Q.(vii) The solution of $\frac{dy}{dx}=\left(\frac{y}{x}\right)^{1/3}$ is $y^{2/3}-x^{2/3}=c$. (State True or False.)

Showing the 12 most recent of 86 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The general solution of the differential equation $(2xy + y^2)dy = (x^2 - y^2)dx$ is (A) $x^3 - 3x^2y - y^3 = c$ (B) $x^3 - 3x^2y + y^3 = c$ (C) $x^3 - 3xy^2 + y^3 = c$ (D) $x^3 - 3xy^2 - y^3 = c$

   ›Reveal solutionSolution

   This is a homogeneous differential equation. Substituting $y = vx$ reduces it to a separable form, and integrating gives the general solution $x^3 - 3xy^2 - y^3 = c$, which matches option (D).

   The given equation is $(2xy + y^2)dy = (x^2 - y^2)dx$. Notice that every term is of degree 2 — $2xy$, $y^2$, $x^2$, $y^2$ — so the equation is homogeneous. For a homogeneous equation, the standard trick is to set $y = vx$, which turns the equation into one where variables separate cleanly.

   1. Rewrite in standard form Bring the $dx$ term to the left:

   $(2xy + y^2)dy - (x^2 - y^2)dx = 0$

   Or equivalently,

   $\frac{dy}{dx} = \frac{x^2 - y^2}{2xy + y^2}$

   2. Substitute $y = vx$ Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. The right-hand side becomes:

   $\frac{x^2 - (vx)^2}{2x(vx) + (vx)^2} = \frac{x^2(1 - v^2)}{x^2(2v + v^2)} = \frac{1 - v^2}{2v + v^2}$

   So the equation is:

   $v + x\frac{dv}{dx} = \frac{1 - v^2}{2v + v^2}$

   3. Separate variables Subtract $v$ from both sides:

   $x\frac{dv}{dx} = \frac{1 - v^2}{2v + v^2} - v = \frac{1 - v^2 - v(2v + v^2)}{2v + v^2}$

   Simplify the numerator:

   $1 - v^2 - 2v^2 - v^3 = 1 - 3v^2 - v^3$

   So:

   $x\frac{dv}{dx} = \frac{1 - 3v^2 - v^3}{2v + v^2}$

   Now separate:

   $\frac{2v + v^2}{1 - 3v^2 - v^3} \, dv = \frac{dx}{x}$

   4. Integrate both sides The left-hand side is set up for a simple substitution. Let $u = 1 - 3v^2 - v^3$. Then $du = (-6v - 3v^2)dv = -3(2v + v^2)dv$. Notice that $2v + v^2$ appears in the numerator, so:

   $\frac{2v + v^2}{1 - 3v^2 - v^3} \, dv = -\frac{1}{3} \frac{du}{u}$

   Integrating:

   $\int -\frac{1}{3} \frac{du}{u} = \int \frac{dx}{x}$

   $-\frac{1}{3} \log|u| = \log|x| + C$

   Multiply by $-3$:

   $\log|u| = -3\log|x| - 3C$ …

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2. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.The general solution of $\dfrac{dy}{dx} + y f'(x) - f(x) f'(x) = 0, \; y \neq f(x)$ is (A) $y = f(x) + 1 + c e^{-f(x)}$ (B) $y = c e^{-f(x)}$ (C) $y = f(x) - 1 + c e^{-f(x)}$ (D) $y = f(x) + c e^{f(x)}$

   ›Reveal solutionSolution

   This is a first-order linear ODE disguised by the presence of $f(x)$ and $f'(x)$. By rewriting it in standard form and using an integrating factor $e^{f(x)}$, the general solution simplifies to $y = f(x) - 1 + c e^{-f(x)}$, which corresponds to option (C).

   We start with the given differential equation:

   $\frac{dy}{dx} + y f'(x) - f(x) f'(x) = 0, \quad y \neq f(x).$

   ---

   Concept & Intuition

   The equation looks messy because of the $f(x)$ and $f'(x)$ terms, but notice that $f'(x)$ appears as a coefficient of $y$ and also multiplied by $f(x)$. This suggests we can rearrange it into the standard linear form $\frac{dy}{dx} + P(x) y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$ only. Once in that form, the method of integrating factor works cleanly. The key trick: treat $f(x)$ as some known function (we don't need its explicit form), and $f'(x)$ as its derivative.

   ---

   Step-by-step solution

   1. Rewrite the equation in standard linear form Bring the term $- f(x) f'(x)$ to the right-hand side:

   $\frac{dy}{dx} + f'(x) \, y = f(x) f'(x).$

   This is now of the form $\frac{dy}{dx} + P(x) y = Q(x)$ with $P(x) = f'(x)$ and $Q(x) = f(x) f'(x)$.

   2. Find the integrating factor The integrating factor $\mu(x)$ is given by $e^{\int P(x) \, dx}$. Here:

   $\int P(x) \, dx = \int f'(x) \, dx = f(x) + C.$

   We only need one integrating factor, so take $\mu(x) = e^{f(x)}$.

   3. Multiply through by the integrating factor

   $e^{f(x)} \frac{dy}{dx} + e^{f(x)} f'(x) \, y = e^{f(x)} f(x) f'(x).$

   The left-hand side is the derivative of $y e^{f(x)}$ with respect to $x$ (by the product rule, since $\frac{d}{dx} e^{f(x)} = e^{f(x)} f'(x)$). So we have:

   $\frac{d}{dx} \left( y e^{f(x)} \right) = e^{f(x)} f(x) f'(x).$

   4. Integrate both sides

   $y e^{f(x)} = \int e^{f(x)} f(x) f'(x) \, dx.$

   Notice that the integrand is set up for a substitution: let $u = f(x)$, then $du = f'(x) \, dx$, so:

   $\int e^{u} u \, du.$

   This is a standard integral. Use integration by parts: let $w = u$, $dv = e^u du$, then $dw = du$, $v = e^u$. So:

   $\int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u + C = e^u (u - 1) + C.$ …

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3. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.The general solution of $\dfrac{dy}{dx} = x + \sin x \cos y + x \cos y + \sin x$ is (A) $\tan \dfrac{x}{2} = \dfrac{y^2}{2} - \cos y + C$ (B) $\tan \dfrac{y}{2} = \dfrac{x^2}{2} - \cos x + C$ (C) $\sec^2 \dfrac{y}{2} = \dfrac{x^2}{2} - \cos x + C$ (D) $\tan \dfrac{y}{2} = \dfrac{x^2}{2} + \cos x + Cx$

   ›Reveal solutionSolution

   The given differential equation is separable after factoring. The solution is found by integrating both sides, leading to $\tan\frac{y}{2} = \frac{x^2}{2} - \cos x + C$, which matches option (B).

   The key is to notice that the right-hand side can be grouped into terms that depend only on $x$ and terms that depend only on $y$. That’s the hallmark of a separable differential equation — and once you see the pattern, the integration is straightforward.

   Let’s rewrite the equation:

   $\frac{dy}{dx} = x + \sin x \cos y + x \cos y + \sin x$

   Group the terms cleverly:

   $\frac{dy}{dx} = (x + \sin x) + (\sin x \cos y + x \cos y)$

   Factor $\cos y$ from the last two terms:

   $\frac{dy}{dx} = (x + \sin x) + \cos y (x + \sin x)$

   Now factor $(x + \sin x)$ out of the whole right-hand side:

   $\frac{dy}{dx} = (x + \sin x)(1 + \cos y)$

   This is clearly separable: the $x$-part is $(x + \sin x)$ and the $y$-part is $(1 + \cos y)$.

   1. Separate the variables Bring all $y$ terms to the left and $x$ terms to the right:

   $\frac{dy}{1 + \cos y} = (x + \sin x)\, dx$

   2. Integrate both sides The left side uses a standard trigonometric identity. Recall:

   $1 + \cos y = 2\cos^2\frac{y}{2}$

   So:

   $\frac{1}{1 + \cos y} = \frac{1}{2\cos^2\frac{y}{2}} = \frac{1}{2}\sec^2\frac{y}{2}$

   Therefore:

   $\int \frac{dy}{1 + \cos y} = \frac{1}{2}\int \sec^2\frac{y}{2}\, dy$

   Let $u = y/2$, then $dy = 2\,du$, and:

   $\frac{1}{2}\int \sec^2 u \cdot 2\,du = \int \sec^2 u\, du = \tan u + C = \tan\frac{y}{2} + C$

   The right side integrates easily:

   $$$$

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4. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.The general solution of the differential equation $(\sec x + \tan x) \dfrac{dy}{dx} + (\sec^2 x + \sec x \tan x) y = 1$ is (A) $(1 + \sin x) y = n \cos x + c$ (B) $(1 + \cos x) y = x \sin x + c$ (C) $(\sec x + \tan x) y = x \sec x + c$ (D) $(\sec x + \tan x) y = x + c$

   ›Reveal solutionSolution

   This is a first-order linear ODE. Rewriting it in standard form and applying the integrating factor method shows that the general solution is $(\sec x + \tan x) y = x + c$, which matches option (D).

   The key concept is recognizing the equation as a first-order linear differential equation of the form

   $\frac{dy}{dx} + P(x) y = Q(x).$

   The standard method is to multiply through by an integrating factor $\mu(x) = e^{\int P\,dx}$, which makes the left side a perfect derivative. Here, the coefficients are cleverly arranged so that the integrating factor simplifies dramatically.

   1. Rewrite in standard form The given equation is

   $(\sec x + \tan x) \frac{dy}{dx} + (\sec^2 x + \sec x \tan x) y = 1.$

   Divide through by $(\sec x + \tan x)$ to isolate $\frac{dy}{dx}$:

   $\frac{dy}{dx} + \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, y = \frac{1}{\sec x + \tan x}.$

   2. Simplify the coefficient of $y$ Factor the numerator: $\sec^2 x + \sec x \tan x = \sec x (\sec x + \tan x)$. Hence

   $\frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} = \sec x.$

   So the ODE becomes

   $\frac{dy}{dx} + (\sec x) y = \frac{1}{\sec x + \tan x}.$

   3. Find the integrating factor

   $\mu(x) = e^{\int \sec x \, dx}.$

   A standard integral: $\int \sec x \, dx = \log |\sec x + \tan x| + C$.

   Thus

   $\mu(x) = e^{\log |\sec x + \tan x|} = \sec x + \tan x.$

   (We take the positive branch for typical intervals.)

   4. Multiply through by $\mu(x)$

   $(\sec x + \tan x) \frac{dy}{dx} + (\sec x + \tan x)(\sec x) y = 1.$

   Notice the left side is exactly the derivative of $(\sec x + \tan x) y$ because

   $\frac{d}{dx} \big[ (\sec x + \tan x) y \big] = (\sec x + \tan x) \frac{dy}{dx} + (\sec x \tan x + \sec^2 x) y,$ …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The integrating factor of the linear differential equation in $x$ given by $\dfrac{dy}{dx} = \dfrac{1}{3x + y + 2}$ is (A) $e^{-3x}$ (B) $e^{-x}$ (C) $e^{-3y}$ (D) $e^{-y}$

   ›Reveal solutionSolution

   The given equation is not linear in $y$, but rewriting it as $\frac{dx}{dy} = 3x + y + 2$ makes it linear in $x$ with integrating factor $e^{-3y}$, so the correct option is (C).

   We are given

   $\frac{dy}{dx} = \frac{1}{3x + y + 2}.$

   At first glance, this looks like a first-order differential equation in $y$ as a function of $x$. But it is not linear in $y$ because the right-hand side is a rational function containing $y$ in the denominator. However, we can flip the relationship: treat $x$ as a function of $y$ instead.

   The key insight: if $\frac{dy}{dx}$ is given, then $\frac{dx}{dy} = 1 / \frac{dy}{dx}$ (provided the derivative is nonzero). This often turns a nonlinear equation in $y$ into a linear equation in $x$.

   ---

   1. Rewrite the equation in terms of $x(y)$ Since $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$, we have

   $\frac{dx}{dy} = 3x + y + 2.$

   This is now a linear first-order differential equation in $x$ with independent variable $y$.

   2. Identify the standard linear form The standard form for a linear ODE in $x$ is

   $\frac{dx}{dy} + P(y)\,x = Q(y).$

   Our equation is

   $\frac{dx}{dy} - 3x = y + 2.$

   So $P(y) = -3$ and $Q(y) = y + 2$.

   3. Recall the integrating factor formula For a linear ODE $\frac{dx}{dy} + P(y)x = Q(y)$, the integrating factor is

   $\mu(y) = e^{\int P(y)\,dy}.$

   Here $P(y) = -3$, so

   $\mu(y) = e^{\int (-3)\,dy} = e^{-3y}.$

   4. Interpret the result …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $\cos x \dfrac{dy}{dx} = y \sin x - 1$, $x \neq (2n+1)\dfrac{\pi}{2}$, $n \in \mathbb{Z}$ is the differential equation corresponding to the curve $y = f(x)$ and $f(0) = 1$ then $f(x) =$ (A) $(1-x)\sec x$ (B) $(1-x)\cos x$ (C) $x + \cos x$ (D) $x + \sec x$

   ›Reveal solutionSolution

   This is a first-order linear ODE. Rewriting it in standard form and using an integrating factor gives $f(x) = (1-x)\sec x$, which matches option (A).

   We start with the given differential equation:

   $\cos x \frac{dy}{dx} = y \sin x - 1$

   The goal is to find $y = f(x)$ satisfying $f(0) = 1$, and then match it to one of the options.

   ---

   Concept and Intuition

   The equation is linear in $y$ but not yet in standard form. The standard form for a first-order linear ODE is:

   $\frac{dy}{dx} + P(x) y = Q(x)$

   Once in this form, we multiply through by an integrating factor $\mu(x) = e^{\int P(x)\,dx}$, which lets us write the left-hand side as the derivative of $\mu(x) y$. Then we integrate both sides.

   Here, dividing by $\cos x$ will give us $P(x) = -\tan x$ and $Q(x) = -\sec x$. The integrating factor simplifies nicely because $\int \tan x\,dx = -\log|\cos x|$, so $\mu(x) = \sec x$.

   ---

   Step-by-step solution

   1. Rewrite in standard form Divide both sides by $\cos x$ (valid since $x \neq (2n+1)\frac{\pi}{2}$):

   $\frac{dy}{dx} = y \tan x - \sec x$

   Bring the $y$ term to the left:

   $\frac{dy}{dx} - (\tan x) y = -\sec x$

   So $P(x) = -\tan x$ and $Q(x) = -\sec x$.

   2. Find the integrating factor

   $\mu(x) = e^{\int P(x)\,dx} = e^{\int -\tan x\,dx}$

   Since $\int \tan x\,dx = -\log|\cos x|$, we have:

   $\int -\tan x\,dx = \log|\cos x|$

   Hence:

   $\mu(x) = e^{\log|\cos x|} = |\cos x|$

   For the domain (where $\cos x > 0$ near $x=0$), we can take $\mu(x) = \cos x$. But it's more standard to use $\sec x$ as the integrating factor when we multiply through — let's check.

   Actually, careful: The standard formula is $\mu = e^{\int P\,dx}$. With $P = -\tan x$, we get $\mu = e^{\log(\cos x)} = \cos x$ (taking positive branch near 0). So the integrating factor is $\cos x$.

   3. Multiply the ODE by $\mu(x) = \cos x$ Original ODE in standard form:

   $\frac{dy}{dx} - (\tan x) y = -\sec x$

   Multiply by $\cos x$:

   $\cos x \frac{dy}{dx} - y \sin x = -1$

   Notice the left side is exactly $\frac{d}{dx}(y \cos x)$ because:

   $$$$

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7. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $l$ and $m$ are respectively the order and the degree of the differential equation $f(x)y'' + g(x)y' = \dfrac{4y}{x}$ whose general solution is $y = ax^2 + b \log x$, then $f(m) + g(m) =$ (A) $2l$ (B) $l$ (C) $3m$ (D) $1 + m$

   ›Reveal solutionSolution

   The key idea is to find the differential equation from its given general solution, then identify its order $l$ and degree $m$, and finally evaluate $f(m)+g(m)$ — the answer is $l$.

   We are told that the general solution of the differential equation

   $f(x)y'' + g(x)y' = \frac{4y}{x}$

   is $y = ax^2 + b \log x$, where $a$ and $b$ are arbitrary constants. The functions $f(x)$ and $g(x)$ are not given explicitly — they are to be determined from the fact that this $y$ satisfies the equation for all $a, b$.

   The problem asks for $f(m) + g(m)$, where $l$ is the order and $m$ is the degree of this differential equation. So we first need to find the differential equation itself.

   1. Find the derivatives of the given solution.

   $y = a x^2 + b \log x$

   Differentiate:

   $y' = 2a x + \frac{b}{x}$

   Differentiate again:

   $y'' = 2a - \frac{b}{x^2}$

   2. Eliminate the arbitrary constants $a$ and $b$.

      We have three equations: $y$, $y'$, $y''$ in terms of $a$ and $b$. We need one equation relating $y$, $y'$, $y''$ and $x$ alone — that is the differential equation.

      From $y'' = 2a - \frac{b}{x^2}$, we can solve for $a$ and $b$ in terms of $y''$ and something else. But a cleaner way:

      From $y' = 2a x + \frac{b}{x}$, multiply by $x$:

   $x y' = 2a x^2 + b$

   From $y = a x^2 + b \log x$, we have $a x^2 = y - b \log x$. Substitute into $x y'$:

   $x y' = 2(y - b \log x) + b = 2y - 2b \log x + b$

   This still has $b$. Instead, let's use $y''$ directly.

   From $y'' = 2a - \frac{b}{x^2}$, multiply by $x^2$:

   $x^2 y'' = 2a x^2 - b$

   But $2a x^2 = 2(y - b \log x)$ from $y = a x^2 + b \log x$. So:

   $x^2 y'' = 2y - 2b \log x - b$

   This still contains $b$. We need another relation to eliminate $b$.

   From $y' = 2a x + \frac{b}{x}$, multiply by $x$:

   $x y' = 2a x^2 + b$

   And $2a x^2 = 2(y - b \log x)$. So:

   $x y' = 2y - 2b \log x + b$

   Now subtract the $x^2 y''$ equation from this? Let's do it systematically.

   We have:

   $x y' = 2y - 2b \log x + b \quad \text{(1)}$

   $x^2 y'' = 2y - 2b \log x - b \quad \text{(2)}$

   Subtract (2) from (1):

   $x y' - x^2 y'' = (2y - 2b \log x + b) - (2y - 2b \log x - b) = 2b$

   So $b = \frac{1}{2}(x y' - x^2 y'')$.

   Now add (1) and (2):

   $x y' + x^2 y'' = (2y - 2b \log x + b) + (2y - 2b \log x - b) = 4y - 4b \log x$

   Substitute $b$:

   $x y' + x^2 y'' = 4y - 4 \left( \frac{1}{2}(x y' - x^2 y'') \right) \log x$

   $x y' + x^2 y'' = 4y - 2(x y' - x^2 y'') \log x$

   Bring terms together:

   $x y' + x^2 y'' + 2(x y' - x^2 y'') \log x = 4y$

   Factor:

   $$$$

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.Consider all functions given in List-1 in the interval $[1,3]$. The List-2 has the values of ‘c’ obtained by applying Lagrange’s mean value theorem on the functions of List-1. Match the functions and values of ‘c’. (A) A – II, B – V, C – IV, D – III (B) A – IV, B – III, C – II, D – V (C) A – II, B – I, C – IV, D – III (D) A – IV, B – V, C – II, D – I

   ›Reveal solutionSolution

   This is a List-I $\to$ List-II matching question on Lagrange's Mean Value Theorem over $[1,3]$; the official key pairs them as A–IV, B–III, C–II, D–V — option (B).

   For each function $f$ in List-I, Lagrange's Mean Value Theorem (LMVT) guarantees a point $c\in(1,3)$ with

   $f'(c)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2},$

   and the value of $c$ so obtained is the List-II entry to be matched. For each function one forms this equation, solves $f'(c)=\tfrac12\big(f(3)-f(1)\big)$, and keeps the root lying in $(1,3)$. …

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9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $y = \sin x + A \cos x$ is the general solution of $\dfrac{dy}{dx} + f(x)y = \sec x$, then an integrating factor of the differential equation is (A) $\sec x$ (B) $\tan x$ (C) $\cos x$ (D) $\sin x$

   ›Reveal solutionSolution

   The given general solution is $y = \sin x + A \cos x$. Differentiating and substituting into the differential equation reveals $f(x) = \tan x$, so the integrating factor is $\sec x$, which is option (A).

   We are told that $y = \sin x + A \cos x$ (where $A$ is an arbitrary constant) is the general solution of

   $\frac{dy}{dx} + f(x) y = \sec x.$

   Our goal is to find the integrating factor (I.F.) of this linear first-order ODE. The standard form is $y' + P(x) y = Q(x)$, and the integrating factor is $\mu(x) = e^{\int P(x)\,dx}$. Here $P(x) = f(x)$.

   The key insight: If we already know the general solution, we can work backwards to find $f(x)$ by differentiating the solution and plugging it into the equation. Then we compute the integrating factor directly.

   ---

   1. Differentiate the given general solution

      $y = \sin x + A \cos x$

      $\frac{dy}{dx} = \cos x - A \sin x$

   2. Substitute into the differential equation

      The equation is $\frac{dy}{dx} + f(x) y = \sec x$.

      So:

   $(\cos x - A \sin x) + f(x)(\sin x + A \cos x) = \sec x$

   3. Group terms involving $A$ and those without $A$ Expand:

   $\cos x - A \sin x + f(x) \sin x + A f(x) \cos x = \sec x$

   Group constant (in $A$) terms: $\cos x + f(x) \sin x$

   Group $A$ terms: $A( -\sin x + f(x) \cos x )$

   Since this must hold for all $A$ (the solution is general), the coefficient of $A$ must be zero. That gives:

   $-\sin x + f(x) \cos x = 0 \quad \Rightarrow \quad f(x) \cos x = \sin x \quad \Rightarrow \quad f(x) = \tan x$

   4. Verify the constant part With $f(x) = \tan x$, the constant part becomes: $\cos x + \tan x \cdot \sin x = \cos x + \frac{\sin^2 x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\cos x} = \frac{1}{\cos x} = \sec x$ …

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.Consider all functions given in List-I in the interval $[1,3]$. The List-2 has the values of ‘c’ obtained by applying Lagrange’s mean value theorem on the functions of List-1. Match the functions and values of ‘c’. (A) A – II, B – V, C – IV, D – III (B) A – II, B – I, C – IV, D – III (C) A – IV, B – V, C – II, D – I (D) A – IV, B – III, C – II, D – V

    ›Reveal solutionSolution

    Apply $f'(c)=\dfrac{f(3)-f(1)}{2}$ to each function. You get $c=\sqrt2$ (any $c$ works for $|x-1|$), $c=\log_3 e^2$, $c=2$ and $c=\log\frac{e^3-e}{2}$ — i.e. A–IV, B–III, C–II, D–V, option (D).

    The concept first

    Lagrange's Mean Value Theorem says something beautifully physical: if you travel from $x=1$ to $x=3$ along a smooth curve, then at some instant your instantaneous rate of change equals your average rate of change. Formally, if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $c \in (a,b)$ with

    $f'(c) = \frac{f(b)-f(a)}{b-a}.$

    So the entire job in each row is: compute the chord slope on the right, set the derivative equal to it, and solve for $c$.

    One subtlety worth noticing: $|x-1|$ is famously not differentiable at $x=1$ — but the kink is at the endpoint of $[1,3]$, and the theorem only needs differentiability on the open interval $(1,3)$. Inside, $x>1$ so $|x-1| = x-1$, perfectly smooth.

    Step-by-step

    1. A: $f(x)=|x-1|$ on $[1,3]$. Here $f(x)=x-1$, $f(1)=0$, $f(3)=2$.

    $\text{chord slope}=\frac{2-0}{2}=1, \qquad f'(x)=1 \ \ \forall x \in (1,3).$

    The equation $f'(c)=1$ is satisfied by every $c$ in $(1,3)$. The only List-2 value lying in $(1,3)$ that is not already claimed by another (uniquely determined) function is $\sqrt2 \approx 1.414$. → A – IV.

    2. B: $f(x)=\log x$ (natural log) on $[1,3]$. $f(1)=0$, $f(3)=\log 3$.

    $f'(c)=\frac1c = \frac{\log 3 - 0}{2} = \frac{\log 3}{2} \ \Longrightarrow\ c = \frac{2}{\log 3}.$ …

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11. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
    Q.If $\dfrac{dy}{dx} - \dfrac{2x}{x^2 + b} y = -2x(x^2 + b),\ y(0) = 12,\ y(1) = 10$, then sum of all possible values of $b$ is (A) $1$ (B) $4$ (C) $-3$ (D) $-1$

    ›Reveal solutionSolution

    This is a first-order linear ODE solved via an integrating factor; the two boundary conditions force a specific value of the parameter $b$, and the sum of all possible $b$ values is $-3$.

    We are given the differential equation

    $\frac{dy}{dx} - \frac{2x}{x^2 + b}\, y = -2x(x^2 + b),$

    with conditions $y(0) = 12$ and $y(1) = 10$. The parameter $b$ is unknown, and we must find all possible $b$ that allow both conditions to hold, then sum them.

    ---

    Concept and intuition

    This is a first-order linear ODE of the form

    $\frac{dy}{dx} + P(x) y = Q(x).$

    The standard method: multiply by an integrating factor $\mu(x) = e^{\int P(x)\,dx}$ to make the left side a perfect derivative. Here $P(x) = -\frac{2x}{x^2+b}$, so the integrating factor will simplify nicely because the numerator is the derivative of the denominator. The right-hand side is a polynomial times $(x^2+b)$, so after multiplication we’ll integrate easily.

    The twist: we have two boundary conditions for a first-order ODE — that usually overdetermines the system. The parameter $b$ must adjust so that both conditions are consistent. We’ll solve the ODE in terms of $b$ and a constant $C$, then impose $y(0)=12$ and $y(1)=10$ to get equations that determine $b$.

    ---

    Step-by-step solution

    1. Identify $P(x)$ and compute the integrating factor.

    Rewrite the ODE as

    $\frac{dy}{dx} + \left(-\frac{2x}{x^2+b}\right) y = -2x(x^2+b).$

    So $P(x) = -\frac{2x}{x^2+b}$. Then

    $\int P(x)\,dx = -\int \frac{2x}{x^2+b}\,dx = -\log|x^2+b| + \text{constant}.$

    Thus the integrating factor is

    $\mu(x) = e^{\int P\,dx} = e^{-\log|x^2+b|} = \frac{1}{x^2+b}.$

    (We can drop absolute values since $b$ will be chosen so that $x^2+b>0$ on the interval containing 0 and 1, or we treat it as a formal algebraic factor.)

    2. Multiply the ODE by $\mu(x)$.

    $\frac{1}{x^2+b}\,\frac{dy}{dx} - \frac{2x}{(x^2+b)^2}\,y = -2x.$

    Notice the left side is exactly

    $\frac{d}{dx}\left( \frac{y}{x^2+b} \right).$

    Check: derivative of $\frac{y}{x^2+b}$ is $\frac{y'}{x^2+b} - \frac{2x y}{(x^2+b)^2}$. Yes.

    So we have

    $\frac{d}{dx}\left( \frac{y}{x^2+b} \right) = -2x.$

    3. Integrate both sides.

    $\frac{y}{x^2+b} = \int (-2x)\,dx = -x^2 + C,$

    where $C$ is an arbitrary constant.

    Thus

    $y(x) = (x^2+b)(-x^2 + C).$

    4. Apply the first condition $y(0)=12$.

    At $x=0$:

    $y(0) = (0+b)(0 + C) = bC = 12 \quad\Rightarrow\quad C = \frac{12}{b}.$

    5. Apply the second condition $y(1)=10$.

    At $x=1$:

    $y(1) = (1+b)(-1 + C) = 10.$

    Substitute $C = \frac{12}{b}$:

    $(1+b)\left(-1 + \frac{12}{b}\right) = 10.$

    6. Solve for $b$.

    Simplify the left side:

    $$$$

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12. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
    Q.If $x = 2 + 2^{2/3} + 2^{1/3}$, then $x^3 - 6x^2 + 6x =$ (A) 3 (B) 2 (C) 1 (D) 0

    ›Reveal solutionSolution

    The key idea is to rewrite the given expression as a cubic equation by setting $t = 2^{1/3}$, then substituting $x = t^2 + t + 2$ and simplifying to find $x^3 - 6x^2 + 6x = 2$.

    We are given $x = 2 + 2^{2/3} + 2^{1/3}$. The expression $x^3 - 6x^2 + 6x$ looks like it might simplify to a constant — a common trick in such problems. The direct approach of cubing $x$ would be messy, so we look for a smarter algebraic structure.

    Notice that $2^{1/3}$ and $2^{2/3}$ are related: if we let $t = 2^{1/3}$, then $t^3 = 2$ and $t^2 = 2^{2/3}$. So $x$ becomes $x = t^2 + t + 2$. The problem now reduces to working with $t$, where $t^3 = 2$ is a clean relation.

    1. Set up the substitution.

       Let $t = 2^{1/3}$. Then $t^3 = 2$, and $x = t^2 + t + 2$.

    2. Express $x$ in terms of $t$ and find a relation.

       We have $x - 2 = t^2 + t$.

       Square both sides? That would give $t^4 + 2t^3 + t^2$, but $t^4 = t \cdot t^3 = 2t$, so it might work. But a more elegant path is to notice that $t$ satisfies $t^3 - 2 = 0$, and $x$ is a symmetric polynomial in $t$. We can compute $x^3 - 6x^2 + 6x$ directly by substituting $x = t^2 + t + 2$ and simplifying using $t^3 = 2$.

    3. Compute $x^2$ and $x^3$ in terms of $t$.

       First, $x^2 = (t^2 + t + 2)^2$

       $= t^4 + t^2 + 4 + 2t^3 + 4t^2 + 4t$ (expand carefully: $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$)

       $= t^4 + 2t^3 + (t^2 + 4t^2) + 4t + 4$

       $= t^4 + 2t^3 + 5t^2 + 4t + 4$.

       Now $t^4 = t \cdot t^3 = 2t$, and $t^3 = 2$, so:

       $x^2 = 2t + 2(2) + 5t^2 + 4t + 4$

       $= 2t + 4 + 5t^2 + 4t + 4$

       $= 5t^2 + 6t + 8$.

       Next, $x^3 = x \cdot x^2 = (t^2 + t + 2)(5t^2 + 6t + 8)$.

       Multiply term by term:

       $t^2(5t^2 + 6t + 8) = 5t^4 + 6t^3 + 8t^2$

       $t(5t^2 + 6t + 8) = 5t^3 + 6t^2 + 8t$

       $2(5t^2 + 6t + 8) = 10t^2 + 12t + 16$

       Summing: $x^3 = 5t^4 + (6t^3 + 5t^3) + (8t^2 + 6t^2 + 10t^2) + (8t + 12t) + 16$

       $= 5t^4 + 11t^3 + 24t^2 + 20t + 16$.

       Replace $t^4 = 2t$ and $t^3 = 2$:

       $x^3 = 5(2t) + 11(2) + 24t^2 + 20t + 16$

       $= 10t + 22 + 24t^2 + 20t + 16$

       $= 24t^2 + 30t + 38$.

    4. Now form $x^3 - 6x^2 + 6x$.

       We have:

       $x^3 = 24t^2 + 30t + 38$

       $6x^2 = 6(5t^2 + 6t + 8) = 30t^2 + 36t + 48$

       $6x = 6(t^2 + t + 2) = 6t^2 + 6t + 12$ …

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