Q.(x) Solution of $\frac{dy}{dx}=\frac{y}{x}+\tan\frac{y}{x}$ is $\sin\left(\frac{y}{x}\right)=cx$. (State True or False.)

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 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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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 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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7. 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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8. 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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9. 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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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 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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12. 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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