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

Application of Derivatives PYQ — JEE Main 2021 · 31 Aug · Shift 1 · Q22 with Solution

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Correct answer: C

We need to find the number of real roots of $e^{4x}+2e^{3x}-e^x-6=0.$
Since exponential terms are positive, use the substitution $t=e^x.$ Then $t>0.$ The equation becomes $t^4+2t^3-t-6=0.$
Now factor the polynomial. Try simple rational roots. For $t=1$ : $1+2-1-6=-4\neq 0.$ For $t=-1$ : $1-2+1-6=-6\neq 0.$ For $t=\sqrt[?]{}$$ not useful directly, so group terms: $$t^4+2t^3-t-6=t^3(t+2)-1(t+6),$$ which does not help. Try factorization of the form $$(t^2+at+b)(t^2+ct+d).$$ But an easier check is$ t=-2 $: $$16-16+2-6=-4\neq 0,$$ and$ t=\sqrt[?]{}$ still not immediate.

Let us test whether $t=\frac{3}{2}$ works:

=\frac{81}{16}+\frac{54}{8}-\frac32-6 =\frac{81}{16}+\frac{108}{16}-\frac{24}{16}-\frac{96}{16} =\frac{69}{16}\neq 0.$$ So instead, analyze the function $$f(t)=t^4+2t^3-t-6, \qquad t>0.$$ 4. Differentiate: $$f'(t)=4t^3+6t^2-1.$$ Again, $$f''(t)=12t^2+12t>0 \quad \text{for } t>0.$$ So $f'(t)$ is strictly increasing on $(0,\infty)$. Now check values: $$f'(0)=-1<0,$$ $$f'(1)=4+6-1=9>0.$$ Hence $f'(t)$ has exactly one root in $(0,\infty)$, so $f(t)$ decreases first and then increases. Therefore, on $t>0$, the equation $f(t)=0$ can have at most two positive roots. 5. Check signs of $f(t)$: $$f(0)=-6<0,$$ $$f(1)=1+2-1-6=-4<0,$$ $$f(2)=16+16-2-6=24>0.$$ So there is at least one positive root in $(1,2)$. Since $f(t)$ decreases first and then increases, and both $f(0)$ and $f(1)$ are negative, the minimum value is also negative. After the minimum, the function increases and crosses the axis only once for $t>0$. Thus there is exactly one positive root. 6. Since $t=e^x>0$, each positive root $t$ gives exactly one real $x=\ln t$. Therefore, the original equation has exactly **one real root**. 7. Hence the correct option is $$\boxed{\text{C: }1}.$$

View written solutionFree

Correct answer: C

We need to find the number of real roots of $e^{4x}+2e^{3x}-e^x-6=0.$
Since exponential terms are positive, use the substitution $t=e^x.$ Then $t>0.$ The equation becomes $t^4+2t^3-t-6=0.$
Now factor the polynomial. Try simple rational roots. For $t=1$ : $1+2-1-6=-4\neq 0.$ For $t=-1$ : $1-2+1-6=-6\neq 0.$ For $t=\sqrt[?]{}$$ not useful directly, so group terms: $$t^4+2t^3-t-6=t^3(t+2)-1(t+6),$$ which does not help. Try factorization of the form $$(t^2+at+b)(t^2+ct+d).$$ But an easier check is$ t=-2 $: $$16-16+2-6=-4\neq 0,$$ and$ t=\sqrt[?]{}$ still not immediate.

Let us test whether $t=\frac{3}{2}$ works:

=\frac{81}{16}+\frac{54}{8}-\frac32-6 =\frac{81}{16}+\frac{108}{16}-\frac{24}{16}-\frac{96}{16} =\frac{69}{16}\neq 0.$$ So instead, analyze the function $$f(t)=t^4+2t^3-t-6, \qquad t>0.$$ 4. Differentiate: $$f'(t)=4t^3+6t^2-1.$$ Again, $$f''(t)=12t^2+12t>0 \quad \text{for } t>0.$$ So $f'(t)$ is strictly increasing on $(0,\infty)$. Now check values: $$f'(0)=-1<0,$$ $$f'(1)=4+6-1=9>0.$$ Hence $f'(t)$ has exactly one root in $(0,\infty)$, so $f(t)$ decreases first and then increases. Therefore, on $t>0$, the equation $f(t)=0$ can have at most two positive roots. 5. Check signs of $f(t)$: $$f(0)=-6<0,$$ $$f(1)=1+2-1-6=-4<0,$$ $$f(2)=16+16-2-6=24>0.$$ So there is at least one positive root in $(1,2)$. Since $f(t)$ decreases first and then increases, and both $f(0)$ and $f(1)$ are negative, the minimum value is also negative. After the minimum, the function increases and crosses the axis only once for $t>0$. Thus there is exactly one positive root. 6. Since $t=e^x>0$, each positive root $t$ gives exactly one real $x=\ln t$. Therefore, the original equation has exactly **one real root**. 7. Hence the correct option is $$\boxed{\text{C: }1}.$$