Q.Verify that the function $y = a\cos x + b\sin x$, where $a, b \in \mathbf{R}$ is a solution of the differential equation $\frac{d^2y}{dx^2} + y = 0$.

Showing the 12 most recent of 14 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.The solution of the equation $2x^3-x^2-22x-24=0$ when two of the roots are in the ratio $3:4$ is (A) $3, 4, \dfrac{1}{2}$ (B) $-\dfrac{3}{2}, -2, 4$ (C) $-\dfrac{1}{2}, \dfrac{3}{2}, 2$ (D) $-\dfrac{3}{2}, 2, \dfrac{5}{2}$

    ›Reveal solutionSolution

    Using Vieta's formulas with two roots in ratio $3:4$, the roots are $-\tfrac32,-2,4$.

    Concept and Intuition

    For $2x^3-x^2-22x-24=0$ (dividing by 2: $x^3-\tfrac12x^2-11x-12=0$), Vieta gives: sum of roots $=\tfrac12$, sum of products of pairs $=-11$, product of roots $=12$. If two roots are in ratio $3:4$, write them as $3k,4k$ and let the third be $r$; then use the three Vieta relations to pin down $k$ and $r$.

    Step-by-Step Solution

    1. Sum: $3k+4k+r=7k+r=\tfrac12$.
    2. Product: $3k\cdot4k\cdot r=12k^2r=12\Rightarrow k^2r=1$.
    3. Pair-sum: $12k^2+3kr+4kr=12k^2+7kr=-11$.
    4. Rather than solve the system directly, test the answer choice that has two entries in ratio $3:4$: in $(B)$, $-\tfrac32$ and $-2$ have ratio $-\tfrac32:-2=3:4$, so $k=-\tfrac12$, giving $3k=-\tfrac32,4k=-2$, and third root $r=4$. …

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11. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.For the equation $x^2 - 5|x| - 14 = 0$, (A) all roots are real (B) all the roots are imaginary (C) Two roots are real (D) No real roots

    ›Reveal solutionSolution

    Substituting $t=|x|$ reduces the equation to a simple quadratic in $t$, of which only the

    nonnegative root is valid, giving $x=\pm7$ — two roots, both real.

    Concept and Intuition

    An equation involving $|x|$ is best handled by substituting $t=|x|\ge0$, solving the resulting

    polynomial in $t$, and then discarding any root that violates $t\ge0$ (since $|x|$ can never be

    negative). Each valid $t$-value then corresponds to (up to) two real $x$-values, $x=\pm t$.

    Step-by-Step Solution

    1. Substitute $t=|x|$, $t\ge 0$: the equation becomes $t^2-5t-14=0$.
    2. Factor: $t^2-5t-14 = (t-7)(t+2) = 0 \Rightarrow t=7$ or $t=-2$.
    3. Reject $t=-2$ since $|x|$ can never be negative; keep $t=7$.
    4. $|x|=7 \Rightarrow x=7$ or $x=-7$.
    5. Both values, $7$ and $-7$, are real numbers — the equation has exactly two roots and both are real.

    Common Mistakes …

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12. 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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