Q.Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $y = Ax$ : $xy' = y \ (x \neq 0)$

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The differential equation among the following, whose general solution is $y = A e^{5x} + B e^{-4x}$ is (A) $5\dfrac{dy}{dx} + 4\dfrac{dx}{dy} = 0$ (B) $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} + 20y = 0$ (C) $\left(\dfrac{dy}{dx}\right)^2 - \dfrac{dy}{dx} - 20y = 0$ (D) $\dfrac{d^2y}{dx^2} - \dfrac{dy}{dx} - 20y = 0$

   ›Reveal solutionSolution

   Since $y=Ae^{5x}+Be^{-4x}$, the characteristic roots are $5$ and $-4$, giving the ODE $y''-y'-20y=0$.

   Concept and Intuition

   For a linear homogeneous ODE with constant coefficients, a general solution $y=Ae^{m_1x}+Be^{m_2x}$ corresponds exactly to characteristic roots $m_1,m_2$, via the characteristic (auxiliary) equation $(m-m_1)(m-m_2)=0$.

   Step-by-Step Solution

   1. The roots implied by $y=Ae^{5x}+Be^{-4x}$ are $m_1=5,\ m_2=-4$.
   2. The characteristic equation is

   $(m-5)(m+4)=0 \Rightarrow m^2-m-20=0$

   3. This corresponds to the differential equation

   $\frac{d^2y}{dx^2}-\frac{dy}{dx}-20y=0$

   4. Verify: substituting $y=e^{5x}$ gives $25-5-20=0$ ✓; substituting $y=e^{-4x}$ gives $16-(-4)-20=0$ ✓.

   Common Mistakes …

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2. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If $t_n = \dfrac{1}{4}(n+2)(n+3)$, $n \in N$, then which one of the following is true? Assertion (A) : $\dfrac{1}{t_1} + \dfrac{1}{t_2} + \ldots + \dfrac{1}{t_{2003}} = \dfrac{2003}{3009}$ Reason (R) : $\dfrac{1}{t_1} + \dfrac{1}{t_2} + \ldots + \dfrac{1}{t_n} = \dfrac{4n}{(2n+3)}$ (A) (A) and (R) are true and (R) is a correct explanation of (A) (B) (A) and (R) are true, but (R) is not the correct explanation of (A) (C) (A) is true, (R) is false (D) (A) is false, (R) is false

   ›Reveal solutionSolution

   Telescoping the series gives $\sum 1/t_n = 4n/(3(n+3))$, which matches neither the Assertion's numeric claim nor the Reason's formula; both are false. Answer: (D).

   Concept and Intuition

   Whenever a general term factors as a product of two linear terms in the denominator, partial fractions convert the sum into a telescoping series where all interior terms cancel, leaving only a couple of boundary terms — this is the standard technique for such series, and it lets us derive the correct closed form to check against both the given Assertion and Reason.

   Step-by-Step Solution

   1. $\dfrac{1}{t_n}=\dfrac{4}{(n+2)(n+3)}=4\left(\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)$ by partial fractions.
   2. Sum from $n=1$ to $N$: $\displaystyle\sum_{n=1}^N \frac1{t_n} = 4\left[\left(\frac13-\frac14\right)+\left(\frac14-\frac15\right)+\cdots+\left(\frac1{N+2}-\frac1{N+3}\right)\right]=4\left(\frac13-\frac1{N+3}\right)=\frac{4N}{3(N+3)}$.
   3. Check the Reason's claimed formula $\dfrac{4n}{2n+3}$ against this at $n=1$: correct value is $\dfrac{4(1)}{3(4)}=\dfrac13$; Reason gives $\dfrac{4}{5}$. These disagree, so the Reason is false. …

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3. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The number of solutions of the equation $\sqrt{3x^2+x+5} = x-3$ is (A) 2 (B) 1 (C) 0 (D) 4

   ›Reveal solutionSolution

   This tests the crucial extra step in radical equations: after squaring, you must check the domain restriction ($x-3\geq0$) that the square root imposes, since squaring can introduce extraneous roots.

   Concept and Intuition

   Since $\sqrt{3x^2+x+5}$ is always non-negative, the equation $\sqrt{3x^2+x+5}=x-3$ can only hold when the right side $x-3$ is also non-negative, i.e. $x\geq3$. Squaring both sides is a valid algebraic step but doesn't preserve this sign restriction automatically — so every candidate root must be checked against $x\geq3$ before being accepted.

   Step-by-Step Solution

   1. Require $x-3\geq0 \Rightarrow x\geq3$ for the equation to possibly hold.
   2. Square both sides: $3x^2+x+5 = (x-3)^2 = x^2-6x+9$.
   3. Rearrange: $3x^2+x+5-x^2+6x-9=0 \Rightarrow 2x^2+7x-4=0$.
   4. Solve using the quadratic formula: $x = \dfrac{-7\pm\sqrt{49+32}}{4} = \dfrac{-7\pm9}{4}$, giving $x=\dfrac12$ or $x=-4$. …

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4. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If c and d are the roots of $x^2+ax+b=0$, then a root of $x^2+(4c+a)x+(b+2ac+4c^2)=0$ is (A) $d+2c$ (B) $d+c$ (C) $d-c$ (D) $d-2c$

   ›Reveal solutionSolution

   Rewriting the new quadratic's coefficients in terms of $c,d$ (using $a=-(c+d)$, $b=cd$) and testing candidates shows $x=d-2c$ satisfies it exactly.

   Concept and Intuition

   The trick is to eliminate $a,b$ using Vieta's relations for the original quadratic, turning the second quadratic into a polynomial purely in $c,d$. Then the four given candidate roots (differing only in the coefficient/sign of $c$) can be tested directly by substitution — the correct one makes the expression vanish identically.

   Step-by-Step Solution

   1. Since $c,d$ are roots of $x^2+ax+b=0$: $c+d=-a$ and $cd=b$, i.e. $a=-(c+d)$, $b=cd$.
   2. Coefficient of $x$ in new equation: $4c+a=4c-(c+d)=3c-d$.
   3. Constant term: $b+2ac+4c^2=cd+2c(-(c+d))+4c^2=cd-2c^2-2cd+4c^2=2c^2-cd$.
   4. New equation: $x^2+(3c-d)x+(2c^2-cd)=0$.
   5. Substitute $x=d-2c$: $x^2=d^2-4cd+4c^2$; $(3c-d)x=(3c-d)(d-2c)=3cd-6c^2-d^2+2cd=5cd-6c^2-d^2$. …

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5. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If the equation $x^4+7x^3+18x^2+20x+8=0$ has a repeated root, then that repeated root is (A) $-2$ (B) $-1$ (C) $-3$ (D) $-4$

   ›Reveal solutionSolution

   $x=-2$ satisfies both the quartic and its derivative, confirming it is the repeated root.

   Concept and Intuition

   A value $r$ is a repeated root of a polynomial $P(x)$ exactly when $P(r)=0$ and $P'(r)=0$ (the tangent to the curve is flat exactly at a double root). Testing small integer divisors of the constant term (candidates from the Rational Root Theorem: $\pm1,\pm2,\pm4,\pm8$) is the fastest route here.

   Step-by-Step Solution

   1. $P(x)=x^4+7x^3+18x^2+20x+8$. Try $x=-2$: $(-2)^4=16$, $7(-2)^3=-56$, $18(-2)^2=72$, $20(-2)=-40$, constant $8$. Sum: $16-56+72-40+8=0$. So $x=-2$ is a root.
   2. $P'(x)=4x^3+21x^2+36x+20$. At $x=-2$: $4(-8)=-32$, $21(4)=84$, $36(-2)=-72$, $+20$. Sum: $-32+84-72+20=0$. …

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6. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If the equation $2x^3+5x^2-4x-12=0$ has a repeated root, then the constant term of the quadratic equation whose roots are the distinct roots of the given equation is (A) $-6$ (B) $-5$ (C) $-4$ (D) $-2$

   ›Reveal solutionSolution

   The cubic factors as $(x+2)^2(2x-3)$; the quadratic built from its two distinct roots ($-2,\frac32$) is $2x^2+x-6$, with constant term $-6$.

   Concept and Intuition

   When a cubic has a repeated root, dividing out one copy of that root's factor leaves a quadratic whose two roots are exactly the distinct values among the cubic's three roots. So the strategy is: find the repeated root, do synthetic division once, and read the resulting quadratic.

   Step-by-Step Solution

   1. Test $x=-2$ in $2x^3+5x^2-4x-12$: $2(-8)+5(4)-4(-2)-12=-16+20+8-12=0$ — a root.
   2. Confirm repetition via the derivative $6x^2+10x-4$ at $x=-2$: $24-20-4=0$ — yes, double root.
   3. Synthetic division of $2x^3+5x^2-4x-12$ by $(x+2)$: coefficients $2,5,-4,-12$ → bring down $2$; $2\times(-2)=-4$, $5-4=1$; $1\times(-2)=-2$, $-4-2=-6$; $-6\times(-2)=12$, $-12+12=0$. Quotient: $2x^2+x-6$.
   4. Factor $2x^2+x-6=(2x-3)(x+2)$, confirming the full cubic is $(x+2)^2(2x-3)$ — repeated root $-2$, distinct root $\frac32$. …

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7. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.A complex number z among the following which does not satisfy $z^3+27i=0$ is (A) $(3\sqrt{3}-3i)/2$ (B) $-3i$ (C) $(3\sqrt{3}+3i)/2$ (D) $(-3\sqrt{3}+3i)/2$

   ›Reveal solutionSolution

   Three of the four options form a clean $120^\circ$-spaced triple of cube roots of $27i$; the fourth stands apart as a root of the sign-flipped equation, so it is the one that fails to belong with the other three.

   Concept and Intuition

   The three cube roots of any nonzero complex number are always spaced $120^\circ$ apart on a common circle. Recognizing which options form such a family tells you instantly which one does not belong, without needing to trust a possibly ambiguous overall sign in the stem.

   Step-by-Step Solution

   1. Write each option in polar form (modulus $3$ throughout):
      • $\frac{3\sqrt3-3i}{2}=3(\cos(-30^\circ)+i\sin(-30^\circ))$
      • $-3i=3(\cos(-90^\circ)+i\sin(-90^\circ))=3(\cos270^\circ+i\sin270^\circ)$
      • $\frac{3\sqrt3+3i}{2}=3(\cos30^\circ+i\sin30^\circ)$
      • $\frac{-3\sqrt3+3i}{2}=3(\cos150^\circ+i\sin150^\circ)$
   2. Options B, C, D sit at $270^\circ,30^\circ,150^\circ$ — exactly $120^\circ$ apart — a genuine cube-root triple. Cubing any one of them (e.g. $(-3i)^3=-27i^3=-27(-i)=27i$) gives $27i$, so B, C, D are the three roots of $z^3=27i$.
   3. Option A sits at $-30^\circ$; cubing gives $3^3(\cos(-90^\circ)+i\sin(-90^\circ))=27(0-i)=-27i$ — the genuine root of $z^3=-27i$, i.e. $z^3+27i=0$. …

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8. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.$n \in N$ then the statement $8n + 16 \le 2^n$ is true for (A) $n = 2$ (B) $n = 3$ (C) $n = 6$ (D) $n = 5$

   ›Reveal solutionSolution

   Direct substitution shows the inequality $8n+16\le2^n$ first holds (with equality) at $n=6$ among the given options.

   Concept and Intuition

   For inequalities comparing a linear function ($8n+16$) to an exponential function ($2^n$), the exponential eventually overtakes the linear term, but only after some threshold value of $n$ — found here simply by testing each candidate.

   Step-by-Step Solution

   1. $n=2$: LHS $=8(2)+16=32$; RHS $=2^2=4$. Is $32\le4$? No.
   2. $n=3$: LHS $=8(3)+16=40$; RHS $=2^3=8$. Is $40\le8$? No.
   3. $n=5$: LHS $=8(5)+16=56$; RHS $=2^5=32$. Is $56\le32$? No.
   4. $n=6$: LHS $=8(6)+16=64$; RHS $=2^6=64$. Is $64\le64$? Yes — equality holds, so the inequality (which is $\le$, not strict) is satisfied. …

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9. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.The solution of the differential equation $\dfrac{d^2y}{dx^2} + y = 0$ is ____ (A) $y = 3\sin x + 4\cos x$ (B) $y = x^2$ (C) $y = x + 2$ (D) $y = \log x$

   ›Reveal solutionSolution

   The equation $y''+y=0$ has general solution $C_1\sin x+C_2\cos x$; only $y=3\sin x+4\cos x$ fits and satisfies the equation.

   Concept and Intuition

   $y''+y=0$ is the classic simple-harmonic-motion differential equation. Its solutions are sinusoids, since the second derivative of $\sin x$ or $\cos x$ is the negative of itself, exactly cancelling the $+y$ term.

   Step-by-Step Solution

   1. Recognize the auxiliary equation for $y''+y=0$: $m^2+1=0 \Rightarrow m=\pm i$, giving general solution $y=C_1\cos x+C_2\sin x$.
   2. Check option (A): $y=3\sin x+4\cos x$. Then $y'=3\cos x-4\sin x$, $y''=-3\sin x-4\cos x = -y$. So $y''+y=0$. ✓ Matches the required form.
   3. Check option (B): $y=x^2\Rightarrow y''=2$, so $y''+y=2+x^2\ne0$. ✗
   4. Check option (C): $y=x+2\Rightarrow y''=0$, so $y''+y=x+2\ne0$. ✗ …

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10. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.$2 + \sqrt{5}, 1$ are roots of the cubic equation given by (A) $x^3 + 3x^2 - 3x - 1 = 0$ (B) $x^3 - 3x^2 + 3x - 1 = 0$ (C) $x^3 - 5x^2 + 3x + 1 = 0$ (D) $x^3 + 5x^2 - 3x + 1 = 0$

    ›Reveal solutionSolution

    A rational-coefficient cubic with irrational root $2+\sqrt5$ must also have $2-\sqrt5$ as a

    root; combined with the given root $1$, Vieta's formulas build the cubic

    $x^3-5x^2+3x+1=0$.

    Concept and Intuition

    For polynomials with rational coefficients, irrational roots of the form $p+\sqrt q$ always occur

    in conjugate pairs $p\pm\sqrt q$ (otherwise the coefficients, built from sums/products of the

    roots, would themselves be irrational). So knowing one irrational root and one rational root of a

    cubic with rational coefficients pins down all three roots.

    Step-by-Step Solution

    1. The three roots are $2+\sqrt5,\ 2-\sqrt5,\ 1$ (the conjugate is forced by the rational coefficients).
    2. Sum of roots: $(2+\sqrt5)+(2-\sqrt5)+1 = 4+1 = 5$.
    3. Sum of pairwise products: $(2+\sqrt5)(2-\sqrt5) + (2+\sqrt5)(1) + (2-\sqrt5)(1)$ $= (4-5) + \big[(2+\sqrt5)+(2-\sqrt5)\big] = -1 + 4 = 3$.
    4. Product of roots: $(2+\sqrt5)(2-\sqrt5)(1) = (4-5)(1) = -1$. …

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11. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.If $f(x) = 2x^3 + mx^2 - 13x + n$ and $2, 3$ are the roots of the equation $f(x) = 0$ then the values of $m$ and $n$ are (A) $-5, -30$ (B) $-5, 30$ (C) $5, 30$ (D) $5, -30$

    ›Reveal solutionSolution

    Substituting the two given roots into $f(x)=0$ gives two linear equations in $m,n$; solving them gives $m=-5,\ n=30$.

    Concept and Intuition

    If $r$ is a root of a polynomial equation $f(x)=0$, then $f(r)=0$ exactly. With two known roots and two unknowns ($m,n$), we get two independent linear equations — enough to solve for both unknowns.

    Step-by-Step Solution

    1. $f(x)=2x^3+mx^2-13x+n$.
    2. Since $x=2$ is a root: $f(2)=2(8)+m(4)-13(2)+n=16+4m-26+n=0 \Rightarrow 4m+n=10$.
    3. Since $x=3$ is a root: $f(3)=2(27)+m(9)-13(3)+n=54+9m-39+n=0 \Rightarrow 9m+n=-15$. …

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12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.Let $\omega = \operatorname{cis}\left(\dfrac{2\pi}{3}\right) = \cos\left(\dfrac{2\pi}{3}\right)+i\sin\left(\dfrac{2\pi}{3}\right)$ and $f(x)=x^7-2x^4-4x^3+8$. Which of the following option is correct? (A) $\left\{2^{\frac{1}{2}}, 2^{\frac{1}{3}}, \omega, 2^{\frac{1}{3}}\omega\right\}$ is a subset of the solution set of $f(x)$. (B) $\left\{2^{\frac{1}{2}}, -2^{\frac{1}{3}}, 2^{\frac{1}{3}}\omega^2, 2^{\frac{1}{2}}i\right\}$ is a subset of the solution set of $f(x)$. (C) $\left\{2^{\frac{1}{3}}, 2^{\frac{1}{2}}, -2^{\frac{1}{2}}i, 2^{\frac{1}{3}}\omega^2\right\}$ is not a subset of the solution set of $f(x)$. (D) $\left\{2^{\frac{1}{3}}, 2^{\frac{1}{3}}\omega, 2^{\frac{1}{2}}i, -2^{\frac{1}{2}}\right\}$ is a subset of the solution set of $f(x)$.

    ›Reveal solutionSolution

    Factoring $f$ into $(x^3-2)(x^2-2)(x^2+2)$ pins down all 7 roots exactly; testing each option's listed set against that root list shows only (D) is entirely made of genuine roots.

    Concept and Intuition

    A degree-7 polynomial has (at most) 7 roots; if we can factor it into lower-degree pieces we know every root exactly, turning a "which set is a subset" question into simple list-membership checking.

    Step-by-Step Solution

    1. Group terms: $f(x)=x^7-2x^4-4x^3+8 = x^4(x^3-2) - 4(x^3-2) = (x^3-2)(x^4-4)$.
    2. Factor further: $x^4-4=(x^2-2)(x^2+2)$. So $f(x)=(x^3-2)(x^2-2)(x^2+2)$.
    3. Roots of $x^3-2=0$: $x=2^{1/3},\ 2^{1/3}\omega,\ 2^{1/3}\omega^2$ (using the given $\omega=\text{cis}(2\pi/3)$).
    4. Roots of $x^2-2=0$: $x=\pm 2^{1/2}$.
    5. Roots of $x^2+2=0$: $x=\pm 2^{1/2}i$.
    6. Full root set (7 roots, matching degree 7): $\{2^{1/3},2^{1/3}\omega,2^{1/3}\omega^2,2^{1/2},-2^{1/2},2^{1/2}i,-2^{1/2}i\}$.
    7. Test (A): contains bare $\omega$ — not in the root list (only $2^{1/3}\omega$ etc. are roots). So (A)'s claimed subset is false.
    8. Test (B): contains $-2^{1/3}$ — not a root (only $2^{1/3},2^{1/3}\omega,2^{1/3}\omega^2$ are). False. …

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