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 86 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If the differential equation having $y = Ae^x + B\sin x$ as its general solution is $f(x)\dfrac{d^2y}{dx^2} + g(x)\dfrac{dy}{dx} + h(x)y = 0$, then $f(x) + g(x) + h(x) =$ (A) $\cos x - \sin x$ (B) $4\sin x$ (C) $2\cos x$ (D) $0$

   ›Reveal solutionSolution

   Eliminating $A,B$ from $y=Ae^x+B\sin x$ gives $(\cos x-\sin x)y''+2\sin x\,y'-(\sin x+\cos x)y=0$, so $f+g+h=0$ — option (D).

   Concept. A two‑parameter family $y=Ae^{x}+B\sin x$ satisfies a second‑order ODE obtained by eliminating $A$ and $B$ from $y,\,y',\,y''$. The Wronskian‑style determinant of $\{y,e^{x},\sin x\}$ vanishing is exactly that eliminant.

   Solution.

   1. Differentiate:

   $y=Ae^{x}+B\sin x,\qquad y'=Ae^{x}+B\cos x,\qquad y''=Ae^{x}-B\sin x.$

   2. A non‑trivial $(A,B)$ exists iff $\begin{vmatrix} y & e^{x} & \sin x\\y' & e^{x} & \cos x\\y'' & e^{x} & -\sin x\end{vmatrix}=0.$ …

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8. 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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9. 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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10. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.If $y = \left( x + \sqrt{x^2 + 1} \right)^5$ then $25y =$ (A) $(x^2 + 1)y_2 - x y_1$ (B) $(x^2 + 1)y_2 + x y_1$ (C) $(x^2 + 1)y_2 - 2x y_1$ (D) $(x^2 + 1)y_2 + 2x y_1$

    ›Reveal solutionSolution

    The function $y = (x + \sqrt{x^2+1})^5$ is a power of an inverse hyperbolic sine, so its derivatives satisfy a simple recurrence. Differentiating twice and rearranging gives $25y = (x^2+1)y_2 + x y_1$, which is option (B).

    We have $y = \left( x + \sqrt{x^2 + 1} \right)^5$. The expression inside the parentheses is the standard form for the inverse hyperbolic sine: $\sinh^{-1} x = \log(x + \sqrt{x^2+1})$, so $x + \sqrt{x^2+1} = e^{\sinh^{-1} x}$. That means $y = e^{5\sinh^{-1} x}$. This is a composition that makes differentiation clean: the derivative of $\sinh^{-1} x$ is $\frac{1}{\sqrt{x^2+1}}$, and the chain rule will produce a pattern that eliminates the square root.

    The key insight: instead of brute-force expanding, we can find a relation between $y$, $y_1 = \frac{dy}{dx}$, and $y_2 = \frac{d^2y}{dx^2}$ by differentiating the defining equation. Notice that $x + \sqrt{x^2+1}$ satisfies a neat property: its reciprocal is $\sqrt{x^2+1} - x$. This will help us isolate derivatives.

    1. First derivative. Let $u = x + \sqrt{x^2+1}$. Then $y = u^5$, and $\frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2+1}} = \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}} = \frac{u}{\sqrt{x^2+1}}$. So by the chain rule:

    $y_1 = 5u^4 \cdot \frac{du}{dx} = 5u^4 \cdot \frac{u}{\sqrt{x^2+1}} = \frac{5u^5}{\sqrt{x^2+1}} = \frac{5y}{\sqrt{x^2+1}}.$

    Hence

    $\sqrt{x^2+1}\, y_1 = 5y. \qquad(1)$

    2. Second derivative. Differentiate (1) with respect to $x$. The left side is a product:

    $\frac{d}{dx}\left( \sqrt{x^2+1}\, y_1 \right) = \frac{x}{\sqrt{x^2+1}}\, y_1 + \sqrt{x^2+1}\, y_2.$

    The right side differentiates to $5y_1$. So: …

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11. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.Assertion (A): The curves $y^2 = 4x$ and $x^2 = -2y$ intersect at $(1, 2)$ orthogonally. Reason (R): If the product of the slopes of the tangents drawn to two curves at their point of intersection is $-1$, then the curves are said to cut each other orthogonally. The correct option among the following is (A) (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

    ›Reveal solutionSolution

    The assertion claims the curves intersect orthogonally at $(1, 2)$, but $(1, 2)$ is not an intersection point of the curves. The reason correctly defines orthogonal intersection. Thus, Assertion (A) is false, and Reason (R) is true.

    When we talk about two curves intersecting orthogonally, we mean that at their point of intersection, the tangents to each curve are perpendicular to each other. The condition for two lines to be perpendicular is that the product of their slopes is $-1$.

    Let's analyze the given assertion and reason.

    1. Analyze Assertion (A): The curves $y^2 = 4x$ and $x^2 = -2y$ intersect at $(1, 2)$ orthogonally.

       First, we must verify if the point $(1, 2)$ is indeed a point of intersection for both curves. A point of intersection must satisfy the equations of both curves.

       • For the curve $y^2 = 4x$:

         Substitute $x=1$ and $y=2$:

         $2^2 = 4(1)$

         $4 = 4$

         This is true, so $(1, 2)$ lies on the curve $y^2 = 4x$.

       • For the curve $x^2 = -2y$:

         Substitute $x=1$ and $y=2$:

         $1^2 = -2(2)$

         $1 = -4$

         This is false. The point $(1, 2)$ does not lie on the curve $x^2 = -2y$.

       Since $(1, 2)$ does not lie on both curves, it cannot be a point of intersection. Therefore, the assertion that the curves intersect at $(1, 2)$ is false.

       Watch out

       It is crucial to first check if the given point is an actual point of intersection. If it's not, then any claim about the nature of intersection (like orthogonality) at that point is automatically false, regardless of what the slopes might suggest if we were to calculate them there.

       Even though the assertion is already false, for a complete understanding, let's briefly check the orthogonality condition if $(1, 2)$ were an intersection point.

       • Slope of tangent to $y^2 = 4x$ at $(1, 2)$:

         Differentiate $y^2 = 4x$ implicitly with respect to $x$:

         $2y \frac{dy}{dx} = 4$

         $\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}$

         At $(1, 2)$, the slope $m_1 = \frac{2}{2} = 1$.

       • Slope of tangent to $x^2 = -2y$ at $(1, 2)$:

         Differentiate $x^2 = -2y$ implicitly with respect to $x$:

         $2x = -2 \frac{dy}{dx}$

         $\frac{dy}{dx} = \frac{2x}{-2} = -x$

         At $(1, 2)$, the slope $m_2 = -1$. …

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12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.Consider the following statements Assertion (A): When $x, y, z$ are positive numbers, then

    [!FORMULA] $\tan^{-1} \left( \frac{x(x+y+z)}{yz} \right) + \tan^{-1} \left( \frac{y(x+y+z)}{xz} \right) + \tan^{-1} \left( \frac{z(x+y+z)}{xy} \right) = \pi$

    Reason (R): $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right)$ if $a > 0$ & $b > 0$ The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

    ›Reveal solutionSolution

    The three inverse tangents really do add to $\pi$, because their tangents satisfy $a+b+c=abc$. But the Reason omits the essential restriction $ab<1$, so it is false as printed. (A) true, (R) false — option (C).

    The concept first

    If $A+B+C=\pi$ then $\tan A+\tan B+\tan C=\tan A\tan B\tan C$; and conversely, if $a,b,c>0$ satisfy

    $a+b+c=abc,$

    then $\tan^{-1}a+\tan^{-1}b+\tan^{-1}c=\pi$ (each term lies in $(0,\pi/2)$, so the sum lies in $(0,3\pi/2)$ and $\pi$ is the only multiple of $\pi$ available).

    The second-hand tool, $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\dfrac{a+b}{1-ab}$, is where students lose marks. For $a,b>0$ the true statement is

    $\tan^{-1}a+\tan^{-1}b=\begin{cases}\tan^{-1}\dfrac{a+b}{1-ab}, & ab<1,\\[2mm] \pi+\tan^{-1}\dfrac{a+b}{1-ab}, & ab>1,\end{cases}$

    because the left side can exceed $\pi/2$, whereas $\tan^{-1}$ can never return more than $\pi/2$.

    Step 1 — Test the Assertion

    Put $s=x+y+z$ and

    $a=\sqrt{\frac{xs}{yz}},\qquad b=\sqrt{\frac{ys}{zx}},\qquad c=\sqrt{\frac{zs}{xy}} .$

    Then

    $abc=\sqrt{\frac{xyz\,s^{3}}{x^{2}y^{2}z^{2}}}=\frac{s^{3/2}}{\sqrt{xyz}},$

    $a+b+c=\frac{\sqrt{s}\,\left(x+y+z\right)}{\sqrt{xyz}}=\frac{s^{3/2}}{\sqrt{xyz}} .$

    They are equal, so $a+b+c=abc$ and therefore the sum of the three inverse tangents is $\pi$. (A) is true. Quick check with $x=y=z=1$: each term is $\tan^{-1}\sqrt{3}=\dfrac{\pi}{3}$, and $3\times\dfrac{\pi}{3}=\pi$. ✓

    Step 2 — Test the Reason …

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