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JEE Mathematics 2025 Question with Solution

The number of solutions of the equation (9x−9x+2)(2x−7x+3)=0\left( \frac{9}{x} - \frac{9}{\sqrt{x}} + 2 \right) \left( \frac{2}{x} - \frac{7}{\sqrt{x}} + 3 \right) = 0 is:

  • A

    33

  • B

    11

  • C

    22

  • D

    44

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

(9x−9x+2)(2x−7x+3)=0\left( \frac{9}{x} - \frac{9}{\sqrt{x}} + 2 \right) \left( \frac{2}{x} - \frac{7}{\sqrt{x}} + 3 \right) = 0

with x>0x > 0.

Find: The number of solutions for xx.

Since the product is zero, solve each factor separately.

Let

y=xy = \sqrt{x}

so that

x=y2,y>0.x = y^2, \quad y > 0.

For the first factor,

9x−9x+2=0\frac{9}{x} - \frac{9}{\sqrt{x}} + 2 = 0

becomes

9y2−9y+2=0.\frac{9}{y^2} - \frac{9}{y} + 2 = 0.

Multiplying by y2y^2,

9−9y+2y2=09 - 9y + 2y^2 = 0

that is,

2y2−9y+9=0.2y^2 - 9y + 9 = 0.

Solving,

y=3ory=32.y = 3 \quad \text{or} \quad y = \frac{3}{2}.

Hence,

x=9orx=94.x = 9 \quad \text{or} \quad x = \frac{9}{4}.

For the second factor,

2x−7x+3=0\frac{2}{x} - \frac{7}{\sqrt{x}} + 3 = 0

becomes

2y2−7y+3=0.\frac{2}{y^2} - \frac{7}{y} + 3 = 0.

Multiplying by y2y^2,

2−7y+3y2=02 - 7y + 3y^2 = 0

that is,

3y2−7y+2=0.3y^2 - 7y + 2 = 0.

Solving,

y=2ory=13.y = 2 \quad \text{or} \quad y = \frac{1}{3}.

Hence,

x=4orx=19.x = 4 \quad \text{or} \quad x = \frac{1}{9}.

Therefore, the complete set of solutions is

x=9,  94,  4,  19.x = 9, \; \frac{9}{4}, \; 4, \; \frac{1}{9}.

There are 44 distinct solutions. The correct option is D.

The solution states option C, but the working clearly gives 44 solutions. Hence the defensible answer is D.

Substitution and factorization

Given:

(9x−9x+2)(2x−7x+3)=0\left( \frac{9}{x} - \frac{9}{\sqrt{x}} + 2 \right) \left( \frac{2}{x} - \frac{7}{\sqrt{x}} + 3 \right) = 0

Find: Number of values of xx satisfying the equation.

Using the substitution

α=1x,x>0,\alpha = \frac{1}{\sqrt{x}}, \quad x > 0,

we get

(9α2−9α+2)(2α2−7α+3)=0.(9\alpha^2 - 9\alpha + 2)(2\alpha^2 - 7\alpha + 3) = 0.

Factor both quadratics:

9α2−9α+2=(3α−1)(3α−2)9\alpha^2 - 9\alpha + 2 = (3\alpha - 1)(3\alpha - 2)

and

2α2−7α+3=(α−3)(2α−1).2\alpha^2 - 7\alpha + 3 = (\alpha - 3)(2\alpha - 1).

So,

(3α−2)(3α−1)(α−3)(2α−1)=0.(3\alpha - 2)(3\alpha - 1)(\alpha - 3)(2\alpha - 1) = 0.

Hence,

α=23,  13,  3,  12.\alpha = \frac{2}{3}, \; \frac{1}{3}, \; 3, \; \frac{1}{2}.

Now use

x=1α2.x = \frac{1}{\alpha^2}.

Thus,

α=13⇒x=9,\alpha = \frac{1}{3} \Rightarrow x = 9, α=12⇒x=4,\alpha = \frac{1}{2} \Rightarrow x = 4, α=23⇒x=94,\alpha = \frac{2}{3} \Rightarrow x = \frac{9}{4}, α=3⇒x=19.\alpha = 3 \Rightarrow x = \frac{1}{9}.

These are four distinct positive values of xx. Therefore, the number of solutions is 44, so the correct option is D.

Common mistakes

  • Treating the product equation as if both factors must be zero. For AB=0AB=0, it is enough that either factor is zero. Solve both factors separately and combine all valid solutions.

  • Ignoring the domain restriction from x\sqrt{x} and the denominators. Since x\sqrt{x} is in the denominator, xx must satisfy x>0x>0. Do not allow x=0x=0 or negative values.

  • Making an error while substituting y=xy=\sqrt{x} or α=1x\alpha=\frac{1}{\sqrt{x}}. After substitution, clear denominators carefully before solving the quadratic.

  • Counting repeated or transformed roots incorrectly. Solve for the substitution variable first, then convert back to xx using x=y2x=y^2 or x=1α2x=\frac{1}{\alpha^2}, and finally count distinct values of xx.

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