The number of real roots of the equation e^6x - e^4x - 2e^3x - 12e^2x + e^x + 1 = 0 is :

Quadratic Equation and Inequalities PYQ — JEE Main 2021 · 25 Jul · Shift 1 · Q36 with Solution

View written solutionFree

Correct answer: A

Substitute to simplify the equation

Let $t=e^x.$ Since $e^x>0$ for all real $x$ , we have $t>0.$

The given equation becomes $t^6-t^4-2t^3-12t^2+t+1=0.$

So we need the number of positive real roots of $P(t)=t^6-t^4-2t^3-12t^2+t+1.$

Try factoring the polynomial

We look for a factorization of the form $P(t)=(t^3+at^2+bt+c)(t^3-at^2+bt-c),$ because the coefficients suggest some symmetry.

Try $P(t)=(t^3+t^2-3t-1)(t^3-t^2-3t+1).$

Now verify by multiplication:

[ \begin{aligned} &(t^3+t^2-3t-1)(t^3-t^2-3t+1) \ &= t^6 - t^5 - 3t^4 + t^3 + t^5 - t^4 - 3t^3 + t^2 \ &\quad -3t^4 + 3t^3 + 9t^2 - 3t - t^3 + t^2 + 3t -1 \ &= t^6 - 7t^4 + 11t^2 -1 \end{aligned} ]

This does not match, so we try another approach.

Observe a useful grouping

Let us test whether the polynomial can be written in terms of $u=t-\frac1t.$ But a simpler approach is direct factorization by checking rational roots.

For $t=1$ : $P(1)=1-1-2-12+1+1=-12\ne0.$

For $t=-1$ : $P(-1)=1-1+2-12-1+1=-10\ne0.$

Now let us factor systematically.

Using algebraic factorization, we get $P(t)=(t^2+t-1)(t^4-t^3-3t^2-9t-1).$

Check quickly: [ \begin{aligned} &(t^2+t-1)(t^4-t^3-3t^2-9t-1) \ &= t^6 - t^5 - 3t^4 - 9t^3 - t^2 \ &\quad + t^5 - t^4 - 3t^3 - 9t^2 - t \ &\quad - t^4 + t^3 + 3t^2 + 9t + 1 \ &= t^6 -5t^4 -11t^3 -7t^2 +8t +1 \end{aligned} ] Again this does not match. So let us proceed more carefully.

Try a substitution using symmetry-like terms

Given $t^6-t^4-2t^3-12t^2+t+1=0,$ divide by $t^3$ (valid since $t>0$ ): $t^3-t-2-\frac{12}{t}+\frac{1}{t^2}+\frac{1}{t^3}=0.$ This is not symmetric enough directly.

So instead, let us factor by inspection with quadratic factors.

Suppose $P(t)=(t^2+at+b)(t^4-at^3+ct^2+dt+e).$ Matching constant term gives $be=1$ .

A cleaner route is to check whether $t^2-3t-1$ is a factor:

Divide $P(t)$ by $t^2-3t-1$ .

Performing the division, $P(t)=(t^2-3t-1)(t^4+3t^3+9t^2+4t-1).$

Now divide the quartic further by $t^2+3t-1$ : $t^4+3t^3+9t^2+4t-1=(t^2+3t-1)(t^2+10).$

Hence $P(t)=(t^2-3t-1)(t^2+3t-1)(t^2+10).$

Let us verify: [ \begin{aligned} (t^2-3t-1)(t^2+3t-1) &=(t^2-1)^2-(3t)^2 \ &= t^4-11t^2+1. \end{aligned} ] Then [ \begin{aligned} (t^4-11t^2+1)(t^2+10) &= t^6+10t^4-11t^4-110t^2+t^2+10 \ &= t^6-t^4-109t^2+10, \end{aligned} ] which again does not match. So this factorization is also incorrect.

Use sign analysis and calculus on the transformed polynomial

We now study

\qquad t>0.$$ We want the number of positive roots. Compute values: - $$P(0)=1>0$$ - $$P(1)=1-1-2-12+1+1=-12<0$$ - $$P(2)=64-16-16-48+2+1=-13<0$$ - $$P(3)=729-81-54-108+3+1=490>0$$ So there is: - one root in $(0,1)$, - one root in $(2,3)$. Now we check whether there can be more positive roots. --- 6. **Use Descartes' Rule of Signs** For $$P(t)=t^6-t^4-2t^3-12t^2+t+1,$$ the sequence of nonzero coefficients is $$1,",-1,",-2,",-12,1,1.$$ Sign changes are: - $1$ to $-1$ : one change - $-12$ to $1$ : second change So the number of positive roots is either $2$ or $0$. Since we already found two intervals where roots exist, the number of positive roots is exactly $$2.$$ Since $t=e^x$ is one-to-one, each positive root of $t$ gives exactly one real root $x$. Therefore, the original equation has exactly $$2$$ real roots. --- 7. **Evaluate options** - **A: 2** ✅ - **B: 4** ❌ - **C: 6** ❌ - **D: 1** ❌ Thus the correct option is **A**.

More from Quadratic Equation and Inequalities

View written solutionFree

Correct answer: A

Substitute to simplify the equation

Let $t=e^x.$ Since $e^x>0$ for all real $x$ , we have $t>0.$

The given equation becomes $t^6-t^4-2t^3-12t^2+t+1=0.$

So we need the number of positive real roots of $P(t)=t^6-t^4-2t^3-12t^2+t+1.$

Try factoring the polynomial

We look for a factorization of the form $P(t)=(t^3+at^2+bt+c)(t^3-at^2+bt-c),$ because the coefficients suggest some symmetry.

Try $P(t)=(t^3+t^2-3t-1)(t^3-t^2-3t+1).$

Now verify by multiplication:

[ \begin{aligned} &(t^3+t^2-3t-1)(t^3-t^2-3t+1) \ &= t^6 - t^5 - 3t^4 + t^3 + t^5 - t^4 - 3t^3 + t^2 \ &\quad -3t^4 + 3t^3 + 9t^2 - 3t - t^3 + t^2 + 3t -1 \ &= t^6 - 7t^4 + 11t^2 -1 \end{aligned} ]

This does not match, so we try another approach.

Observe a useful grouping

Let us test whether the polynomial can be written in terms of $u=t-\frac1t.$ But a simpler approach is direct factorization by checking rational roots.

For $t=1$ : $P(1)=1-1-2-12+1+1=-12\ne0.$

For $t=-1$ : $P(-1)=1-1+2-12-1+1=-10\ne0.$

Now let us factor systematically.

Using algebraic factorization, we get $P(t)=(t^2+t-1)(t^4-t^3-3t^2-9t-1).$

Try a substitution using symmetry-like terms

Given $t^6-t^4-2t^3-12t^2+t+1=0,$ divide by $t^3$ (valid since $t>0$ ): $t^3-t-2-\frac{12}{t}+\frac{1}{t^2}+\frac{1}{t^3}=0.$ This is not symmetric enough directly.

So instead, let us factor by inspection with quadratic factors.

Suppose $P(t)=(t^2+at+b)(t^4-at^3+ct^2+dt+e).$ Matching constant term gives $be=1$ .

A cleaner route is to check whether $t^2-3t-1$ is a factor:

Divide $P(t)$ by $t^2-3t-1$ .

Performing the division, $P(t)=(t^2-3t-1)(t^4+3t^3+9t^2+4t-1).$

Now divide the quartic further by $t^2+3t-1$ : $t^4+3t^3+9t^2+4t-1=(t^2+3t-1)(t^2+10).$

Hence $P(t)=(t^2-3t-1)(t^2+3t-1)(t^2+10).$

Use sign analysis and calculus on the transformed polynomial

We now study

\qquad t>0.$$ We want the number of positive roots. Compute values: - $$P(0)=1>0$$ - $$P(1)=1-1-2-12+1+1=-12<0$$ - $$P(2)=64-16-16-48+2+1=-13<0$$ - $$P(3)=729-81-54-108+3+1=490>0$$ So there is: - one root in $(0,1)$, - one root in $(2,3)$. Now we check whether there can be more positive roots. --- 6. **Use Descartes' Rule of Signs** For $$P(t)=t^6-t^4-2t^3-12t^2+t+1,$$ the sequence of nonzero coefficients is $$1,",-1,",-2,",-12,1,1.$$ Sign changes are: - $1$ to $-1$ : one change - $-12$ to $1$ : second change So the number of positive roots is either $2$ or $0$. Since we already found two intervals where roots exist, the number of positive roots is exactly $$2.$$ Since $t=e^x$ is one-to-one, each positive root of $t$ gives exactly one real root $x$. Therefore, the original equation has exactly $$2$$ real roots. --- 7. **Evaluate options** - **A: 2** ✅ - **B: 4** ❌ - **C: 6** ❌ - **D: 1** ❌ Thus the correct option is **A**.

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

More from Quadratic Equation and Inequalities

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

More from Quadratic Equation and Inequalities