Q.Evaluate: $\int \dfrac{e^{6\log x}-e^{5\log x}}{e^{4\log x}-e^{3\log x}}\,dx$

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $\sinh x = \dfrac{-1}{2}$, $\cosh y = 2$, and $x + y = \log p$, then $p =$ (A) $\dfrac{4+2\sqrt{3}}{\sqrt{5}+1}$ (B) $\dfrac{\sqrt{5}-1}{4\sqrt{3}+2}$ (C) $(\sqrt{5}-1)(2+\sqrt{3})$ (D) $\dfrac{4-2\sqrt{3}}{\sqrt{5}+1}$

   ›Reveal solutionSolution

   Solve each hyperbolic equation for $e^x$ and $e^y$ as quadratics in $t=e^{(\cdot)}$, then multiply to get $p=e^{x+y}$.

   Concept and Intuition

   $\sinh x$ and $\cosh y$ are defined purely in terms of $e^x,e^y$, so each condition is really a quadratic in $t=e^x$ (or $e^y$) once you clear denominators. Taking the positive root (since $e^{(\cdot)}>0$ always) pins down each exponential uniquely (using the principal/positive value of $y$), and then $p=e^{x+y}=e^x e^y$ is just a product.

   Step-by-Step Solution

   1. $\sinh x = \dfrac{e^x-e^{-x}}{2}=-\dfrac12 \Rightarrow e^x - e^{-x} = -1$.
   2. Let $t=e^x$: $t-\dfrac1t=-1 \Rightarrow t^2+t-1=0 \Rightarrow t=\dfrac{-1\pm\sqrt5}{2}$. Since $t>0$, $t=\dfrac{\sqrt5-1}{2}$.
   3. $\cosh y=\dfrac{e^y+e^{-y}}{2}=2 \Rightarrow e^y+e^{-y}=4$.
   4. Let $s=e^y$: $s+\dfrac1s=4 \Rightarrow s^2-4s+1=0 \Rightarrow s=2\pm\sqrt3$. Taking the principal (positive) $y=\cosh^{-1}2$, $s=e^y=2+\sqrt3$.
   5. $p=e^{x+y}=e^x\cdot e^y = \dfrac{\sqrt5-1}{2}\cdot(2+\sqrt3)$. …

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2. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.The equation $x^{\frac{3}{4}(\log_2 x)^2+\log_2 x-\frac{5}{4}}=\sqrt{2}$ has (A) no real roots (B) only one real solution (C) exactly two real solutions (D) exactly three real solutions

   ›Reveal solutionSolution

   Substituting $t=\log_2x$ turns the equation into a cubic $3t^3+4t^2-5t-2=0$, which factors into three distinct real roots — each corresponding to one valid positive $x$. So the equation has exactly three real solutions.

   Concept and Intuition

   Whenever the variable appears both as the base and inside the exponent (here $x$ appears as the base and $\log_2x$ appears inside the exponent), substituting $t=\log_2x$ is the natural move: it converts everything into a polynomial equation in $t$, and since $x=2^t$ is a strictly increasing bijection from $\mathbb R$ to $(0,\infty)$, every real solution $t$ corresponds to exactly one valid positive $x$ (no solutions are lost or spuriously created).

   Step-by-Step Solution

   1. Domain: $x>0$ (so $\log_2x$ is defined). Let $t=\log_2x$.
   2. The given equation is $x^{\frac34t^2+t-\frac54}=\sqrt2$. Take $\log_2$ of both sides (valid since both sides are positive):

   $\left(\tfrac34t^2+t-\tfrac54\right)\cdot\log_2x = \log_2\sqrt2 = \tfrac12.$

   3. Since $\log_2 x = t$: $\left(\tfrac34t^2+t-\tfrac54\right)t = \tfrac12$.
   4. Expand: $\tfrac34t^3+t^2-\tfrac54t-\tfrac12=0$. Multiply through by $4$: $3t^3+4t^2-5t-2=0$.
   5. Test $t=1$: $3(1)+4(1)-5(1)-2 = 3+4-5-2=0$. ✓ So $(t-1)$ is a factor.
   6. Divide: $3t^3+4t^2-5t-2 = (t-1)(3t^2+7t+2)$.
   7. Solve $3t^2+7t+2=0$ via the quadratic formula: $t = \dfrac{-7\pm\sqrt{7^2-4(3)(2)}}{2(3)} = \dfrac{-7\pm\sqrt{49-24}}{6}=\dfrac{-7\pm5}{6}$.
   8. This gives $t=\dfrac{-7+5}{6}=-\dfrac13$ and $t=\dfrac{-7-5}{6}=-2$.
   9. So the three roots are $t=1,\ -\dfrac13,\ -2$ — all distinct real numbers. …

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3. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$\displaystyle\lim_{x\to\infty}\left[\left(1+\dfrac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\dfrac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\dfrac{27}{n^3}\right)^{\frac{9}{n^3}}\ldots(2)^{\frac{1}{n}}\right]=$ (A) $\log2-\dfrac{1}{2}$ (B) $e^{\left(\log2-\frac{1}{2}\right)}$ (C) $e^{\left(\frac{2\log2-1}{3}\right)}$ (D) $\dfrac{1}{3}(2\log2-1)$

   ›Reveal solutionSolution

   Taking the log converts the product into a Riemann sum that becomes $\int_0^1 x^2\log(1+x^3)dx=\frac13(2\log2-1)$; exponentiating gives option (C).

   Concept and Intuition

   A product of the form $\prod_k\left(1+\frac{k^3}{n^3}\right)^{k^2/n^3}$ is a classic "log turns product into Riemann sum" limit: taking $\log$ converts the exponents into a sum that, in the limit $n\to\infty$, becomes a definite integral in $x=k/n$.

   Step-by-Step Solution

   1. Identify the general factor: for $k=1,2,\dots,n$, the term is $\left(1+\dfrac{k^3}{n^3}\right)^{k^2/n^3}$; at $k=n$ this is $(1+1)^{n^2/n^3}=2^{1/n}$, matching the last given factor.
   2. Let $L=\log(\text{product})=\displaystyle\sum_{k=1}^n\frac{k^2}{n^3}\log\left(1+\frac{k^3}{n^3}\right)=\frac1n\sum_{k=1}^n\left(\frac kn\right)^2\log\left(1+\left(\frac kn\right)^3\right)$.
   3. As $n\to\infty$ this Riemann sum $\to\displaystyle\int_0^1 x^2\log(1+x^3)\,dx$. …

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4. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.$\cosh(\log 4) =$ (A) $\dfrac{8}{17}$ (B) $\dfrac{17}{8}$ (C) $0$ (D) $\dfrac{9}{8}$

   ›Reveal solutionSolution

   A direct application of the hyperbolic cosine definition; $\cosh(\log4)=\dfrac{17}{8}$.

   Concept and Intuition

   $\cosh u$ is defined purely in terms of the exponential function, so as soon as $e^u$ is known (here $u=\log4$ means natural log, so $e^u=4$ exactly), the value follows immediately — no trig or hyperbolic identities beyond the definition are needed.

   Step-by-Step Solution

   1. $\cosh u=\dfrac{e^u+e^{-u}}{2}$.
   2. With $u=\log4$ (natural log), $e^u=e^{\log4}=4$.
   3. $e^{-u}=\dfrac1{e^u}=\dfrac14$.
   4. $\cosh(\log4)=\dfrac{4+\frac14}{2}=\dfrac{\frac{16}4+\frac14}{2}=\dfrac{\frac{17}4}{2}=\dfrac{17}{8}$.

   Common Mistakes …

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5. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $\alpha=\log_e(2+\sqrt{3})$, then $\dfrac{\cosh\alpha}{1-\tanh\alpha}+\dfrac{\sinh\alpha}{1-\coth\alpha}=$ (A) $4+2\sqrt{3}$ (B) $7+4\sqrt{3}$ (C) $\dfrac{\sqrt{3}+1}{2}$ (D) $2+\sqrt{3}$

   ›Reveal solutionSolution

   The given expression simplifies to $e^\alpha$ using $\cosh^2\alpha-\sinh^2\alpha=1$ and $\cosh\alpha-\sinh\alpha=e^{-\alpha}$; with $\alpha=\log(2+\sqrt3)$, this is just $2+\sqrt3$.

   Concept and Intuition

   Expressions combining $\tanh,\coth$ with $\cosh,\sinh$ often simplify dramatically by writing everything over a common denominator involving $\cosh\alpha-\sinh\alpha$ or $\cosh\alpha+\sinh\alpha$ — these combinations are exactly $e^{\mp\alpha}$, which is the whole point of hyperbolic functions being defined via exponentials.

   Step-by-Step Solution

   1. $\dfrac{\cosh\alpha}{1-\tanh\alpha} = \dfrac{\cosh\alpha}{1-\frac{\sinh\alpha}{\cosh\alpha}} = \dfrac{\cosh^2\alpha}{\cosh\alpha-\sinh\alpha}$.
   2. $\dfrac{\sinh\alpha}{1-\coth\alpha} = \dfrac{\sinh\alpha}{1-\frac{\cosh\alpha}{\sinh\alpha}} = \dfrac{\sinh^2\alpha}{\sinh\alpha-\cosh\alpha} = \dfrac{-\sinh^2\alpha}{\cosh\alpha-\sinh\alpha}$.
   3. Adding: $\dfrac{\cosh^2\alpha-\sinh^2\alpha}{\cosh\alpha-\sinh\alpha} = \dfrac{1}{\cosh\alpha-\sinh\alpha}$ (using the identity $\cosh^2\alpha-\sinh^2\alpha=1$). …

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6. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $2\sinh x = \cosh x$, then $x=$ (A) $\dfrac{1}{3}\log 2$ (B) $2\log 3$ (C) $\dfrac{1}{2}\log 3$ (D) $\log 9$

   ›Reveal solutionSolution

   Writing $\sinh x,\cosh x$ in exponential form and solving the resulting linear equation in $e^x$ gives $e^{2x}=3$, so $x=\frac12\ln3$.

   Concept and Intuition

   $\sinh x$ and $\cosh x$ are defined directly via $e^x$: $\sinh x=\dfrac{e^x-e^{-x}}2$, $\cosh x=\dfrac{e^x+e^{-x}}2$. Substituting these definitions converts a hyperbolic equation into ordinary exponential algebra.

   Step-by-Step Solution

   1. $2\sinh x=\cosh x\Rightarrow2\cdot\dfrac{e^x-e^{-x}}2=\dfrac{e^x+e^{-x}}2$.
   2. Simplify left side: $e^x-e^{-x}=\dfrac{e^x+e^{-x}}2$.
   3. Multiply both sides by 2: $2e^x-2e^{-x}=e^x+e^{-x}$.
   4. Rearranging: $2e^x-e^x=e^{-x}+2e^{-x}\Rightarrow e^x=3e^{-x}$.
   5. Multiply both sides by $e^x$: $e^{2x}=3$. …

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7. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.If $\alpha$ and $\beta$ are the roots of the equation $2^{6x}-3(2^{3x+2})+32=0$ with $\beta<1$, then $2\alpha+3\beta=$ (A) $-3$ (B) $-4$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Tests solving an exponential equation via substitution, then correctly assigning roots based on the given size constraint. Answer: $2\alpha+3\beta=4$ (option D).

   Concept and Intuition

   Equations with repeated exponential terms like $2^{6x}$ and $2^{3x+2}$ can be reduced to a quadratic by substituting $t=2^{3x}$ (noting $2^{6x}=t^2$ and $2^{3x+2}=4\cdot2^{3x}=4t$). After solving for $t$, convert back to $x$ using logarithms (or by recognizing powers of 2), then use the given ordering constraint to correctly assign which root is $\alpha$ and which is $\beta$.

   Step-by-Step Solution

   1. Equation: $2^{6x}-3(2^{3x+2})+32=0$. Let $t=2^{3x}$, so $2^{6x}=t^2$ and $2^{3x+2}=4t$.
   2. Substitute: $t^2-12t+32=0$.
   3. Solve: $t=\dfrac{12\pm\sqrt{144-128}}{2}=\dfrac{12\pm4}{2}$, giving $t=8$ or $t=4$.
   4. $t=8=2^3 \Rightarrow 2^{3x}=2^3\Rightarrow 3x=3\Rightarrow x=1$.
   5. $t=4=2^2 \Rightarrow 2^{3x}=2^2\Rightarrow 3x=2\Rightarrow x=\tfrac23$. …

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8. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If $4+6(e^{2x}+1)\tanh x = 11\cosh x + 11\sinh x$, then $x =$ (A) $\log 10$ (B) $\log 4$ (C) $\log 5$ (D) $\log 2$

   ›Reveal solutionSolution

   Rewriting the hyperbolic expression in terms of $y=e^x$ reduces the equation to a solvable quadratic $6y^2-11y-2=0$, giving $x=\log 2$.

   Concept and Intuition

   Hyperbolic-function equations usually collapse into ordinary algebraic (often quadratic) equations once everything is expressed via $e^x$, since $\cosh x, \sinh x, \tanh x$ are all built from $e^x$ and $e^{-x}$.

   Step-by-Step Solution

   1. Note $e^{2x}+1 = e^x(e^x+e^{-x}) = 2e^x\cosh x$.
   2. So $6(e^{2x}+1)\tanh x = 6\cdot 2e^x\cosh x\cdot\frac{\sinh x}{\cosh x} = 12e^x\sinh x$.
   3. The equation becomes $4+12e^x\sinh x = 11(\cosh x+\sinh x) = 11e^x$ (since $\cosh x+\sinh x=e^x$).
   4. Since $e^x\sinh x = e^x\cdot\frac{e^x-e^{-x}}{2} = \frac{e^{2x}-1}{2}$, we get $4+6(e^{2x}-1) = 11e^x$, i.e. $6e^{2x}-2=11e^x$. …

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9. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.If $4^x - 3^{x-1/2} = 3^{x+1/2} - 2^{2x-1}$ then the value of $x$ is (A) $7/2$ (B) $5/2$ (C) $1/2$ (D) $3/2$

   ›Reveal solutionSolution

   Grouping powers of 2 on one side and powers of 3 on the other reduces the equation to $(4/3)^x = 8/(3\sqrt3)$, solved by $x=3/2$.

   Concept and Intuition

   Exponential equations mixing two different bases (here 2 and 3, disguised as 4 and 3) can often be solved by collecting all terms of each base onto one side, factoring out the common power, and then comparing/solving the resulting single-base-ratio equation.

   Step-by-Step Solution

   1. Rewrite $4^x=2^{2x}$ and rearrange the given equation $4^x-3^{x-1/2}=3^{x+1/2}-2^{2x-1}$ as $2^{2x}+2^{2x-1} = 3^{x+1/2}+3^{x-1/2}$.
   2. LHS: $2^{2x}+2^{2x-1}=2^{2x}(1+\tfrac12)=2^{2x}\cdot\tfrac32$.
   3. RHS: $3^{x+1/2}+3^{x-1/2}=3^x(\sqrt3+\tfrac1{\sqrt3})=3^x\cdot\dfrac{4}{\sqrt3}$.
   4. So $2^{2x}\cdot\dfrac32 = 3^x\cdot\dfrac{4}{\sqrt3} \Rightarrow \dfrac{2^{2x}}{3^x}=\dfrac{8}{3\sqrt3} \Rightarrow \left(\dfrac43\right)^x = \dfrac{8}{3\sqrt3}$. …

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10. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.If $\lim_{n \to \infty} x^n \log_e x = 0$, then $\log_x 12 =$ (A) Negative (B) Positive (C) Zero (D) any value between -1 and 1

    ›Reveal solutionSolution

    The limit condition forces $0<x<1$; with a base strictly between $0$ and $1$, $\log_x$ is decreasing, so $\log_x 12$ (evaluated at an argument bigger than $1$) must be negative.

    Concept and Intuition

    $x^n\to0$ as $n\to\infty$ only when $|x|<1$; combined with the domain of $\log$ requiring $x>0$, this pins down $0<x<1$. For such a base, the logarithm function is decreasing (unlike bases $>1$, where it's increasing), which flips the usual intuition about signs.

    Step-by-Step Solution

    1. For the limit $\lim_{n\to\infty} x^n \ln x = 0$ to hold with $\ln x$ a fixed (nonzero, for $x\neq1$) real number, we need $x^n \to 0$.
    2. $x^n\to0$ as $n\to\infty$ precisely when $0<x<1$ (for $x>1$, $x^n\to\infty$; for $x=1$, $x^n=1$ always, and the whole expression is identically $0$ trivially but $x=1$ is excluded from any logarithm base anyway).
    3. So the condition effectively forces $0 < x < 1$. …

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11. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.$4^x - 3^{x-\frac{1}{2}} = 3^{x+\frac{1}{2}} - 2^{2x-1} \Rightarrow x =$ (A) $\frac{5}{2}$ (B) $\frac{1}{2}$ (C) $\frac{3}{2}$ (D) $\frac{7}{2}$

    ›Reveal solutionSolution

    Regrouping all the base-2 terms on one side and all the base-3 terms on the other turns the equation into a single balance that is exactly satisfied at $x=\frac32$.

    Concept and Intuition

    Exponential equations mixing two different bases (2 and 3 here, via $4=2^2$) usually can't be solved by taking a single logarithm cleanly; instead, collect all terms of one base together (using $a^{p}+a^{q}=a^{\min}(a^{|p-q|}+1)$-type factoring) and all terms of the other base together, then look for the value of $x$ that balances both sides — often it's a "nice" number found by testing, then confirmed algebraically.

    Step-by-Step Solution

    1. Rewrite every term with base 2 or base 3: $4^x=2^{2x}$, $2^{2x-1}=\dfrac{2^{2x}}{2}$.
    2. The equation $4^x-3^{x-1/2}=3^{x+1/2}-2^{2x-1}$ becomes $2^{2x}-3^{x-1/2} = 3^{x+1/2}-\dfrac{2^{2x}}{2}$.
    3. Move all base-2 terms to the left, all base-3 terms to the right: $2^{2x}+\dfrac{2^{2x}}{2} = 3^{x+1/2}+3^{x-1/2}$, i.e. $\dfrac{3}{2}\cdot 2^{2x} = 3^{x+1/2}+3^{x-1/2}$.
    4. Factor the right side: $3^{x+1/2}+3^{x-1/2} = 3^{x-1/2}(3^1+3^0)=3^{x-1/2}\cdot 4$. …

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12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.If $2\cosh 2x + 10\sinh 2x = 5$, then $x =$ (A) $\frac{1}{2}\log\frac{4}{3}$ (B) $\frac{1}{2}\log\frac{2}{3}$ (C) $\frac{1}{2}\log\frac{3}{2}$ (D) $\frac{1}{2}\log\frac{3}{4}$

    ›Reveal solutionSolution

    Rewrite the hyperbolic equation in terms of $e^{2x}$, reducing it to a quadratic.

    Concept and Intuition

    $\cosh$ and $\sinh$ are combinations of $e^{2x}$ and $e^{-2x}$; substituting these turns the transcendental equation into an algebraic (quadratic) one in $u=e^{2x}$.

    Step-by-Step Solution

    1. $\cosh2x = \dfrac{e^{2x}+e^{-2x}}{2}$, $\sinh2x=\dfrac{e^{2x}-e^{-2x}}{2}$.
    2. $2\cosh2x = e^{2x}+e^{-2x}$; $10\sinh2x = 5(e^{2x}-e^{-2x})$.
    3. Sum: $e^{2x}+e^{-2x}+5e^{2x}-5e^{-2x} = 5 \Rightarrow 6e^{2x} - 4e^{-2x}=5$.
    4. Let $u=e^{2x}$ (so $e^{-2x}=1/u$): $6u - \dfrac4u = 5 \Rightarrow 6u^2 - 5u - 4=0$.
    5. Solve: $u = \dfrac{5\pm\sqrt{25+96}}{12} = \dfrac{5\pm11}{12}$, giving $u=\dfrac{16}{12}=\dfrac43$ or $u=-\dfrac12$ (rejected, since $u=e^{2x}>0$). …

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