JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Determining the number and signs of real roots of a reciprocal exponential equation

Q: Why do we divide by t^2 and use u = t - 1/t instead of u = t + 1/t?

The coefficients of the cubic and linear terms are +8 and -8. Factoring out 8 gives 8(t - 1/t), which dictates using u = t - 1/t.

Q: Why does 0 < t < 1 mean x is negative?

Here, x = ln(t). The natural logarithm function is strictly increasing, and ln(1) = 0. Therefore, ln(t) is strictly negative when 0 < t < 1.

The equation e^(4 x)+8 e^(3 x)+13 e^(2 x)-8 e^x+1=0, x ∈ ℝ has:

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Afour solutions two of which are negativeBtwo solutions and only one of them is negativeCtwo solutions and both are negativeDno solution

Step-by-step solutionView

Correct answer

Substituting $t = e^{x}$ and dividing by $t^{2}$ allows the equation to be transformed via $u = t - \frac{1}{t}$ , yielding two valid positive values of $t$ , both of which lie strictly between $0$ and $1$ , corresponding to two negative real solutions for $x$ .

Option analysis

Why each option works or fails

A · four solutions two of which are negative

The student treats all four roots of the quartic polynomial in $t = e^{x}$ as valid real values for $x$ , forgetting that $t$ must be strictly positive. Check the condition $t > 0$ for $t = e^{x}$ ; negative values of $t$ cannot produce real values of $x$ .

B · two solutions and only one of them is negative

The student identifies two positive roots for $t = e^{x}$ but assumes that one root is greater than $1$ (yielding a positive $x$ ) and one is less than $1$ (yielding a negative $x$ ). Compare each positive root $t$ to $1$ : because both satisfy $0 < t < 1$ , both corresponding solutions $x = \ln (t)$ are strictly negative.

C · two solutions and both are negative

The student correctly transforms the equation, enforces $t = e^{x} > 0$ , and checks whether $t > 1$ or $0 < t < 1$ . Correctly observed that the substitution gives two valid roots for $t$ , both lying in the interval $(0, 1)$ , giving two negative values for $x$ .

D · no solution

The student makes an algebraic sign error when substituting $u = t - 1 / t$ or $u = t + 1 / t$ , concluding that the quadratic has no real roots. Carefully square $t - 1 / t = u$ to get $t^{2} + 1 / t^{2} = u^{2} + 2$ , and solve the resulting quadratic $u^{2} + 8 u + 15 = 0$ .

Reviewed route

Solution

StepWorking

01Given

The equation is $e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0$ for $x \in R$ .

02Goal

Determine the number of real solutions $x$ and their signs.

03Approach

Substitute $t = e^{x} > 0$ . Since the polynomial $t^{4} + 8 t^{3} + 13 t^{2} - 8 t + 1 = 0$ has anti-symmetric inner coefficients ( $+ 8$ and $- 8$ ), divide by $t^{2}$ and use the auxiliary substitution $u = t - \frac{1}{t}$ . Solve the resulting quadratic in $u$ , find the corresponding positive roots $t$ , and deduce the signs of $x = \ln t$ .

04Execute

Substitute $t = e^{x}$ with $t > 0$ : $t^{4} + 8 t^{3} + 13 t^{2} - 8 t + 1 = 0$ Dividing by $t^{2} \neq 0$ : $(t^{2} + \frac{1}{t^{2}}) + 8 (t - \frac{1}{t}) + 13 = 0$ Let $u = t - \frac{1}{t}$ . Then $u^{2} = t^{2} - 2 + \frac{1}{t^{2}} ⟹ t^{2} + \frac{1}{t^{2}} = u^{2} + 2$ . The equation becomes: $(u^{2} + 2) + 8 u + 13 = 0 ⟹ u^{2} + 8 u + 15 = 0$ Factoring gives: $(u + 3) (u + 5) = 0 ⟹ u = - 3 or u = - 5$

05Execute

Now solve for $t > 0$ from each value of $u$ : Case 1: $t - \frac{1}{t} = - 3 ⟹ t^{2} + 3 t - 1 = 0$ The roots are $t = \frac{- 3 \pm \sqrt{9 - 4 (1) (- 1)}}{2} = \frac{- 3 \pm \sqrt{13}}{2}$ . Since $t > 0$ , we have one valid root: $t_{1} = \frac{\sqrt{13} - 3}{2}$ . Since $3 = \sqrt{9} < \sqrt{13} < \sqrt{25} = 5$ , we have $0 < \sqrt{13} - 3 < 2$ , so $0 < t_{1} < 1$ . Case 2: $t - \frac{1}{t} = - 5 ⟹ t^{2} + 5 t - 1 = 0$ The roots are $t = \frac{- 5 \pm \sqrt{25 - 4 (1) (- 1)}}{2} = \frac{- 5 \pm \sqrt{29}}{2}$ . Since $t > 0$ , we have one valid root: $t_{2} = \frac{\sqrt{29} - 5}{2}$ . Since $5 = \sqrt{25} < \sqrt{29} < \sqrt{49} = 7$ , we have $0 < \sqrt{29} - 5 < 2$ , so $0 < t_{2} < 1$ .

06Execute

Since $x = \ln t$ , for both roots we have $0 < t < 1$ , which gives $\ln t < 0$ . Thus, $x_{1} = \ln t_{1} < 0$ and $x_{2} = \ln t_{2} < 0$ . Therefore, there are exactly two real solutions, and both are negative.

✓Verify

Each quadratic $t^{2} + k u - 1 = 0$ has constant term $- 1$ , meaning its two roots have product $- 1$ . Hence, exactly one root is positive and one is negative for each equation, yielding precisely two positive values of $t$ . Since $t^{2} + k t - 1 = 0$ evaluated at $t = 1$ gives $k$ , and for $k \in {3, 5}$ we have $1 + k - 1 = k > 0$ while at $t = 0$ it is $- 1 < 0$ , the single positive root strictly lies in $(0, 1)$ . Thus $\ln t < 0$ for both, confirming both solutions $x$ are negative.

Hints that build this answer step by step

Letting $t = e^{x} > 0$ , the equation becomes $t^{4} + 8 t^{3} + 13 t^{2} - 8 t + 1 = 0$ . What is the most effective algebraic reduction?
Divide by $t^{2}$ and substitute $u = t - \frac{1}{t}$
Using $u = t - \frac{1}{t}$ , we have $t^{2} + \frac{1}{t^{2}} = u^{2} + 2$ . What are the values of $u$ ?
$u = - 3$ and $u = - 5$
For $u = - 3$ and $u = - 5$ , how many positive solutions exist for $t$ in $t - \frac{1}{t} = u$ , and where do they lie relative to $1$ ?
Each value of $u$ gives exactly one positive solution $t$ , and both satisfy $0 < t < 1$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we divide by t^2 and use u = t - 1/t instead of u = t + 1/t?

The coefficients of the cubic and linear terms are +8 and -8. Factoring out 8 gives 8(t - 1/t), which dictates using u = t - 1/t.

Why does 0 < t < 1 mean x is negative?

Here, $x = \ln (t)$ . The natural logarithm function is strictly increasing, and $\ln (1) = 0$ . Therefore, $\ln (t)$ is strictly negative when $0 < t < 1$ .

Answer

Substituting $t = e^{x}$ and dividing by $t^{2}$ allows the equation to be transformed via $u = t - \frac{1}{t}$ , yielding two valid positive values of $t$ , both of which lie strictly between $0$ and $1$ , corresponding to two negative real solutions for $x$ .

Why each option works or fails

A: four solutions two of which are negative - The student treats all four roots of the quartic polynomial in $t = e^{x}$ as valid real values for $x$ , forgetting that $t$ must be strictly positive. Check the condition $t > 0$ for $t = e^{x}$ ; negative values of $t$ cannot produce real values of $x$ .
B: two solutions and only one of them is negative - The student identifies two positive roots for $t = e^{x}$ but assumes that one root is greater than $1$ (yielding a positive $x$ ) and one is less than $1$ (yielding a negative $x$ ). Compare each positive root $t$ to $1$ : because both satisfy $0 < t < 1$ , both corresponding solutions $x = \ln (t)$ are strictly negative.
C · correct: two solutions and both are negative - The student correctly transforms the equation, enforces $t = e^{x} > 0$ , and checks whether $t > 1$ or $0 < t < 1$ . Correctly observed that the substitution gives two valid roots for $t$ , both lying in the interval $(0, 1)$ , giving two negative values for $x$ .
D: no solution - The student makes an algebraic sign error when substituting $u = t - 1 / t$ or $u = t + 1 / t$ , concluding that the quadratic has no real roots. Carefully square $t - 1 / t = u$ to get $t^{2} + 1 / t^{2} = u^{2} + 2$ , and solve the resulting quadratic $u^{2} + 8 u + 15 = 0$ .

Step-by-step solution

Given: The equation is $e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0$ for $x \in R$ .
Goal: Determine the number of real solutions $x$ and their signs.
Approach: Substitute $t = e^{x} > 0$ . Since the polynomial $t^{4} + 8 t^{3} + 13 t^{2} - 8 t + 1 = 0$ has anti-symmetric inner coefficients ( $+ 8$ and $- 8$ ), divide by $t^{2}$ and use the auxiliary substitution $u = t - \frac{1}{t}$ . Solve the resulting quadratic in $u$ , find the corresponding positive roots $t$ , and deduce the signs of $x = \ln t$ .
Execute: Substitute $t = e^{x}$ with $t > 0$ : $t^{4} + 8 t^{3} + 13 t^{2} - 8 t + 1 = 0$ Dividing by $t^{2} \neq 0$ : $(t^{2} + \frac{1}{t^{2}}) + 8 (t - \frac{1}{t}) + 13 = 0$ Let $u = t - \frac{1}{t}$ . Then $u^{2} = t^{2} - 2 + \frac{1}{t^{2}} ⟹ t^{2} + \frac{1}{t^{2}} = u^{2} + 2$ . The equation becomes: $(u^{2} + 2) + 8 u + 13 = 0 ⟹ u^{2} + 8 u + 15 = 0$ Factoring gives: $(u + 3) (u + 5) = 0 ⟹ u = - 3 or u = - 5$
Execute: Now solve for $t > 0$ from each value of $u$ : Case 1: $t - \frac{1}{t} = - 3 ⟹ t^{2} + 3 t - 1 = 0$ The roots are $t = \frac{- 3 \pm \sqrt{9 - 4 (1) (- 1)}}{2} = \frac{- 3 \pm \sqrt{13}}{2}$ . Since $t > 0$ , we have one valid root: $t_{1} = \frac{\sqrt{13} - 3}{2}$ . Since $3 = \sqrt{9} < \sqrt{13} < \sqrt{25} = 5$ , we have $0 < \sqrt{13} - 3 < 2$ , so $0 < t_{1} < 1$ . Case 2: $t - \frac{1}{t} = - 5 ⟹ t^{2} + 5 t - 1 = 0$ The roots are $t = \frac{- 5 \pm \sqrt{25 - 4 (1) (- 1)}}{2} = \frac{- 5 \pm \sqrt{29}}{2}$ . Since $t > 0$ , we have one valid root: $t_{2} = \frac{\sqrt{29} - 5}{2}$ . Since $5 = \sqrt{25} < \sqrt{29} < \sqrt{49} = 7$ , we have $0 < \sqrt{29} - 5 < 2$ , so $0 < t_{2} < 1$ .
Execute: Since $x = \ln t$ , for both roots we have $0 < t < 1$ , which gives $\ln t < 0$ . Thus, $x_{1} = \ln t_{1} < 0$ and $x_{2} = \ln t_{2} < 0$ . Therefore, there are exactly two real solutions, and both are negative.
Verify: Each quadratic $t^{2} + k u - 1 = 0$ has constant term $- 1$ , meaning its two roots have product $- 1$ . Hence, exactly one root is positive and one is negative for each equation, yielding precisely two positive values of $t$ . Since $t^{2} + k t - 1 = 0$ evaluated at $t = 1$ gives $k$ , and for $k \in {3, 5}$ we have $1 + k - 1 = k > 0$ while at $t = 0$ it is $- 1 < 0$ , the single positive root strictly lies in $(0, 1)$ . Thus $\ln t < 0$ for both, confirming both solutions $x$ are negative.

Determining the number and signs of real roots of a reciprocal exponential equation

What feels right?

Substituting $t = e^{x}$ and dividing by $t^{2}$ allows the equation to be transformed via $u = t - \frac{1}{t}$ , yielding two valid positive values of $t$ , both of which lie strictly between $0$ and $1$ , corresponding to two negative real solutions for $x$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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