Integral Calculus JEE 2022: Full Solution and Root Count

What was the JEE Advanced 2022 integral calculus question?

The integral calculus JEE 2022 problem has exactly two admissible parameter values, one on each allowed interval, and both are irrational. The correct multi-select answer is C and D. This question appeared in JEE Advanced 2022 Paper 1 Mathematics under Integral Calculus. Its hard-tier classification supports an expected solving time of about 300 seconds.

The parameter belongs to the disconnected domain: $a \in (- \infty, 0) \cup (1, \infty)$

The equation is:

\int_{1}^{e} \frac{(\ln x)^{2}}{x {[a - (\ln x)^{3 / 2}]}^{2}} d x = 1

The four claims are:

This is a multi-correct question. No diagram is needed because the problem is purely algebraic.

How do we convert the definite integral into a function of the parameter?

The official substitution reduces the integral calculus JEE 2022 expression to a rational integral in the new variable. It keeps the parameter unchanged and produces a function whose roots can be counted using derivatives.

Set: $t = (\ln x)^{3 / 2}$

Therefore: $\ln x = t^{2 / 3}$

Differentiating gives:

\frac{d x}{x} = \frac{2}{3} t^{- 1 / 3} d t

Now combine the numerator and differential:

(\ln x)^{2} \frac{d x}{x} = t^{4 / 3} \cdot \frac{2}{3} t^{- 1 / 3} d t = \frac{2}{3} t d t

The limits become: $x = 1 \Rightarrow t = 0$ $x = e \Rightarrow t = 1$

Thus:

I (a) = \frac{2}{3} \int_{0}^{1} \frac{t}{(a - t)^{2}} d t

Split the integrand:

\frac{t}{(a - t)^{2}} = \frac{a}{(a - t)^{2}} - \frac{1}{a - t}

An antiderivative is:

\int \frac{t}{(a - t)^{2}} d t = \frac{a}{a - t} + \ln | a - t | + C

Apply the limits:

I (a) = \frac{2}{3} {[\frac{a}{a - t} + \ln | a - t |]}_{0}^{1}

I (a) = \frac{2}{3} (\frac{a}{a - 1} + \ln | a - 1 | - 1 - \ln | a |)

Since:

\frac{a}{a - 1} - 1 = \frac{1}{a - 1}

we get:

I (a) = \frac{2}{3} [\frac{1}{a - 1} + \ln | \frac{a - 1}{a} |]

Therefore, the equation becomes:

F (a) = \frac{1}{a - 1} + \ln | \frac{a - 1}{a} | = \frac{3}{2}

How many solutions lie in the two allowed intervals?

There is exactly one solution in each allowed interval, so exactly two admissible values exist. The intervals must be studied separately because the derivative has opposite signs on them.

Define:

F (a) = \frac{1}{a - 1} + \ln | \frac{a - 1}{a} |

Differentiation gives:

F^{'} (a) = - \frac{1}{a (a - 1)^{2}}

How many solutions lie in the interval from 1 to infinity?

Exactly one solution lies in this interval. The function is continuous, strictly decreasing, and its values fall from infinity to zero.

For: $a \in (1, \infty)$ we have: $F^{'} (a) < 0$

The endpoint limits are:

{lim}_{a \to 1^{+}} F (a) = \infty

{lim}_{a \to \infty} F (a) = 0

By continuity and strict decrease, exactly one solution of the following equation lies in this interval:

F (a) = \frac{3}{2}

How many solutions lie in the interval from negative infinity to 0?

Exactly one solution lies in this interval. Here the function is continuous, strictly increasing, and its values rise from zero to infinity.

For: $a \in (- \infty, 0)$ we have: $F^{'} (a) > 0$

The endpoint limits are:

{lim}_{a \to - \infty} F (a) = 0

{lim}_{a \to 0^{-}} F (a) = \infty

By continuity and strict increase, exactly one solution lies in the negative interval. Hence there are exactly two admissible values, so D is true and A is false.

Why are both solutions irrational rather than integers?

No integer satisfies the equation, while both existing roots are irrational. Monotonic comparison eliminates every integer, and a standard transcendence result rules out every rational solution.

Why can no positive integer satisfy the equation?

For every integer satisfying: $a \geq 2$ the decreasing nature of the function gives:

F (a) \leq F (2) = 1 - \ln 2 < \frac{3}{2}

Therefore, no admissible positive integer satisfies the equation.

Why can no negative integer satisfy the equation?

For every integer satisfying: $a \leq - 1$ the increasing nature of the function gives:

F (a) \leq F (- 1) = \ln 2 - \frac{1}{2} < \frac{3}{2}

Thus no integer solves the equation, making B false.

How does transcendence prove that both roots are irrational?

A change of variable turns the equation into a statement about the exponential of a nonzero rational number. Assuming that a root is rational then forces one number to be both rational and transcendental.

Set:

u = \frac{a - 1}{a}

On both allowed intervals:

u > 0, u \neq 1

Also:

\frac{1}{a - 1} = \frac{1 - u}{u}

Substitution into the equation gives:

\frac{1 - u}{u} + \ln u = \frac{3}{2}

Hence:

\ln u + \frac{1}{u} = \frac{5}{2}

Assume that a satisfying value of the parameter is rational. Then the corresponding value of the new variable is rational, and:

\ln u = \frac{5}{2} - \frac{1}{u} = r

The number on the right is rational. It is nonzero because a zero logarithm would force: $u = 1$

This value is excluded. Therefore: $u = e^{r}$

For every nonzero algebraic number: $r$ the number: $e^{r}$ is transcendental. This contradicts the rationality of: $u = e^{r}$

Both satisfying parameter values are irrational, so C is true. The final multi-correct answer is C and D.

Why can incorrect global monotonicity lead to option A?

Option A can result from treating the allowed domain as one continuous interval. The actual domain has two separate intervals, and the function has different monotonic behaviour on each one.

The domain is: $(- \infty, 0) \cup (1, \infty)$

The sign of the derivative changes because the factor containing the parameter changes sign:

F^{'} (a) = - \frac{1}{a (a - 1)^{2}}

Therefore, the function increases when the parameter is negative but decreases when it is greater than one. One monotonicity conclusion cannot be applied across the excluded region.

Checking only the behaviour as the magnitude of the parameter tends to infinity is also insufficient. Although the function tends to zero at both distant ends, it tends to infinity at the two finite boundary points:

{lim}_{a \to 0^{-}} F (a) = \infty

{lim}_{a \to 1^{+}} F (a) = \infty

The correct general method is:

Split the parameter domain at every excluded point.
Determine monotonicity separately on each interval.
Check both endpoint limits.
Count the roots on each interval before combining the results.

Which related parameter-based definite integrals should you solve next?

These three problems practise the same substitution, domain splitting, monotonicity, endpoint limits and arithmetic classification. Reduce each integral to a parameter function before attempting to count its roots.

For: $a \in (- \infty, 0) \cup (1, \infty)$

evaluate:

\int_{1}^{e} \frac{(\ln x)^{2}}{x {[a - (\ln x)^{3 / 2}]}^{2}} d x

and determine for which positive constants the equation: $I (a) = k$ has two solutions.

For a parameter outside the closed interval from zero to one, analyse the number of solutions of:

\int_{0}^{1} \frac{t}{(a - t)^{2}} d t = c, c > 0

using derivatives and endpoint limits.

Suppose:

u > 0, u \neq 1

and: $\ln u + \frac{1}{u}$ is rational. Investigate why assuming the new variable is rational forces a contradiction, except in the excluded zero-log case.

For more integral calculus JEE 2022 practice, use the free past-paper archive to search every JEE Advanced paper from 2007 by year, subject or chapter, each with a worked solution.

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).

If that step was the hard part, work through Redox Reactions and Electrochemistry JEE 2013: Full Solution.

Frequently asked questions

What is the correct answer to the JEE Advanced 2022 integral calculus question?

The correct multi-select answer is C and D. The transformed parameter function has one root in each allowed interval, so D is true. A transcendence argument shows both roots are irrational, so C is true.

Which substitution is used in the integral calculus JEE 2022 problem?

Use t = (ln x)^(3/2). This converts the given integral into (2/3) times the integral of t/(a − t)² from 0 to 1. The resulting expression can then be analysed as a function of a.

How many values of a satisfy the integral equation?

Exactly two admissible values satisfy the equation. One lies in (−∞, 0), where the parameter function is strictly increasing, and the other lies in (1, ∞), where it is strictly decreasing.

Why are both parameter values irrational?

Assuming a rational root makes u = (a − 1)/a rational and gives ln u as a nonzero rational number r. Then u = e^r would be transcendental, contradicting the rationality of u. Therefore, neither root can be rational.

Why must the parameter domain be split into two intervals?

The admissible domain is (−∞, 0) ∪ (1, ∞), not one continuous interval. Since F′(a) has opposite signs on these intervals, applying one global monotonicity conclusion can produce the incorrect option A.

Integral Calculus JEE 2022: Why the Correct Answer Is C and D