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Limit, Continuity and Differentiability JEE 2010: Integral PYQ

JEE Advanced 2010 Mathematics Limit, Continuity and Differentiability Differentiability of functions defined by integrals

By Founder, JEEnius - IIT Kanpur Alumni · Sep 13, 2026 · 4 min read

Hard 3 min target

Let f be a real-valued function defined on the interval (0,∞) by

f(x)=lnx+∫0x1+sintdt.

Then which of the following statement(s) is (are) true?

Show answerAnswer

B) f′(x) exists for all x∈(0,∞) and f′ is continuous on (0,∞), but not differentiable on (0,∞)

C) There exists α>1 such that |f′(x)|<|f(x)| for all x∈(α,∞)

Explanation

Given

f(x)=lnx+∫0x1+sintdt

Let

g(x)=1+sinx

Since 1+sinx≥0 for all x, g(x) is defined for all real x. Also, g is continuous everywhere because it is a composition of continuous functions.

By the Fundamental Theorem of Calculus,

f′(x)=1x+1+sinx

for every x∈(0,∞).

So f′(x) exists for all x∈(0,∞).

Also, 1x is continuous on (0,∞) and 1+sinx is continuous everywhere. Hence f′ is continuous on (0,∞).

Now check differentiability of f′.

For points where 1+sinx>0, the derivative of 1+sinx exists and equals

cosx21+sinx

But at points where

1+sinx=0

we have

sinx=−1

So

x=3π2+2nπ

for integers n such that x>0.

Near such a point a=3π2+2nπ, put x=a+h.

Then

1+sin(a+h)=1−cosh

Using the identity

1−cosh=2sin2h2

we get

1+sin(a+h)=2|sinh2|

For small h,

|sinh2|~|h|2

Thus

1+sin(a+h)~|h|2

This has a corner at h=0, so it is not differentiable at a.

Therefore f′ is continuous on (0,∞) but not differentiable on all of (0,∞). Hence option B is true, and option A is false.

Now consider option C.

For x>1,

f′(x)=1x+1+sinx

Since

0≤1+sinx≤2

and

1x<1

we get

0<f′(x)<1+2

So |f′(x)| is bounded for x>1.

Now show that f(x) becomes arbitrarily large as x→∞.

The function 1+sint is non-negative and periodic, and it is not identically zero. Its integral over one full period is positive.

Indeed,

1+sint=|sint2+cost2|

So over every interval of length 2π, the integral contributes a positive fixed amount. Therefore

∫0x1+sintdt→∞

as

x→∞

Also,

lnx→∞

Therefore

f(x)→∞

as

x→∞

Hence there exists some α>1 such that for every x>α,

f(x)>1+2

Since

|f′(x)|<1+2

we get

|f′(x)|<|f(x)|

for all

x∈(α,∞)

Thus option C is true.

Now consider option D.

Option D says that |f(x)|+|f′(x)| is bounded on (0,∞).

But we have already shown that

f(x)→∞

as

x→∞

So |f(x)| is unbounded. Therefore |f(x)|+|f′(x)| cannot be bounded above by a constant β on (0,∞).

Hence option D is false.

Therefore the correct options are B and C.

Mathematics artwork for the article: Limit, Continuity and Differentiability JEE 2010: Integral PYQ

Which options are correct in the JEE 2010 integral-defined function question?

B and C are correct in the JEE 2010 integral-defined function question. The first derivative is continuous throughout the domain but has corners at the square-root zeros. The function grows without bound while its derivative stays bounded beyond one.

This is a multiple-correct Mathematics question from the JEE 2010 Advanced archive, Paper 1, in Limit, Continuity and Differentiability. The function is

f(x)=lnx+∫0x1+sintdt,x>0.

Decide which claims hold:

∃α>1:|f′(x)|<|f(x)|for every x>α.
  • D: One positive constant bounds the following sum throughout the domain:
∃β>0:|f(x)|+|f′(x)|≤βfor every x>0.

Why does the first derivative exist and remain continuous?

The Fundamental Theorem of Calculus applies because the integrand is continuous, including at its zeros. Differentiability of the integrand is not required to obtain the first derivative of its integral.

Define g(x)=1+sinx.

The sine range gives

−1≤sinx≤1⟹1+sinx≥0(x∈R).

The square-root function is continuous on the non-negative real numbers. Therefore this composition is defined and continuous everywhere, even when its value is zero.

In particular, it is continuous on every integration interval:

[0,x],x>0.

The Fundamental Theorem of Calculus gives

f′(x)=1x+1+sinx,x>0.

Both terms are continuous on the positive domain. Hence the first derivative exists and is continuous throughout that domain.

Where does the second derivative fail, and what does that prove about A and B?

The second derivative fails exactly at the positive zeros of the radicand. The proof requires unequal one-sided derivatives, not merely a zero denominator.

Away from those zeros, differentiation is valid:

f″(x)=−1x2+cosx21+sinx,1+sinx>0.

The excluded positive points are

1+sina=0⟹a=3π2+2nπ,n=0,1,2,…

Examine a small displacement from any such point:

x=a+h,sin(a+h)=sinacosh+cosasinh=−cosh.

Therefore

g(a+h)=1−cosh=2|sinh2|,g(a)=0.

Using the standard limit,

sin(h/2)h/2⟶1,

the difference quotient gives

g−′(a)=limh→0−2|sin(h/2)|h=−12,
g+′(a)=limh→0+2|sin(h/2)|h=12.

The reciprocal term contributes the same derivative from both sides:

1a+h−1ah=−1a(a+h)⟶−1a2.

Thus the one-sided derivatives of the first derivative are

(f′)−′(a)=−1a2−12,(f′)+′(a)=−1a2+12.

They are unequal, so A is false and B is true. B means the first derivative is not differentiable everywhere, not that it is nowhere differentiable.

Why does the eventual inequality hold, while a global bound fails?

Beyond one, the first derivative has a fixed upper bound, while the function tends to positive infinity. This proves C for every sufficiently large input, not just selected large values, and rules out D.

First,

x>1:0≤g(x)≤2,0<1x<1,

so 0<f′(x)<1+2.

For growth, use the official identity:

1+sint=|sint2+cost2|.

The integrand is non-negative, continuous and not identically zero. It is periodic with period 2π.

Its integral over a full period is therefore positive:

I=∫02πg(t)dt>0.

No explicit antiderivative is needed. Split the integration interval into complete periods and a remainder:

x=2πN+r,N∈{0,1,2,…},0≤r<2π.

Periodicity and non-negativity give

∫0xg(t)dt=NI+∫0rg(t)dt≥NI⟶∞(x→∞).

Together with logarithmic growth, this gives

lnx⟶∞⟹f(x)⟶∞.

By the definition of this limit, choose a threshold satisfying

α>1,f(x)>1+2for every x>α.

Both the function and its derivative are then positive, so

|f′(x)|<1+2<|f(x)|for every x>α.

Thus C is true. Since the function’s absolute value is unbounded as the input tends to infinity, no constant can bound its sum with the absolute derivative throughout the domain, so D is false.

Verdict: A false, B true, C true, D false.

Why is applying the chain rule everywhere an invalid method?

The chain-rule expression is justified only where the radicand is positive. Using it to claim that the second derivative exists at every positive input ignores this restriction:

g′(x)=cosx21+sinx.

At a radicand zero, an undefined formula alone does not prove non-differentiability. Return to the difference quotient.

The unequal one-sided derivatives come from the absolute value:

sin2(h/2)=|sin(h/2)|,

not the sine expression without modulus. Check the differentiated expression’s domain, then test excluded interior points separately.

How can you apply this method to two related questions?

Use continuity to obtain the first derivative, then test its differentiability separately. For eventual inequalities, compare a bounded derivative with a function tending to infinity.

Both questions below are original chapter practice, not verified PYQs or additional JEE 2010 questions.

Does integrating an absolute value remove its corner completely?

The first derivative exists at the corner, but the second does not. Question 1: Determine whether both derivatives exist at the stated point:

H(x)=∫0x|t−1|dt,x>0;x=1.

Continuity of the integrand gives

H′(x)=|x−1|,H′(1)=0.

The second-derivative test is

H′(1+h)−H′(1)h=|h|h⟶{−1,h→0−,+1,h→0+.

Hence the second derivative does not exist there. For a numerical check,

H(1)=∫01(1−t)dt=[t−t22]01=12,
|−0.01|−0.01=−1,|0.01|0.01=1.

What changes if the square root is removed?

The second derivative exists everywhere in the positive domain, and the eventual absolute-value inequality still holds. Question 2: Check both claims for

P(x)=lnx+∫0x(1+sint)dt,x>0.

Direct integration and differentiation give

P(x)=lnx+x+1−cosx,
P′(x)=1x+1+sinx,P″(x)=−1x2+cosx.

The second-derivative formula is defined at every positive input. For the inequality,

x>1:0<P′(x)<3,P(x)≥lnx+x⟶∞.

By this limit, the function exceeds three for every sufficiently large input, proving

|P′(x)|<|P(x)|.

In your next integral-defined problem, check continuity of the integrand first, then its differentiability. Integration gives a first derivative from a continuous integrand; the square root in the original problem creates the obstruction to the second derivative.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Keep going with Motion in a Straight Line JEE 2005: Two-Particle Solution.

Frequently asked questions

Which options are correct in the JEE 2010 integral-defined function question?

B and C are correct; A and D are false. The first derivative is continuous for every x > 0 but is not differentiable at the positive zeros of 1 + sin x. The function tends to infinity while its derivative stays bounded for x > 1, proving the eventual inequality and ruling out a global bound.

Why is the first derivative continuous when the square-root integrand has corners?

The Fundamental Theorem of Calculus requires continuity of the integrand, not its differentiability. Since √(1 + sin x) is continuous everywhere, the JEE function has derivative f'(x) = 1/x + √(1 + sin x) for x > 0. Both terms are continuous throughout this domain, including at the square-root zeros.

Where does the second derivative fail in the JEE 2010 question?

The second derivative fails exactly at a = 3π/2 + 2nπ, where n = 0, 1, 2, …. At each such point, the left and right derivatives of f' are −1/a² − 1/√2 and −1/a² + 1/√2, respectively. These unequal values prove non-differentiability; an undefined chain-rule formula alone does not.

How do you prove option C without evaluating the integral?

For the JEE function, 0 < f'(x) < 1 + √2 whenever x > 1. The integral is non-negative, so f(x) ≥ ln x and therefore f(x) tends to infinity. Choose α > 1 such that f(x) > 1 + √2 for every x > α; then |f'(x)| < |f(x)| throughout that range.

continuitydefinite integralsdifferentiabilityjee 2010one sided derivatives

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