally, a function f is continuous at a point a if:
1. f(a) is defined (i.e. a is in the domain of f).
2. The limit of f(x) as x approaches a exists (i.e. the left and right-hand limits are equal).
3. The limit of f(x) as x approaches a is equal to f(a).
If a function is continuous at every point in its domain, it is called a continuous function. Continuous functions have many useful properties and are often used in mathematical models and real-world applications.
Derivatve of f(x) 
is given by
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Derivatve of f(x) is given by
is given bya)
b)
c)
d)
|
Neha Sharma answered |
y
The derivative of y = ef(x)is dy/dx = f'(x)ef(x)
In this case, f(x) = x2 , and the derivative of x2 = 2x
Therefore, f'(x)= 2x,
dy/dx = 2x
The derivative of y = ef(x)is dy/dx = f'(x)ef(x)
In this case, f(x) = x2 , and the derivative of x2 = 2x
Therefore, f'(x)= 2x,
dy/dx = 2x
Which of the following functions are not continuous.- a)
- b)
- c)ex
- d)
Correct answer is option 'A'. Can you explain this answer?
Which of the following functions are not continuous.
a)
b)
c)
ex
d)
|
Anu answered |
b,c,and d are continuous and [x] is discontinuous at integer points.
If 3 sin(xy) + 4 cos (xy) = 5, then 
= .....- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If 3 sin(xy) + 4 cos (xy) = 5, then = .....
= .....a)
b)
c)
d)
|
Krishna Iyer answered |
3sinxy + 4cosxy = 5
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
so let, 3/5 = cosA
⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
⇒ (cosAsinxy + sinAcosxy) = 1
⇒ sin(A+xy) = 1
⇒ A + xy = 2πk + π/2 (k is any integer)
⇒ sin⁻¹(4/5) + xy = 2πk + π/2
differenciating both sides with respect to x
0 + xdy/dx + y = 0
dy/dx = -y/x
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
so let, 3/5 = cosA
⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
⇒ (cosAsinxy + sinAcosxy) = 1
⇒ sin(A+xy) = 1
⇒ A + xy = 2πk + π/2 (k is any integer)
⇒ sin⁻¹(4/5) + xy = 2πk + π/2
differenciating both sides with respect to x
0 + xdy/dx + y = 0
dy/dx = -y/x
y = log(sec + tan x)
- a)sec x tan x – 1
- b)sec x
- c)sec x tan x + 1
- d)tan x
Correct answer is option 'B'. Can you explain this answer?
y = log(sec + tan x)
a)
sec x tan x – 1
b)
sec x
c)
sec x tan x + 1
d)
tan x
|
|
Lavanya Menon answered |
y = log(secx + tanx)
dy/dx = 1/(secx + tanx){(secxtanx) + sec2x}
= secx(secx + tanx)/(secx + tanx)
= secx
dy/dx = 1/(secx + tanx){(secxtanx) + sec2x}
= secx(secx + tanx)/(secx + tanx)
= secx
For what values of a and b, f is a continuous function.- a)a=2,b=0
- b)a=1,b=0
- c)a=0,b=2
- d)a=0,b=0
Correct answer is 'A'. Can you explain this answer?
For what values of a and b, f is a continuous function.a)
a=2,b=0
b)
a=1,b=0
c)
a=0,b=2
d)
a=0,b=0
|
|
Tejas Verma answered |
For continuity: LHL=RHL
at x=2,
LHL: x < 2 ⇒ f(x) = 2*a
RHL: x ≥ 2 ⇒ f(x) = 4
For continuity: LHL = RHL
⇒ 2a = 4 ⇒ a = 2
LHL: x < 2 ⇒ f(x) = 2*a
RHL: x ≥ 2 ⇒ f(x) = 4
For continuity: LHL = RHL
⇒ 2a = 4 ⇒ a = 2
at x = 0,
LHL: x < 0 ⇒ f(x) = b
RHL: x ≥ 0 ⇒ f(x) = 0 * a
For continuity: LHL = RHL
⇒ b = 0
LHL: x < 0 ⇒ f(x) = b
RHL: x ≥ 0 ⇒ f(x) = 0 * a
For continuity: LHL = RHL
⇒ b = 0
F(x) = tan (log x)
F'(x) =- a)sec2 (logx)
- b)
- c)
- d)-x sec2 (logx)
Correct answer is option 'B'. Can you explain this answer?
F(x) = tan (log x)
F'(x) =
F'(x) =
a)
sec2 (logx)
b)
c)
d)
-x sec2 (logx)
|
Virendra Singh answered |
I have chosen (b) but it is showing (d) is correct
Correct answer is option 'A'. Can you explain this answer?
|
|
Aryan Khanna answered |
y = tan-1(1-cosx)/sinx
y = tan-1{2sin2(x/2)/(2sin(x/2)cos(x/2)}
y = tan-1{tan x/2}
y = x/2 => dy/dx = 1/2
y = tan-1{2sin2(x/2)/(2sin(x/2)cos(x/2)}
y = tan-1{tan x/2}
y = x/2 => dy/dx = 1/2
Function f(x) = log x + 
is continuous at- a)(0,1)
- b)[-1,1]
- c)(0,∞)
- d)(0,1]
Correct answer is option 'D'. Can you explain this answer?
Function f(x) = log x + is continuous at
is continuous ata)
(0,1)
b)
[-1,1]
c)
(0,∞)
d)
(0,1]
|
Om Desai answered |
- [-1,1] cannot be continuous interval because log is not defined at 0.
- The value of x cannot be greater than 1 because then the function will become complex.
- (0,1) will not be considered because its continuous at 1 as well. Hence D is the correct option.
Find the derivate of y = sin4x + cos4x- a)– sin 2x
- b)4 sin3 x + 4 cos3 x
- c)– sin 4x
- d)4 sin x cos x cos 2x
Correct answer is option 'C'. Can you explain this answer?
Find the derivate of y = sin4x + cos4x
a)
– sin 2x
b)
4 sin3 x + 4 cos3 x
c)
– sin 4x
d)
4 sin x cos x cos 2x
|
Mohit Rajpoot answered |
y=sin4x, and z=cos4x
So by using chain rule
df(x)/dx = dsin4x/dx + dcos4x/dx
=dy4/dy * dy/dx + dz4/dz * dzdx
=dy4/dy * dsinxdx + dz4/dz * dcosx/dx
=4y(4−1)⋅cosx+4z(4−1)⋅(−sinx)
=4sin3xcosx − 4cos3xsinx
=4sinxcosx(sin2x − cos2x)
=2sin2x(−cos2x)
=−2sin2xcos2x
=−sin4x
So by using chain rule
df(x)/dx = dsin4x/dx + dcos4x/dx
=dy4/dy * dy/dx + dz4/dz * dzdx
=dy4/dy * dsinxdx + dz4/dz * dcosx/dx
=4y(4−1)⋅cosx+4z(4−1)⋅(−sinx)
=4sin3xcosx − 4cos3xsinx
=4sinxcosx(sin2x − cos2x)
=2sin2x(−cos2x)
=−2sin2xcos2x
=−sin4x
The derivative of 2x tan x is- a)2x log 2 [ sec2 x + tan x]
- b)2x tan x [sec x + log 2]
- c)2x [sec2 x + log 2 tan x]
- d)2x [sec2 x + tan x]
Correct answer is option 'C'. Can you explain this answer?
The derivative of 2x tan x is
a)
2x log 2 [ sec2 x + tan x]
b)
2x tan x [sec x + log 2]
c)
2x [sec2 x + log 2 tan x]
d)
2x [sec2 x + tan x]
|
Sushil Kumar answered |
dy/dx = 2x(tanx)' + tanx (2x)'
dy/dx = 2x (secx)2 + 2x tanx log2 , {since , (ax)' = ax loga (x)'}
dy/dx = 2x[(Secx)2 + log2 tanx]
dy/dx = 2x (secx)2 + 2x tanx log2 , {since , (ax)' = ax loga (x)'}
dy/dx = 2x[(Secx)2 + log2 tanx]
Can you explain the answer of this question below:
- A:
1/2
- B:
0
- C:
1
- D:
none of these
The answer is b.
1/2
0
1
none of these
|
Shivani answered |
Apply L-hospital rule.....u will get the ans....
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
a)
b)
c)
d)
|
Poonam Reddy answered |
y + sin y = 5x
dy/dx + cos ydy/dx = 5
dy/dx = 5/(1+cos y)
dy/dx + cos ydy/dx = 5
dy/dx = 5/(1+cos y)
If x sin (a + y) = sin y, then 
is equal to- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If x sin (a + y) = sin y, then is equal to
is equal toa)
b)
c)
d)
|
Swati Verma answered |
x sin(a+y) = sin y
⇒
⇒
:
Direction: Read the following text and answer the following questions on the basis of the same:Ms. Remka of city school is teaching chain rule to her students with the help of a flow-chartThe chain rule says that if h and g are functions and f(x) = g(h(x)), then
Let f(x) = sin x and g(x) = x3d/dx sin x3 _______.- a)cos (x3)
- b)–cos (x3)
- c)3x2 sin (x3)
- d)3x2 cos (x3)
Correct answer is option 'D'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
Ms. Remka of city school is teaching chain rule to her students with the help of a flow-chart
The chain rule says that if h and g are functions and f(x) = g(h(x)), then
Let f(x) = sin x and g(x) = x3
d/dx sin x3 _______.
a)
cos (x3)
b)
–cos (x3)
c)
3x2 sin (x3)
d)
3x2 cos (x3)
|
Varun Kapoor answered |
= 3x2 cosx3
Geometrically the Mean Value theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is- a)Parallel to the x axis
- b)Parallel to the y axis
- c)Parallel to the line joining the end points of the curve
- d)Parallel to the line y = x
Correct answer is option 'C'. Can you explain this answer?
Geometrically the Mean Value theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is
a)
Parallel to the x axis
b)
Parallel to the y axis
c)
Parallel to the line joining the end points of the curve
d)
Parallel to the line y = x
|
Ujjwal Gupta answered |
No, option C is correct answer.
Examine the continuity of function 
- a)Discontinuous at x=1,2
- b)Discontinuous at x=1
- c)Continuous everywhere.
- d)Discontinuous at x=2
Correct answer is option 'C'. Can you explain this answer?
Examine the continuity of function
a)
Discontinuous at x=1,2
b)
Discontinuous at x=1
c)
Continuous everywhere.
d)
Discontinuous at x=2
|
|
Om Desai answered |
Lim f (x) = lim (x-1)(x-2) at x tend to k
► So it get k2-3k+2
► Now f (k) = k2 -3k+2
► So f (x) =f (k) so continous at everywhere
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
a)
b)
c)
d)
|
Suhani Dangarh answered |
Put x=tan thita. then you will get. 2 tan inverse x then differentiate
If f(x) = x + cot x, 
- a)-4
- b)2
- c)4
- d)-2
Correct answer is option 'C'. Can you explain this answer?
If f(x) = x + cot x,
a)
-4
b)
2
c)
4
d)
-2
|
|
Aryan Khanna answered |
f(x) = x + cot x
f’(x) = 1 + (-cosec2 x)
f”(x) = 0 - 2cosec x(-cosec x cot x)
= 2 cosec2 x cot x
f”(π/4) = 2 cosec2 (π/4) cot(π/4)
= 2 [(2)^½]2 (1)
= 4
f’(x) = 1 + (-cosec2 x)
f”(x) = 0 - 2cosec x(-cosec x cot x)
= 2 cosec2 x cot x
f”(π/4) = 2 cosec2 (π/4) cot(π/4)
= 2 [(2)^½]2 (1)
= 4
If f(x) = ex, then the value of f'(-3) is- a)log (3)
- b)e2
- c)log (-3)
- d)e-3
Correct answer is option 'D'. Can you explain this answer?
If f(x) = ex, then the value of f'(-3) is
a)
log (3)
b)
e
2
c)
log (-3)
d)
e-3
|
Naincy Tripathi answered |
As , f (x) = e^x now put (-3)in place of x as it is asking f (-3) f (-3) = e^-3
Differentiate 
with respect to x.- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Differentiate with respect to x.
with respect to x.a)
b)
c)
d)
|
|
Tejas Verma answered |
y = e^(-x)2.................(1)
Put u = (-x)2
du/dx = -2x dx
Differentiating eq(1) y = eu
dy/du = e^u
⇒ dy/dx = (dy/du) * (du/dx)
= (eu) * (-2x)
⇒ - 2xe(-x2)
Put u = (-x)2
du/dx = -2x dx
Differentiating eq(1) y = eu
dy/du = e^u
⇒ dy/dx = (dy/du) * (du/dx)
= (eu) * (-2x)
⇒ - 2xe(-x2)
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
Rocky Gupta answered |
X^a y^b = (x + y)^(a + b)
taking ln on both sides :-
alnx + b lny = (a + b) ln(x + y)
diff both sides w.r.t x :-
a/x + by'/y = ( (a + b)/(x + y) ) + (((a + b) y'))/(x + y)
or,
a/x - (a +b)/(x + y) = y'[((a + b) / (x + y)) - b/y]
or,
(ax + ay - ax - bx)/x = y' [ (ay + by - bx - by)/y] (cancel (x + y))
or,
y' = dy/dx = ( y (ay - bx) )/(x( ay - bx)) = y/x
therefore we can easily say that the option (D) is the correct answer
taking ln on both sides :-
alnx + b lny = (a + b) ln(x + y)
diff both sides w.r.t x :-
a/x + by'/y = ( (a + b)/(x + y) ) + (((a + b) y'))/(x + y)
or,
a/x - (a +b)/(x + y) = y'[((a + b) / (x + y)) - b/y]
or,
(ax + ay - ax - bx)/x = y' [ (ay + by - bx - by)/y] (cancel (x + y))
or,
y' = dy/dx = ( y (ay - bx) )/(x( ay - bx)) = y/x
therefore we can easily say that the option (D) is the correct answer
- a)2t
- b)1/2a
- c)-t/2a
- d)t/4a
Correct answer is option 'B'. Can you explain this answer?
a)
2t
b)
1/2a
c)
-t/2a
d)
t/4a
|
|
Om Desai answered |
y = at4, x = 2at2
dy/dt = 4at3 dx/dt = 4at => dt/dx = 1/4at
Divide dy/dt by dx/dt, we get
dy/dx = t2
d2 y/dx2 = 2t dt/dx……………….(1)
Put the value of dt/dx in eq(1)
d2 y /dx2 = 2t(1/4at)
= 1/2a
dy/dt = 4at3 dx/dt = 4at => dt/dx = 1/4at
Divide dy/dt by dx/dt, we get
dy/dx = t2
d2 y/dx2 = 2t dt/dx……………….(1)
Put the value of dt/dx in eq(1)
d2 y /dx2 = 2t(1/4at)
= 1/2a
The differential coefficient of (log x)tanx is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The differential coefficient of (log x)tanx is:
a)
b)
c)
d)
|
Gaurav Kumar answered |
y = (lnx)tanx
ln(y) = tanx(ln(lnx))
d(lny)/dx = d(tanx(ln(lnx)))/dx
Using product rule -
(1/y)dy/dx = (secx)2(ln(lnx)) + (1÷xlnx)tanx
dy/dx = [(secx)2(ln(lnx)) + (1÷xlnx)tanx ]×y
dy/dx = [(secx)2(ln(lnx)) + (1÷xlnx)tanx ]×[(lnx)tanx]
ln(y) = tanx(ln(lnx))
d(lny)/dx = d(tanx(ln(lnx)))/dx
Using product rule -
(1/y)dy/dx = (secx)2(ln(lnx)) + (1÷xlnx)tanx
dy/dx = [(secx)2(ln(lnx)) + (1÷xlnx)tanx ]×y
dy/dx = [(secx)2(ln(lnx)) + (1÷xlnx)tanx ]×[(lnx)tanx]
f (x) = max {x, x3},then the number of points where f (x) is not differentiable, are- a)1
- b) 2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
f (x) = max {x, x3},then the number of points where f (x) is not differentiable, are
a)
1
b)
2
c)
3
d)
4
|
|
Gaurav Rane answered |
Solution:
To find the points where the given function is not differentiable, we need to find the points where the function is not continuous or where the derivative does not exist.
Step 1: Finding the points of non-differentiability where the function is not continuous.
Since the function is the maximum of two functions, we need to consider the points where these two functions are equal.
Let x = x³, then x³ - x = 0, which gives x = 0, 1.
Therefore, the function is not continuous at x = 0 and x = 1.
Step 2: Finding the points of non-differentiability where the derivative does not exist.
At x = 0, the left-hand derivative is 1 and the right-hand derivative is 0.
At x = 1, the left-hand derivative is 3 and the right-hand derivative is 0.
Therefore, the function is not differentiable at x = 0 and x = 1.
Hence, the function f(x) = max{x, x³} is not differentiable at three points, namely x = 0, x = 1, and wherever the function changes from x to x³ or vice versa.
Therefore, the correct option is (C) 3.
To find the points where the given function is not differentiable, we need to find the points where the function is not continuous or where the derivative does not exist.
Step 1: Finding the points of non-differentiability where the function is not continuous.
Since the function is the maximum of two functions, we need to consider the points where these two functions are equal.
Let x = x³, then x³ - x = 0, which gives x = 0, 1.
Therefore, the function is not continuous at x = 0 and x = 1.
Step 2: Finding the points of non-differentiability where the derivative does not exist.
At x = 0, the left-hand derivative is 1 and the right-hand derivative is 0.
At x = 1, the left-hand derivative is 3 and the right-hand derivative is 0.
Therefore, the function is not differentiable at x = 0 and x = 1.
Hence, the function f(x) = max{x, x³} is not differentiable at three points, namely x = 0, x = 1, and wherever the function changes from x to x³ or vice versa.
Therefore, the correct option is (C) 3.
If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is
- a) Continuous at x = 0
- b) Continuous in (-1, 0)
- c) Differentiable at x = 1
- d) Differentiable in (-1, 1)
Correct answer is option 'A'. Can you explain this answer?
If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is
a)
Continuous at x = 0b)
Continuous in (-1, 0)c)
Differentiable at x = 1d)
Differentiable in (-1, 1)
|
Asha Choudhury answered |
Solution:
To determine the continuity and differentiability of f(x), we need to evaluate left-hand limit, right-hand limit and the derivative of f(x) at every point in the domain of f(x).
Left-hand limit of f(x):
As x approaches 0 from the left, [x] approaches -1 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between -x and x. Thus, the left-hand limit of f(x) as x approaches 0 is:
lim x→0- f(x) = lim x→0- [x sin(px)] = lim x→0- (-x) = 0.
Right-hand limit of f(x):
As x approaches 0 from the right, [x] approaches 0 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between 0 and x. Thus, the right-hand limit of f(x) as x approaches 0 is:
lim x→0+ f(x) = lim x→0+ [x sin(px)] = lim x→0+ (0) = 0.
Since the left-hand limit and right-hand limit are equal, f(x) is continuous at x=0.
Derivative of f(x):
For x≠0, f(x) is piecewise linear with slope sin(px) and jump discontinuity at every integer multiple of 1/p. Therefore, f(x) is not differentiable at these points.
Since f(x) is continuous at x=0 and not differentiable at any other point, the correct answer is option 'A'.
To determine the continuity and differentiability of f(x), we need to evaluate left-hand limit, right-hand limit and the derivative of f(x) at every point in the domain of f(x).
Left-hand limit of f(x):
As x approaches 0 from the left, [x] approaches -1 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between -x and x. Thus, the left-hand limit of f(x) as x approaches 0 is:
lim x→0- f(x) = lim x→0- [x sin(px)] = lim x→0- (-x) = 0.
Right-hand limit of f(x):
As x approaches 0 from the right, [x] approaches 0 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between 0 and x. Thus, the right-hand limit of f(x) as x approaches 0 is:
lim x→0+ f(x) = lim x→0+ [x sin(px)] = lim x→0+ (0) = 0.
Since the left-hand limit and right-hand limit are equal, f(x) is continuous at x=0.
Derivative of f(x):
For x≠0, f(x) is piecewise linear with slope sin(px) and jump discontinuity at every integer multiple of 1/p. Therefore, f(x) is not differentiable at these points.
Since f(x) is continuous at x=0 and not differentiable at any other point, the correct answer is option 'A'.
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