Trigonometry JEE 2025: Inverse-Cosine and Log Domains

Q: Why do I get 88 instead of 96 in this domain question?

One possible cause is incorrectly turning 2x + 5 < 3 into 2x < 5 - 3, which gives the wrong upper boundary x = 1. The correct subtraction gives 2x < 3 - 5, so x < -1. Using the wrong boundary produces 88; changing open brackets to closed brackets alone does not change the endpoint sum.

What is the correct answer to the trigonometry JEE 2025 domain question?

The correct answer is A) 96. This trigonometry JEE 2025 problem requires two separate domain calculations, followed by substitution of their endpoints. The source is JEE Main, April 2025, Shift-02.

The question bank labels it hard, tag 4, with an expected solving time of 120 seconds. These are bank benchmarks, not an official NTA difficulty classification or measured student performance.

Find the domains of these functions, then use their boundary values to evaluate the expression below:

f (x) = \arccos (\frac{4 x + 5}{3 x - 7}), g (x) = \log_{2} (2 - 6 \log_{27} (2 x + 5)) .

The respective domains are written as:

D_{f} = [α, β], D_{g} = (γ, δ) .

The required value is:

| 7 (α + β) + 4 (γ + δ) | .

How do you find the inverse-cosine domain without a sign mistake?

Start with the complete inverse-cosine restriction:

- 1 \leq \frac{4 x + 5}{3 x - 7} \leq 1, x \neq \frac{7}{3} .

Solve the two bounds separately and intersect their solution sets. Inverse cosine accepts both boundary inputs, negative one and one, so equality is allowed. A denominator zero is always excluded.

For the first bound, move everything to one side. This gives a rational inequality whose numerator and denominator signs must be checked separately:

\frac{4 x + 5}{3 x - 7} + 1 \geq 0 ⟹ \frac{7 x - 2}{3 x - 7} \geq 0 .

Numerator zero: \frac{2}{7}; denominator pole: \frac{7}{3} .

The signs across the three intervals are:

Below the numerator zero, both numerator and denominator are negative, so the quotient is positive:

x \in (- \infty, \frac{2}{7}) : \frac{-}{-} gives + .

Between the zero and pole, the numerator is positive and denominator negative:

x \in (\frac{2}{7}, \frac{7}{3}) : \frac{+}{-} gives - .

Above the pole, both are positive:

x \in (\frac{7}{3}, \infty) : \frac{+}{+} gives + .

Keep positive values and the numerator zero. Exclude the pole:

S_{1} = (- \infty, \frac{2}{7}] \cup (\frac{7}{3}, \infty) .

For the second bound, subtract one. Again, check signs on either side of the numerator zero and denominator pole:

\frac{4 x + 5}{3 x - 7} - 1 \leq 0 ⟹ \frac{x + 12}{3 x - 7} \leq 0 .

Below the numerator zero:

x \in (- \infty, - 12) : \frac{-}{-} gives + .

Between the zero and pole:

x \in (- 12, \frac{7}{3}) : \frac{+}{-} gives - .

Above the pole:

x \in (\frac{7}{3}, \infty) : \frac{+}{+} gives + .

Keep negative values and the numerator zero. Intersect with the first solution set:

S_{2} = [- 12, \frac{7}{3}), S_{1} \cap S_{2} = [α, β] = [- 12, \frac{2}{7}] .

α = - 12, β = \frac{2}{7} .

Direct substitution confirms both closed brackets. The inverse-cosine argument reaches its accepted boundary inputs:

{\frac{4 x + 5}{3 x - 7} |}_{x = - 12} = \frac{- 43}{- 43} = 1,

{\frac{4 x + 5}{3 x - 7} |}_{x = 2 / 7} = \frac{43 / 7}{- 43 / 7} = - 1 .

Do not cross-multiply without establishing the denominator’s sign. It changes sign at its pole, and multiplication by a negative value reverses the inequality.

How do you apply both logarithm restrictions?

Both logarithm arguments must be strictly positive. The inner logarithm supplies the lower boundary; the outer logarithm supplies the upper boundary. Intersect these restrictions rather than stopping after the inner-log check.

The inner logarithm requires:

2 x + 5 > 0 ⟹ x > - \frac{5}{2} .

The outer logarithm requires:

2 - 6 \log_{27} (2 x + 5) > 0,

- 6 \log_{27} (2 x + 5) > - 2 ⟹ \log_{27} (2 x + 5) < \frac{1}{3} .

The inequality reverses when we divide by negative six. Since the logarithm’s base is greater than one, exponentiation preserves the new direction:

27 > 1, 2 x + 5 < 27^{1 / 3} = 3 .

2 x < - 2 ⟹ x < - 1 .

Intersecting the lower and upper restrictions gives the domain and its boundary values:

(γ, δ) = (- \frac{5}{2}, - 1), γ = - \frac{5}{2}, δ = - 1 .

At the lower boundary, the inner logarithm has argument zero. At the upper boundary, the outer logarithm has argument zero:

2 (- \frac{5}{2}) + 5 = 0, 2 - 6 \log_{27} (3) = 2 - 6 (\frac{1}{3}) = 0 .

Both endpoints are excluded. The endpoint symbols still denote these boundary values, even though neither value belongs to the domain.

How do the four endpoints give option A?

Substituting the boundary values gives A) 96. The expression combines endpoints from two separately determined domains; it does not ask for the common domain of the two functions.

First calculate the endpoint sums, then apply the coefficients:

α + β = - 12 + \frac{2}{7} = - \frac{82}{7},

γ + δ = - \frac{5}{2} - 1 = - \frac{7}{2} .

7 (α + β) = - 82, 4 (γ + δ) = - 14 .

| 7 (α + β) + 4 (γ + δ) | = | - 82 - 14 | = | - 96 | = 96 .

Keep the negative signs until the final absolute-value operation. Taking separate absolute values earlier is not the operation specified in the question.

How can a transposition error produce option B, 88?

One possible wrong route is to reverse the subtraction when isolating the variable. This explains how 88 can arise, not how students were observed to answer. Keep the inverse-cosine domain and inner-log lower bound correct, but make this invalid step:

2 x + 5 < 3 ⇏ 2 x < 5 - 3 \Rightarrow x < 1 .

Subtracting five from both sides gives three minus five, not five minus three. Use this rule:

a x + c < d ⟹ a x < d - c .

The wrong logarithmic interval and endpoint sum become:

(- \frac{5}{2}, 1), - \frac{5}{2} + 1 = - \frac{3}{2} .

They produce the supplied distractor:

| - 82 + 4 (- \frac{3}{2}) | = | - 88 | = 88 (option B) .

For an independent rejection test, substitute zero, which lies inside the wrong interval. The outer logarithm’s argument is negative:

x = 0 : \log_{27} (5) > \log_{27} (3) = \frac{1}{3} ⟹ 2 - 6 \log_{27} (5) < 0 .

The outer logarithm therefore fails. Changing open brackets to closed brackets alone cannot explain 88, because bracket choice does not change the endpoint sum.

Which two practice questions check the same method?

These original practice variations, not additional JEE 2025 questions, check rational signs and endpoint inclusion. Attempt both before reading the worked answers.

Question 1: Find the domain of:

h (x) = \arcsin (\frac{2 x + 1}{x - 2}) .

Impose the inverse-sine restriction and exclude the pole. Split the bounds into two rational inequalities:

- 1 \leq \frac{2 x + 1}{x - 2} \leq 1, x \neq 2 .

\frac{3 x - 1}{x - 2} \geq 0, \frac{x + 3}{x - 2} \leq 0 .

Each quotient has positive, negative, positive signs across its ordered numerator zero and denominator pole. Their solution sets intersect as follows:

D_{h} = [(- \infty, \frac{1}{3}] \cup (2, \infty)] \cap [- 3, 2) = [- 3, \frac{1}{3}] .

Question 2: Find the domain of:

k (x) = \arccos (\log_{3} (2 x + 5)) .

Require a positive inner-log argument and an outer inverse-cosine input between its accepted bounds. Since the logarithm’s base exceeds one, exponentiation preserves both inequalities:

2 x + 5 > 0, - 1 \leq \log_{3} (2 x + 5) \leq 1 .

\frac{1}{3} \leq 2 x + 5 \leq 3 ⟹ D_{k} = [- \frac{7}{3}, - 1] .

Here both endpoints are included: the outer inverse cosine accepts negative one and one, while the inner logarithm remains defined at both boundaries. In the original nested-log function, each boundary makes a logarithm’s argument zero.

For your next revision pass, solve these again without looking. Verify every bracket by substituting its endpoint into the original function.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

If that step was the hard part, work through Chemical Thermodynamics JEE 2025: Coefficient Ratio.

Frequently asked questions

What is the answer to the trigonometry JEE 2025 domain question?

The correct answer is A) 96. The inverse-cosine domain is [-12, 2/7], and the nested-log domain is (-5/2, -1). Substituting their boundary values into |7(α + β) + 4(γ + δ)| gives |-82 - 14| = 96.

How do I find the domain of arccos((4x+5)/(3x-7))?

Require -1 ≤ (4x+5)/(3x-7) ≤ 1 and exclude x = 7/3. Solve the two bounds separately using rational sign charts, then intersect their solution sets to obtain [-12, 2/7]. Both endpoints are included because the inverse-cosine argument equals 1 and -1 there.

Why are both endpoints excluded from the nested-log domain?

For g(x) = log₂(2 - 6 log₂₇(2x+5)), both logarithm arguments must be strictly positive, giving -5/2 < x < -1. At x = -5/2, the inner logarithm has argument zero; at x = -1, the outer logarithm has argument zero. Neither endpoint belongs to the domain, but both remain its boundary values for the required calculation.

Why do I get 88 instead of 96 in this domain question?