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MCQMediumJEE 2024Quadratic Equations in Complex Numbers

JEE Mathematics 2024 Question with Solution

Let SS = {x∈Rx \in \mathbb{R} : (3+2)x+(3−2)x=10(\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10}. Then the number of elements in SS is:

  • A

    44

  • B

    00

  • C

    22

  • D

    11

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

(3+2)x+(3−2)x=10(\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10

Find: The number of real values of xx, that is, the number of elements in SS.

Let

t=(3+2)xt = (\sqrt{3} + \sqrt{2})^x

Since

(3−2)=13+2(\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}}

we get

(3−2)x=1t(\sqrt{3} - \sqrt{2})^x = \frac{1}{t}

Substituting in the given equation,

t+1t=10t + \frac{1}{t} = 10

Multiplying by tt,

t2−10t+1=0t^2 - 10t + 1 = 0

Solving the quadratic equation,

t=10±100−42=10±962=10±462=5±26t = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6}

Now,

(3+2)2=3+2+26=5+26(\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}

So one case gives

(3+2)x=(3+2)2  ⟹  x=2(\sqrt{3} + \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^2 \implies x = 2

Also,

(3−2)2=3+2−26=5−26(\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6}

and

(3−2)2=1(3+2)2=(3+2)−2(\sqrt{3} - \sqrt{2})^2 = \frac{1}{(\sqrt{3} + \sqrt{2})^2} = (\sqrt{3} + \sqrt{2})^{-2}

Thus,

(3+2)x=(3+2)−2  ⟹  x=−2(\sqrt{3} + \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^{-2} \implies x = -2

Hence the real solutions are x=2x = 2 and x=−2x = -2. Therefore, the set SS has 22 elements. The correct option is C.

Using reciprocal roots

Given:

(3+2)x+(3−2)x=10(\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10

Find: How many real solutions xx satisfy the equation.

Observe that

(3+2)(3−2)=3−2=1(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1

So the two bases are reciprocals of each other. If we take

t=(3+2)xt = (\sqrt{3} + \sqrt{2})^x

then automatically

(3−2)x=1t(\sqrt{3} - \sqrt{2})^x = \frac{1}{t}

Therefore,

t+1t=10t + \frac{1}{t} = 10

which gives

t2−10t+1=0t^2 - 10t + 1 = 0

Hence,

t=5±26t = 5 \pm 2\sqrt{6}

Both values are positive, and they are reciprocals because

(5+26)(5−26)=1(5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 1

Now,

5+26=(3+2)25 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2

and

5−26=(3+2)−25 - 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^{-2}

So the two corresponding values of xx are

x=2andx=−2x = 2 \quad \text{and} \quad x = -2

Thus, the number of elements in SS is 22. Therefore, the correct option is C.

Common mistakes

  • Taking only t=5+26t = 5 + 2\sqrt{6} and ignoring t=5−26t = 5 - 2\sqrt{6}. This is wrong because both roots of t2−10t+1=0t^2 - 10t + 1 = 0 are positive. You must convert both valid tt values back into corresponding real values of xx.

  • Missing the identity (3−2)=13+2(\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}}. Without this, the substitution does not simplify correctly. First use the product of the conjugates to show the bases are reciprocals.

  • Writing 5−26=(3+2)25 - 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2. This is incorrect because (3+2)2=5+26(\sqrt{3} + \sqrt{2})^2 = 5 + 2\sqrt{6}. Instead, recognize 5−26=(3+2)−25 - 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^{-2} or (3−2)2(\sqrt{3} - \sqrt{2})^2.

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