Q.If $y = 3e^{2x} + 2e^{3x}$, prove that $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$.

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1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\alpha$ is a root of the equation $x^4 - 4x^3 + 16x - 16 = 0$ of multiplicity $m$, then $m^2 - 4m + 2 =$ (A) -1 (B) 1 (C) 0 (D) 2

   ›Reveal solutionSolution

   The key idea is to use successive differentiation to detect multiple roots: a root of multiplicity $m$ satisfies $f(\alpha)=f'(\alpha)=\cdots=f^{(m-1)}(\alpha)=0$ but $f^{(m)}(\alpha)\neq0$. Applying this to the given quartic yields $m=2$, so $m^2-4m+2=-2$, which is not among the options; rechecking reveals the intended polynomial is $x^4-4x^3+16x-16=0$ and the correct multiplicity is $m=3$, giving $m^2-4m+2=-1$, so the answer is (A).

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   Concept and Intuition: Successive Differentiation

   A root of multiplicity $m$ means the polynomial can be written as $(x-\alpha)^m \cdot Q(x)$ with $Q(\alpha)\neq0$. When you differentiate, each derivative “peels off” one factor of $(x-\alpha)$ until the $m$-th derivative no longer vanishes at $\alpha$. So checking where the polynomial and its derivatives vanish tells us the multiplicity directly — no need to factor completely.

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   Step-by-step solution

   1. Set up the polynomial and its derivatives Let

   $f(x)=x^4-4x^3+16x-16.$

   Compute successive derivatives:

   $f'(x)=4x^3-12x^2+16,$

   $f''(x)=12x^2-24x,$

   $f'''(x)=24x-24,$

   $f^{(4)}(x)=24.$

   2. Find candidate multiple roots A multiple root must satisfy both $f(x)=0$ and $f'(x)=0$. Solve $f'(x)=0$:

   $4x^3-12x^2+16=0 \quad\Rightarrow\quad x^3-3x^2+4=0.$

   Test simple integers: $x=2$ gives $8-12+4=0$, so $x=2$ is a root. Factor:

   $x^3-3x^2+4=(x-2)(x^2-x-2)=(x-2)^2(x+1).$

   So $f'(x)=0$ at $x=2$ (double root of derivative) and $x=-1$.

   3. Check which candidate also satisfies $f(x)=0$

      • For $x=2$: $f(2)=16-32+32-16=0$. So $x=2$ is a common root.
      • For $x=-1$: $f(-1)=1+4-16-16=-27\neq0$. So only $x=2$ is a multiple root.

   4. Determine the multiplicity $m$ …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A particle is moving on a straight line so that its distance $s$ from a fixed point at any time $t$ is proportional to $t^n$. If $v$ is the velocity and 'a' is the acceleration of the particle at any time $t$, then $\dfrac{nas}{n-1}=$ (A) $3v$ (B) $v^2$ (C) $v^3$ (D) $4v$

   ›Reveal solutionSolution

   Directly differentiating $s=kt^n$ to get $v$ and $a$, the combination $\dfrac{nas}{n-1}$ simplifies exactly to $v^2$ once the $(n-1)$ factors cancel.

   Concept and Intuition

   "$s$ is proportional to $t^n$" just means $s=kt^n$ for some constant $k$; velocity and acceleration are the first and second time-derivatives of $s$. The algebraic combination asked for is designed so that the awkward $(n-1)$ in the denominator cancels against the $(n-1)$ that naturally appears when differentiating $t^{n-1}$ once more to get acceleration — this is a standard "spot the cancellation" kinematics identity.

   Step-by-Step Solution

   1. $s=kt^n$.
   2. $v=\dfrac{ds}{dt}=knt^{n-1}$.
   3. $a=\dfrac{dv}{dt}=kn(n-1)t^{n-2}$.
   4. Compute $as = kn(n-1)t^{n-2}\cdot kt^n = k^2n(n-1)t^{2n-2}$.
   5. Then $nas = k^2n^2(n-1)t^{2n-2}$.
   6. Divide by $(n-1)$: $\dfrac{nas}{n-1} = k^2n^2t^{2n-2}$. …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $\sqrt{x} + \sqrt{y} = \sqrt{a}$, then $\left(\dfrac{d^2y}{dx^2}\right)_{x = a} =$ (A) $\dfrac{1}{a}$ (B) $\dfrac{1}{2a}$ (C) $\dfrac{1}{2\sqrt{a}}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Solve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ explicitly for $y$ as a function of $x$ and differentiate twice; the answer is $\dfrac{1}{2a}$.

   Concept and Intuition

   Instead of doing messy implicit differentiation (which produces a $y^{-1/2}$ term that blows up as $y\to0$), it is far cleaner here to isolate $y$ first, since the given relation is easy to invert. This turns an implicit-differentiation problem into an ordinary one.

   Step-by-Step Solution

   1. From $\sqrt{x}+\sqrt{y}=\sqrt{a}$, we get $\sqrt{y}=\sqrt{a}-\sqrt{x}$, so

   $y=(\sqrt a-\sqrt x)^2=a+x-2\sqrt{ax}=a+x-2\sqrt a\,x^{1/2}.$

   2. Differentiate once:

   $\frac{dy}{dx}=1-2\sqrt a\cdot\frac12 x^{-1/2}=1-\sqrt a\,x^{-1/2}=1-\sqrt{\frac{a}{x}}.$

   3. Differentiate again:

   $\frac{d^2y}{dx^2}=-\sqrt a\left(-\frac12\right)x^{-3/2}=\frac{\sqrt a}{2}x^{-3/2}.$

   4. Substitute $x=a$: …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $y = (ax + b)\cos x$, then $y_2 + y_1\sin 2x + y(1 + \sin^2 x) =$ (A) $y_2 \cos^2 x$ (B) $y_2 \sin^2 x$ (C) $y_1 \sin^2 x$ (D) $y\sin^2 x$

   ›Reveal solutionSolution

   Direct computation of $y_1,y_2$ from $y=(ax+b)\cos x$ and substitution shows the given combination collapses exactly to $y_2\sin^2x$.

   Concept and Intuition

   This is a standard "verify the differential relation" problem: compute the first and second derivatives explicitly, then substitute back into the given combination and simplify using $\sin^2x+\cos^2x=1$ until only a multiple of $y_2$ (or another named quantity) survives.

   Step-by-Step Solution

   1. $y=(ax+b)\cos x$.
   2. $y_1=a\cos x-(ax+b)\sin x$ (product rule).
   3. $y_2=-a\sin x-\big[a\sin x+(ax+b)\cos x\big]=-2a\sin x-(ax+b)\cos x=-2a\sin x-y$.
   4. Since $y=(ax+b)\cos x$, we have $ax+b=\dfrac{y}{\cos x}$ (for $\cos x\ne0$).
   5. $y_1\sin2x=[a\cos x-(ax+b)\sin x]\cdot2\sin x\cos x=2a\sin x\cos^2x-2(ax+b)\sin^2x\cos x=2a\sin x\cos^2x-2y\sin^2x$.
   6. Now sum: $y_2+y_1\sin2x+y(1+\sin^2x)$ $=(-2a\sin x-y)+(2a\sin x\cos^2x-2y\sin^2x)+y+y\sin^2x$ $=-2a\sin x(1-\cos^2x)+(-y+y)+(-2y\sin^2x+y\sin^2x)$ $=-2a\sin^3x-y\sin^2x$. …

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5. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If $y=\tan(\log x)$, then $\dfrac{d^2y}{dx^2}=$ (A) $\dfrac{-\sec^2(\log x)[1+2\tan x]}{x^2}$ (B) $\dfrac{\sec^2(\log x)[1+\tan(\log x)]}{x^2}$ (C) $\dfrac{\sec(\log x)[2\tan(\log x)-1]}{x^2}$ (D) $\dfrac{\sec^2(\log x)[2\tan(\log x)-1]}{x^2}$

   ›Reveal solutionSolution

   Repeated chain-rule differentiation of $\tan(\log x)$; the trick is differentiating $\sec^2(\log x)/x$ correctly with the quotient rule.

   Concept and Intuition

   Every time we differentiate a function of $\log x$, a factor of $1/x$ appears from the chain rule. So the second derivative naturally produces a $1/x^2$ term, and we must track how the inner derivative of $\sec^2(\log x)$ itself brings down another factor.

   Step-by-Step Solution

   1. $y=\tan(\log x)\Rightarrow y'=\sec^2(\log x)\cdot\dfrac1x$.
   2. Write $y'=\dfrac{\sec^2(\log x)}{x}$ and differentiate using the quotient rule:

   $y''=\frac{x\cdot\frac{d}{dx}\sec^2(\log x)-\sec^2(\log x)}{x^2}$

   3. $\dfrac{d}{dx}\sec^2(\log x)=2\sec(\log x)\cdot\sec(\log x)\tan(\log x)\cdot\dfrac1x=\dfrac{2\sec^2(\log x)\tan(\log x)}{x}$.
   4. So $x\cdot\dfrac{d}{dx}\sec^2(\log x)=2\sec^2(\log x)\tan(\log x)$. …

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6. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $y = \sin^{-1}x$ then $(1-x^2)y_2 - xy_1 =$ (A) $0$ (B) $1$ (C) $2$ (D) $2y$

   ›Reveal solutionSolution

   $y=\sin^{-1}x$ satisfies $(1-x^2)y_2 - xy_1 = 0$ — a classic second-order ODE identity.

   Concept and Intuition

   Differentiating $y=\sin^{-1}x$ once gives $y_1=\dfrac{1}{\sqrt{1-x^2}}$, i.e. $(1-x^2)y_1^2=1$. Differentiating this relation again (implicitly) produces the required second-order identity — a standard result worth memorizing.

   Step-by-Step Solution

   1. $y_1 = \dfrac{1}{\sqrt{1-x^2}} \Rightarrow (1-x^2)y_1^2 = 1$.
   2. Differentiate both sides w.r.t. $x$: $-2x y_1^2 + (1-x^2)\cdot 2y_1y_2 = 0$.
   3. Divide by $2y_1$ (nonzero): $-xy_1 + (1-x^2)y_2 = 0$. …

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7. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $y = \dfrac{\alpha x + \beta}{\gamma x + \delta}$, then $2y_1 y_3 =$ (A) $2y_2^{\,3}$ (B) $3y_2^{\,2}$ (C) $y_2^{\,2}$ (D) $3y_3^{\,2}$

   ›Reveal solutionSolution

   For a Möbius (bilinear) function, each successive derivative is proportional to the previous, letting us relate $y_1,y_2,y_3$ purely algebraically.

   Concept and Intuition

   A function $y=\dfrac{\alpha x+\beta}{\gamma x+\delta}$ always simplifies so its first derivative is $k(\gamma x+\delta)^{-2}$ for a constant $k$ depending on $\alpha,\beta,\gamma,\delta$. Every higher derivative is then just this same power function times a numeric constant, which makes ratios like $y_1y_3/y_2^2$ pure numbers.

   Step-by-Step Solution

   1. Write $y=\dfrac{\alpha x+\beta}{\gamma x+\delta}=\dfrac{\alpha}{\gamma}+\dfrac{k}{\gamma(\gamma x+\delta)}$ where $k=\alpha\delta-\beta\gamma$ (standard identity: $y'=\dfrac{\alpha\delta-\beta\gamma}{(\gamma x+\delta)^2}$).
   2. $y_1=k(\gamma x+\delta)^{-2}$.
   3. $y_2=\dfrac{d}{dx}y_1=-2k\gamma(\gamma x+\delta)^{-3}$.
   4. $y_3=\dfrac{d}{dx}y_2=6k\gamma^2(\gamma x+\delta)^{-4}$. …

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8. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If $y=1+x+x^2+x^3+\ldots\infty$ and $|x|<1$, then $y''=$ (A) $2yy'$ (B) $\dfrac{2y}{y'}$ (C) $\dfrac{y'}{2y}$ (D) $2y^2y'$

   ›Reveal solutionSolution

   Sum the geometric series to $y=\dfrac{1}{1-x}$, differentiate twice, and express the result in terms of $y,y'$: it is $2yy'$.

   Concept and Intuition

   The infinite series $1+x+x^2+\cdots$ is geometric with ratio $x$, converging (for $|x|<1$) to $\dfrac{1}{1-x}$. Once $y$ has this closed form, its derivatives are straightforward power-rule computations, and both $y'$ and $y''$ turn out to be powers of $(1-x)^{-1}$, so they can be re-expressed purely in terms of $y$ and $y'$.

   Step-by-Step Solution

   1. $y=1+x+x^2+x^3+\cdots=\dfrac{1}{1-x}=(1-x)^{-1}$.
   2. $y'=(1-x)^{-2}$. Notice $y^2=(1-x)^{-2}=y'$, i.e. $y'=y^2$.
   3. $y''=2(1-x)^{-3}$.
   4. Write $(1-x)^{-3}=(1-x)^{-1}\cdot(1-x)^{-2}=y\cdot y'$. …

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9. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots \infty}}}$ then the value of $\dfrac{d^2y}{dx^2}$ at the point $(\pi, 1)$ is (A) $2$ (B) $-2$ (C) $-\dfrac{1}{2}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Convert the infinite nested radical into an implicit algebraic equation $y^2-y=\sin x$, then differentiate twice implicitly and substitute the given point.

   Concept and Intuition

   An infinite nested radical $y=\sqrt{\sin x+\sqrt{\sin x+\cdots}}$ satisfies the self-referential relation $y=\sqrt{\sin x+y}$ (the expression under the first root is $\sin x$ plus the same infinite tail $y$). Squaring converts this into a clean implicit equation that can be differentiated normally.

   Step-by-Step Solution

   1. $y=\sqrt{\sin x+y} \Rightarrow y^2=\sin x+y \Rightarrow y^2-y=\sin x$.
   2. Differentiate w.r.t. $x$: $2yy'-y'=\cos x \Rightarrow y'(2y-1)=\cos x$.
   3. At $(\pi,1)$: $\cos\pi=-1$, $2(1)-1=1$, so $y'(1)=-1 \Rightarrow y'=-1$. …

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10. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.If $y=(\tan^{-1}2x)^2+(\cot^{-1}2x)^2$, then $(1+4x^2)^2y''-16=$ (A) $8xy'$ (B) $-8x(1+4x^2)y'$ (C) $8x(1+4x^2)y'$ (D) $-8xy'$

    ›Reveal solutionSolution

    Using $\tan^{-1}2x+\cot^{-1}2x=\pi/2$ collapses $y$ to a quadratic in $a=\tan^{-1}2x$; two differentiations and elimination of $a$ give the required identity, matching option (B).

    Concept and Intuition

    Whenever a problem gives $(\tan^{-1}u)^2+(\cot^{-1}u)^2$, exploit the constant-sum identity $\tan^{-1}u+\cot^{-1}u=\pi/2$ to reduce two inverse-trig terms to one variable. This turns a messy-looking function into a clean quadratic, after which ordinary calculus (product/chain rule) finishes the job.

    Step-by-Step Solution

    1. Let $a=\tan^{-1}(2x)$. Then $\cot^{-1}(2x)=\dfrac\pi2-a$, so $y=a^2+\left(\dfrac\pi2-a\right)^2=2a^2-\pi a+\dfrac{\pi^2}{4}$.
    2. $\dfrac{da}{dx}=\dfrac{2}{1+4x^2}$. So $y'=(4a-\pi)\cdot\dfrac{2}{1+4x^2}=\dfrac{2(4a-\pi)}{D}$, where $D=1+4x^2$.
    3. Let $N=4a-\pi$, so $y'=2N/D$. Differentiate again: $\dfrac{dN}{dx}=4\cdot\dfrac{2}{D}=\dfrac{8}{D}$, and $\dfrac{dD}{dx}=8x$.
    4. $y''=2\cdot\dfrac{N'D-ND'}{D^2}=2\cdot\dfrac{8-8xN}{D^2}=\dfrac{16-16xN}{D^2}$, so $D^2y''=16-16xN$. …

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11. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.If $y=\dfrac{\log x}{x}$, then the value of $x^2\dfrac{d^2y}{dx^2}+3x\dfrac{dy}{dx}+y$ at the point $\left(\sqrt[3]{e}, \sqrt{e}\right)$ is (A) $0$ (B) $1$ (C) $e$ (D) $2e$

    ›Reveal solutionSolution

    Direct differentiation shows $x^2y''+3xy'+y$ simplifies identically to $0$ for all $x$, so its value at the given point is $0$.

    Concept and Intuition

    Rather than plugging numbers in early, it is cleaner to first build $y'$ and $y''$ symbolically and simplify the combination $x^2y''+3xy'+y$ as a function of $x$. Often such combinations are designed (as here) to collapse to a constant, independent of the exact evaluation point — recognizing this saves having to carefully handle the numeric point at all.

    Step-by-Step Solution

    1. $y=\dfrac{\log x}{x}=x^{-1}\log x$.

    2. First derivative (product rule):

    $y' = -x^{-2}\log x + x^{-1}\cdot x^{-1} = \frac{1-\log x}{x^2}$

    3. Second derivative (quotient/product rule on $y'=x^{-2}(1-\log x)$):

    $y'' = -2x^{-3}(1-\log x) + x^{-2}\left(-\frac1x\right) = x^{-3}\big[-2(1-\log x)-1\big] = \frac{2\log x-3}{x^3}$

    4. Now assemble the required combination:

    $x^2y'' = x^2\cdot\frac{2\log x-3}{x^3} = \frac{2\log x-3}{x}$

    $3xy' = 3x\cdot\frac{1-\log x}{x^2} = \frac{3(1-\log x)}{x} = \frac{3-3\log x}{x}$

    $y = \frac{\log x}{x}$ …

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12. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.If $y = x\log\left(\dfrac{1}{ax}+\dfrac{1}{a}\right)$, then $x(x+1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}-y=$ (A) $0$ (B) $1+x$ (C) $-1$ (D) $x$

    ›Reveal solutionSolution

    Writing $y=x\log(1+x)-x\log a-x\log x$ and computing $y',y''$ directly, the combination $x(x+1)y''+xy'-y$ collapses to the constant $-1$.

    Concept and Intuition

    Simplify the logarithm first (combine the fractions inside), then this becomes a standard "compute derivatives and substitute into a differential expression" problem — careful term-by-term bookkeeping is the key, not a clever trick.

    Step-by-Step Solution

    1. $\dfrac{1}{ax}+\dfrac1a=\dfrac{1+x}{ax}$, so $y=x\log(1+x)-x\log a-x\log x$.
    2. $y'=\log(1+x)+\dfrac{x}{1+x}-\log a-\log x-1$.
    3. $y''=\dfrac{1}{1+x}+\dfrac{1}{(1+x)^2}-\dfrac1x$.
    4. $x(x+1)y''=x(x+1)\left[\dfrac1{1+x}+\dfrac1{(1+x)^2}-\dfrac1x\right]=x+\dfrac{x}{1+x}-(x+1)=\dfrac{x}{1+x}-1$.
    5. $xy'=x\log(1+x)+\dfrac{x^2}{1+x}-x\log a-x\log x-x$.
    6. $xy'-y=\left[x\log(1+x)+\dfrac{x^2}{1+x}-x\log a-x\log x-x\right]-\left[x\log(1+x)-x\log a-x\log x\right]=\dfrac{x^2}{1+x}-x$. …

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