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JEE Mathematics 2023 Question with Solution

The number of points where the curve f(x)=e8x−e6x−3e4x−e2x+1,x∈Rf(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1, x \in \mathbb{R} cuts the xx-axis, is equal to _____.

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given:

f(x)=e8x−e6x−3e4x−e2x+1f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1

Find: The number of real points where the curve cuts the xx-axis.

Let

t=e2xt = e^{2x}

Since x∈Rx \in \mathbb{R}, we have t>0t > 0. Then the equation f(x)=0f(x)=0 becomes

t4−t3−3t2−t+1=0t^4 - t^3 - 3t^2 - t + 1 = 0

Divide by t2t^2:

t2−t−3−1t+1t2=0t^2 - t - 3 - \frac{1}{t} + \frac{1}{t^2} = 0

Rearrange as

(t2+1t2)−(t+1t)−3=0\left(t^2 + \frac{1}{t^2}\right) - \left(t + \frac{1}{t}\right) - 3 = 0

Now use

t2+1t2=(t+1t)2−2t^2 + \frac{1}{t^2} = \left(t + \frac{1}{t}\right)^2 - 2

Let

u=t+1tu = t + \frac{1}{t}

Then

ν2−ν−5=0\nu^2 - \nu - 5 = 0

Solve the quadratic:

ν=1±212\nu = \frac{1 \pm \sqrt{21}}{2}

But for real t>0t > 0,

t+1t≥2t + \frac{1}{t} \ge 2

Hence only

ν=1+212\nu = \frac{1 + \sqrt{21}}{2}

is admissible.

For this value of ν\nu, the equation

t+1t=νt + \frac{1}{t} = \nu

gives

t2−νt+1=0t^2 - \nu t + 1 = 0

Since ν>2\nu > 2, this quadratic has two distinct positive real roots for tt. Each positive value of t=e2xt = e^{2x} gives exactly one real value of xx.

Therefore, the curve cuts the xx-axis at 22 points.

Admissible Root Check

The transformed quadratic in ν\nu gives two values, but both do not produce real values of tt. Because t=e2x>0t = e^{2x} > 0, the quantity t+1tt + \frac{1}{t} cannot be less than 22. Therefore ν=1−212\nu = \frac{1 - \sqrt{21}}{2} must be rejected. Only one admissible value of ν\nu remains, and it leads to two positive roots of tt, hence two real values of xx. This matches the final answer 22.

Common mistakes

  • Taking both roots of ν=1±212\nu = \frac{1 \pm \sqrt{21}}{2} as valid. This is wrong because ν=t+1t\nu = t + \frac{1}{t} with t>0t>0 must satisfy ν≥2\nu \ge 2. Always apply the domain restriction after substitution.

  • Forgetting that t=e2xt = e^{2x} is always positive. This can lead to counting invalid negative or zero values of tt. Use t>0t>0 from the start.

  • Assuming two roots in tt automatically mean more than two roots in xx. Since t=e2xt = e^{2x} is one-to-one for real xx, each positive tt corresponds to exactly one real xx. Count roots carefully after back-substitution.

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