Q.Differentiate w.r.t. $x$: $(x + 1)^2 (x + 2)^3 (x + 3)^4$.

Showing the 12 most recent of 50 on this concept.

1. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.If $\dfrac{3x+5}{(x+1)(2x^2+3)} = \dfrac{A}{x+1} + \dfrac{Bx+C}{2x^2+3}$ and $f(x) = Ax^3 + Bx^2 + 7x + C$, then $5C - f'(-2) =$ (A) 19 (B) 15 (C) 4 (D) 34

   ›Reveal solutionSolution

   We first decompose the given rational function into partial fractions to find the constants $A$, $B$, and $C$. Then, we use these constants to define the polynomial $f(x)$, differentiate it to find $f'(x)$, and evaluate $f'(-2)$. Finally, we compute the required expression $5C - f'(-2)$, which evaluates to $\boxed{4}$.

   The problem combines two key calculus concepts: partial fraction decomposition and differentiation of polynomials. The core idea is to first determine the unknown constants $A$, $B$, and $C$ by equating the given rational function with its partial fraction form. Once these constants are known, we can fully define the polynomial function $f(x)$. The next step involves finding the derivative of $f(x)$, denoted as $f'(x)$, and evaluating it at a specific point, $x=-2$. Finally, we substitute the calculated values into the expression $5C - f'(-2)$ to arrive at the final answer.

   Here's a step-by-step breakdown:

   1. Determine the constants $A$, $B$, and $C$ using partial fraction decomposition. We are given the identity:

   $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$

   To find $A$, $B$, and $C$, we combine the terms on the right-hand side:
   

   $\frac{A(2x^2+3) + (Bx+C)(x+1)}{(x+1)(2x^2+3)}$

   Equating the numerators from both sides, we get:
   

   $3x+5 = A(2x^2+3) + (Bx+C)(x+1)$

   This equation must hold for all values of $x$. We can find the constants using two methods:
   
   *   **Method of Substitution (for $A$):**
       To find $A$, we can choose a value of $x$ that makes the term $(Bx+C)(x+1)$ zero. This happens when $x+1=0$, i.e., $x=-1$.
       Substitute $x=-1$ into the equation:
   

   $3(-1)+5 = A(2(-1)^2+3) + (B(-1)+C)(-1+1)$

   $-3+5 = A(2(1)+3) + (C-B)(0)$

   $2 = A(2+3) + 0$

   $2 = 5A$

   $A = \frac{2}{5}$

   *   **Method of Comparing Coefficients (for $B$ and $C$):**
       Expand the right side of the equation $3x+5 = A(2x^2+3) + (Bx+C)(x+1)$:
   

   $3x+5 = 2Ax^2 + 3A + Bx^2 + Bx + Cx + C$

       Group terms by powers of $x$:
   

   $3x+5 = (2A+B)x^2 + (B+C)x + (3A+C)$

       Now, compare the coefficients of $x^2$, $x$, and the constant terms on both sides:
       *   **Coefficient of $x^2$:**
   

   $0 = 2A+B$

           Since $A = \frac{2}{5}$:
   

   $0 = 2\left(\frac{2}{5}\right) + B$

   $0 = \frac{4}{5} + B \implies B = -\frac{4}{5}$

       *   **Coefficient of $x$:**
   

   $3 = B+C$

           Since $B = -\frac{4}{5}$:
   

   $3 = -\frac{4}{5} + C$

   $C = 3 + \frac{4}{5} = \frac{15+4}{5} = \frac{19}{5}$

       *   **Constant term (for verification):**
   

   $5 = 3A+C$

           Substitute $A = \frac{2}{5}$ and $C = \frac{19}{5}$:
   

   $5 = 3\left(\frac{2}{5}\right) + \frac{19}{5}$

   $5 = \frac{6}{5} + \frac{19}{5} = \frac{25}{5}$

   $5 = 5$

           This confirms our values for $A$, $B$, and $C$.
           So, we have $A = \frac{2}{5}$, $B = -\frac{4}{5}$, and $C = \frac{19}{5}$.
   

   2. Define $f(x)$ and find its derivative $f'(x)$.

   We are given $f(x) = Ax^3 + Bx^2 + 7x + C$. …

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2. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $2x^2 - 3xy + 4y^2 + 2x - 3y + 4 = 0$, then $\left(\dfrac{dy}{dx}\right)_{(3,2)} =$ (A) $-5$ (B) $\dfrac{5}{7}$ (C) $-2$ (D) $\dfrac{2}{7}$

   ›Reveal solutionSolution

   Implicit differentiation gives $\dfrac{dy}{dx}=\dfrac{-(4x-3y+2)}{-3x+8y-3}$, and at $(3,2)$ this evaluates to $\dfrac{-8}{4}=-2$. Answer: (C).

   Concept

   For a curve given implicitly, differentiate every term with respect to $x$ treating $y=y(x)$ (chain and product rules), then collect and isolate $\dfrac{dy}{dx}$.

   Solution

   1. Differentiate the relation $2x^2-3xy+4y^2+2x-3y+4=0$:

   $4x-3\!\left(y+x\frac{dy}{dx}\right)+8y\frac{dy}{dx}+2-3\frac{dy}{dx}=0.$

   2. Collect the derivative terms.

   $(-3x+8y-3)\frac{dy}{dx}+(4x-3y+2)=0,$

   $\frac{dy}{dx}=\frac{-(4x-3y+2)}{-3x+8y-3}.$

   3. Evaluate at $(3,2)$.

   $\text{Numerator: } -(4\cdot3-3\cdot2+2)=-(12-6+2)=-8,$

   $\text{Denominator: } -3\cdot3+8\cdot2-3=-9+16-3=4,$

   $\left.\frac{dy}{dx}\right|_{(3,2)}=\frac{-8}{4}=-2.$ …

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3. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.For any quadratic polynomial $f(x)$, it is true that

   [!FORMULA] $f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2,$

   where $a$ is any real number. If

   [!FORMULA] $\frac{3x^2 + 4x + 7}{(x-2)^3} = \frac{A}{(x-2)^3} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)}$

   and $g(x) = 3x^2 + 4x + 7$ then $A + B + C =$ (A) $g(2) + g'(2) + g''(2)$ (B) $g''(2) + 2g(2) + \frac{g'(1)}{2!}$ (C) $g(2) + g'(2) + \frac{g''(2)}{2!}$ (D) $2g(2) + 2g'(2) + \frac{g''(2)}{2!}$

   ›Reveal solutionSolution

   The partial-fraction expansion of $\frac{g(x)}{(x-2)^3}$ is exactly the Taylor expansion of $g(x)$ about $x=2$, divided by $(x-2)^3$. Matching coefficients gives $A = g(2)$, $B = g'(2)$, $C = \frac{g''(2)}{2!}$, so $A+B+C = g(2) + g'(2) + \frac{g''(2)}{2!}$, which is option (C).

   The given identity is a partial-fraction decomposition of a rational function whose denominator is a pure power of $(x-2)$. The numerator $g(x)=3x^2+4x+7$ is a quadratic, so the decomposition has exactly three terms: one with denominator $(x-2)^3$, one with $(x-2)^2$, and one with $(x-2)$. The constants $A, B, C$ are what we need to find.

   The key insight is that the form of the decomposition is identical to the Taylor expansion of $g(x)$ about $x=2$, written as a polynomial in $(x-2)$. Because $g(x)$ is a quadratic, its Taylor series terminates after the second-degree term:

   $g(x) = g(2) + g'(2)(x-2) + \frac{g''(2)}{2!}(x-2)^2.$

   Now divide both sides by $(x-2)^3$:

   $\frac{g(x)}{(x-2)^3} = \frac{g(2)}{(x-2)^3} + \frac{g'(2)}{(x-2)^2} + \frac{g''(2)/2!}{(x-2)}.$

   Comparing this with the given decomposition

   $\frac{g(x)}{(x-2)^3} = \frac{A}{(x-2)^3} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)},$

   we see that the coefficients match term by term:

   1. The coefficient of $\frac{1}{(x-2)^3}$ is $A = g(2)$.
   2. The coefficient of $\frac{1}{(x-2)^2}$ is $B = g'(2)$.
   3. The coefficient of $\frac{1}{(x-2)}$ is $C = \frac{g''(2)}{2!}$.

   Therefore

   $A + B + C = g(2) + g'(2) + \frac{g''(2)}{2!}.$ …

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4. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.

   [!FORMULA] $\frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2} =$

   (A) $\sinh^2 x + \coth^2 x$ (B) $\sinh^2 x + \operatorname{sech}^2 x$ (C) $\cosh^2 x + \operatorname{sech}^2 x$ (D) $\cosh^2 x + \tanh^2 x$

   ›Reveal solutionSolution

   The problem asks us to simplify an expression involving exponentials and match it with an equivalent expression in terms of hyperbolic functions. By expressing the numerator and denominator in terms of $(e^x - e^{-x})^2$ and then using the definitions of $\sinh x$ and $\coth x$, we find the expression simplifies to $\sinh^2 x + \coth^2 x$.

   The core idea here is to recognize that expressions involving $e^x$ and $e^{-x}$ can often be simplified using the definitions of hyperbolic functions. The given expression has terms like $e^x - e^{-x}$ in the denominator and $e^{4x} + e^{-4x}$ in the numerator. Our strategy will be to transform these exponential terms into their hyperbolic function equivalents.

   The fundamental definitions of hyperbolic sine and cosine are:

   $\sinh x = \frac{e^x - e^{-x}}{2}$

   $\cosh x = \frac{e^x + e^{-x}}{2}$

   From these, other hyperbolic functions are defined:

   $\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

   $\coth x = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$

   $\operatorname{sech} x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}$

   $\operatorname{cosech} x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}$

   We will also use the fundamental identity relating $\coth x$ and $\operatorname{cosech} x$:

   $\coth^2 x - \operatorname{cosech}^2 x = 1$

   Let's simplify the given expression step-by-step.

   1. Simplify the denominator:

      The denominator is $4(e^x - e^{-x})^2$.

      From the definition of $\sinh x$, we know $e^x - e^{-x} = 2\sinh x$.

      Substituting this into the denominator:

      $4(e^x - e^{-x})^2 = 4(2\sinh x)^2 = 4(4\sinh^2 x) = 16\sinh^2 x$.

   2. Simplify the numerator:

      The numerator is $e^{4x} + e^{-4x} + 14$.

      We want to express $e^{4x} + e^{-4x}$ in terms of $(e^x - e^{-x})^2$.

      Let $A = e^x$ and $B = e^{-x}$. Then the numerator is $A^4 + B^4 + 14$.

      We know that $A^4 + B^4 = (A^2 + B^2)^2 - 2A^2B^2$. Since $A B = e^x e^{-x} = e^0 = 1$, we have $A^2B^2 = 1$.

      So, $A^4 + B^4 = (A^2 + B^2)^2 - 2$.

      Now, let's express $A^2 + B^2$ in terms of $(A-B)^2$:

      $A^2 + B^2 = (A-B)^2 + 2AB = (A-B)^2 + 2$.

      Substitute this back into the expression for $A^4 + B^4$:

      $A^4 + B^4 = \left((A-B)^2 + 2\right)^2 - 2$.

      Let $X = A-B = e^x - e^{-x}$.

      Then $e^{4x} + e^{-4x} = (X^2 + 2)^2 - 2$.

      Expand this: $(X^2 + 2)^2 - 2 = (X^4 + 4X^2 + 4) - 2 = X^4 + 4X^2 + 2$.

      Now, add the constant term from the numerator:

      Numerator $= (X^4 + 4X^2 + 2) + 14 = X^4 + 4X^2 + 16$.

   3. Combine and simplify the expression:

      The original expression is $\frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$. …

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5. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\dfrac{dy}{dx} = (x^3 - x) - (1 - 3x^2)\tan x + (x^3 - x)\tan^2 x$ and $y(1) = 0$, then $\dfrac{64}{\pi} y\left(\dfrac{\pi}{4}\right) =$ (A) $1$ (B) $\pi^2 + 16$ (C) $\pi^2 - 16$ (D) $16\pi^2$

   ›Reveal solutionSolution

   The key is to notice that the derivative simplifies to a perfect derivative of a product involving $\tan x$ and a polynomial, so we can integrate directly and then evaluate at $x = \pi/4$. The final value is $\pi^2 - 16$, which corresponds to option (C).

   ---

   We are given:

   $\frac{dy}{dx} = (x^3 - x) - (1 - 3x^2)\tan x + (x^3 - x)\tan^2 x$

   with $y(1) = 0$, and we need $\frac{64}{\pi} y(\pi/4)$.

   Concept and intuition:

   The expression looks messy, but the presence of $\tan x$ and $\tan^2 x$ alongside a polynomial suggests it might be the derivative of something like $(\text{polynomial}) \cdot \tan x$ plus something simpler. When we differentiate a product $u(x) \tan x$, we get $u'(x) \tan x + u(x) \sec^2 x$. Since $\sec^2 x = 1 + \tan^2 x$, this can produce terms like $u(x) + u(x) \tan^2 x$ plus a $u'(x) \tan x$ term. That matches our structure perfectly.

   Let’s try to match it.

   ---

   1. Guess the form Suppose $y(x) = (x^3 - x) \tan x + \text{something}$. Differentiate:

   $\frac{d}{dx}\big[(x^3 - x)\tan x\big] = (3x^2 - 1)\tan x + (x^3 - x)\sec^2 x.$

   Since $\sec^2 x = 1 + \tan^2 x$, this becomes:

   $(3x^2 - 1)\tan x + (x^3 - x) + (x^3 - x)\tan^2 x.$

   2. Compare with given $\frac{dy}{dx}$ The given derivative is:

   $(x^3 - x) - (1 - 3x^2)\tan x + (x^3 - x)\tan^2 x.$

   Notice that $-(1 - 3x^2) = 3x^2 - 1$. So the given derivative is exactly:

   $(x^3 - x) + (3x^2 - 1)\tan x + (x^3 - x)\tan^2 x.$

   This matches the derivative we computed for $(x^3 - x)\tan x$ term by term.

   3. Conclusion about $y(x)$ Hence,

   $\frac{dy}{dx} = \frac{d}{dx}\big[(x^3 - x)\tan x\big].$

   Integrating both sides:

   $y(x) = (x^3 - x)\tan x + C.$

   4. Use the initial condition $y(1) = 0$ gives:

   $0 = (1^3 - 1)\tan 1 + C = 0 \cdot \tan 1 + C = C.$

   So $C = 0$, and

   $y(x) = (x^3 - x)\tan x.$

   5. Evaluate at $x = \pi/4$ $$$$

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6. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.If the tangent at a point P on the curve $y = 4x^4 + x$ is perpendicular to the tangent to the same curve at (0, 0), then the point P is (A) $\left( \dfrac{-1}{2}, \dfrac{-1}{4} \right)$ (B) $\left( \dfrac{1}{2}, \dfrac{3}{4} \right)$ (C) $(1, 5)$ (D) $(-1, 3)$

   ›Reveal solutionSolution

   To find the point P, we first determine the slope of the tangent at (0, 0) using the derivative. Since the tangent at P is perpendicular to this, its slope must be the negative reciprocal. We then set the general derivative equal to this required slope to find the x-coordinate of P, and finally use the curve equation to find the y-coordinate. The point P is $\boxed{\left( \dfrac{-1}{2}, \dfrac{-1}{4} \right)}$.

   The slope of the tangent to a curve at any point is given by the derivative of the curve's equation with respect to $x$, evaluated at that point. This derivative, $\frac{dy}{dx}$, represents the instantaneous rate of change of $y$ with respect to $x$, which is precisely the slope of the line tangent to the curve.

   For two lines to be perpendicular, the product of their slopes must be $-1$. If one line has a slope $m_1$ and the other has a slope $m_2$, then $m_1 \cdot m_2 = -1$. This condition is key to solving the problem.

   Here's how we apply these concepts:

   1. Find the derivative of the curve: The given curve is $y = 4x^4 + x$. We differentiate $y$ with respect to $x$ to find the general expression for the slope of the tangent at any point $(x, y)$:

   $\frac{dy}{dx} = \frac{d}{dx}(4x^4 + x) = 16x^3 + 1$

   2. Calculate the slope of the tangent at (0, 0): Let $m_1$ be the slope of the tangent at the point $(0, 0)$. We substitute $x=0$ into the derivative:

   $m_1 = \left. \frac{dy}{dx} \right|_{x=0} = 16(0)^3 + 1 = 1$

   3. Determine the required slope for the tangent at point P: Let $m_2$ be the slope of the tangent at point P. We are given that the tangent at P is perpendicular to the tangent at (0, 0).

      For two perpendicular lines with slopes $m_1$ and $m_2$, we have $m_1 \cdot m_2 = -1$.

      Using this condition:

   $1 \cdot m_2 = -1$

   $m_2 = -1$

   So, the tangent at point P must have a slope of $-1$.
   

   4. Find the x-coordinate of point P:

   We know that the slope of the tangent at any point $(x, y)$ is $16x^3 + 1$. We set this equal to the required slope $m_2 = -1$:

   $16x^3 + 1 = -1$

   $16x^3 = -2$

   $x^3 = \frac{-2}{16}$

   $$ x^3 = \frac{-1}{8} $$ …
   

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7. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $\int e^x \cos x \, dx = \dfrac{e^x}{2} (\cos x + \sin x)$ and $\int \dfrac{\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)}{(3-2x)^2} dx = \dfrac{f(x)}{24} \left[ \cos (g(x)) + \sin (g(x)) \right] + c$ then $g(1) =$ (A) $5$ (B) $\log f(2)$ (C) $\log f(1)$ (D) $0$

   ›Reveal solutionSolution

   This problem requires recognizing a pattern in the given integral formula to evaluate a more complex integral. By identifying the correct substitution, the complex integral transforms into the standard form, allowing us to determine $g(x)$ and $f(x)$. We find $g(1) = \log 5$, which matches $\log f(1)$. The correct option is (C).

   The core idea here is to recognize that the second integral's structure is a generalized form of the first integral. The given formula $\int e^x \cos x \, dx = \dfrac{e^x}{2} (\cos x + \sin x)$ provides a template. When we see the result of the second integral in the form $\dfrac{f(x)}{24} \left[ \cos (g(x)) + \sin (g(x)) \right]$, it strongly suggests that $g(x)$ is the argument of the cosine function in the integrand, and $f(x)$ is related to $e^{g(x)}$. Our strategy will be to make a suitable substitution to transform the second integral into the standard form $\int e^t \cos t \, dt$.

   1. Identify the potential $g(x)$:

      The argument of the cosine function in the second integral is $\log \left( \dfrac{2x+3}{3-2x} \right)$. Given the form of the result, it is highly probable that this expression is $g(x)$.

      Let $g(x) = \log \left( \dfrac{2x+3}{3-2x} \right)$.

   2. Calculate the derivative of $g(x)$:

      We can rewrite $g(x)$ using logarithm properties: $g(x) = \log(2x+3) - \log(3-2x)$.

      Now, differentiate $g(x)$ with respect to $x$:

      $g'(x) = \dfrac{d}{dx} (\log(2x+3)) - \dfrac{d}{dx} (\log(3-2x))$

      $g'(x) = \dfrac{1}{2x+3} \cdot (2) - \dfrac{1}{3-2x} \cdot (-2)$

      $g'(x) = \dfrac{2}{2x+3} + \dfrac{2}{3-2x}$

      Combine the terms:

      $g'(x) = 2 \left( \dfrac{1}{2x+3} + \dfrac{1}{3-2x} \right)$

      $g'(x) = 2 \left( \dfrac{(3-2x) + (2x+3)}{(2x+3)(3-2x)} \right)$

      $g'(x) = 2 \left( \dfrac{6}{(2x+3)(3-2x)} \right)$

      $g'(x) = \dfrac{12}{(2x+3)(3-2x)}$

   3. Determine $e^{g(x)}$:

      From $g(x) = \log \left( \dfrac{2x+3}{3-2x} \right)$, we have:

      $e^{g(x)} = e^{\log \left( \dfrac{2x+3}{3-2x} \right)} = \dfrac{2x+3}{3-2x}$.

   4. Relate $e^{g(x)} g'(x)$ to the integrand:

      Let's multiply $e^{g(x)}$ and $g'(x)$:

      $e^{g(x)} g'(x) = \left( \dfrac{2x+3}{3-2x} \right) \cdot \left( \dfrac{12}{(2x+3)(3-2x)} \right)$

      $e^{g(x)} g'(x) = \dfrac{12}{(3-2x)^2}$.

      Now, observe the integrand of the second integral: $\dfrac{\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)}{(3-2x)^2}$.

      We can rewrite this using $g(x)$ and the expression for $e^{g(x)} g'(x)$:

      $\dfrac{\cos(g(x))}{(3-2x)^2} = \cos(g(x)) \cdot \dfrac{1}{12} \left( \dfrac{12}{(3-2x)^2} \right)$

      $\dfrac{\cos(g(x))}{(3-2x)^2} = \dfrac{1}{12} \cos(g(x)) e^{g(x)} g'(x)$.

   5. Perform the integration:

      The integral becomes:

      $\int \dfrac{1}{12} e^{g(x)} \cos(g(x)) g'(x) \, dx$

      Let $t = g(x)$. Then $dt = g'(x) dx$.

      The integral transforms to:

      $\dfrac{1}{12} \int e^t \cos t \, dt$.

      Using the given formula $\int e^x \cos x \, dx = \dfrac{e^x}{2} (\cos x + \sin x)$: …

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8. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.A particle moves in a straight line such that its displacement S (in mts) at a time t (in sec) is given by $S(t) = t^3 - 4t^2 + 7t$. The instantaneous velocity $v$ at $t = 4$ is (A) 21 m/sec (B) 23 m/sec (C) 20 m/sec (D) 19 m/sec

   ›Reveal solutionSolution

   Instantaneous velocity is found by differentiating the displacement function with respect to time. For $S(t) = t^3 - 4t^2 + 7t$, the instantaneous velocity at $t=4$ seconds is $\boxed{23 \text{ m/sec}}$.

   In physics, when a particle moves, its position changes over time. This change in position is described by its displacement. Velocity, on the other hand, tells us how quickly this displacement is changing and in what direction.

   There are two main types of velocity we consider:

   1. Average velocity: This is the total displacement divided by the total time taken. It gives us an overall idea of the motion over an interval.
   2. Instantaneous velocity: This is the velocity of the particle at a specific instant in time. It describes the rate of change of displacement at that precise moment.

   To find the instantaneous velocity, we use the concept of differentiation from calculus. If the displacement of a particle is given by a function $S(t)$ with respect to time $t$, then its instantaneous velocity $v(t)$ at any time $t$ is the first derivative of the displacement function with respect to time.

   Instantaneous velocity $v(t) = \frac{dS}{dt}$

   This derivative represents the slope of the tangent line to the displacement-time graph at that particular instant, which is precisely the rate of change of displacement at that moment.

   Let's apply this to the given problem.

   1. Identify the displacement function:

      The displacement $S$ (in meters) at time $t$ (in seconds) is given by the function:

      $S(t) = t^3 - 4t^2 + 7t$

   2. Find the velocity function by differentiation:

      To find the instantaneous velocity $v(t)$, we differentiate $S(t)$ with respect to $t$. We use the power rule for differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$.

      $v(t) = \frac{dS}{dt} = \frac{d}{dt}(t^3 - 4t^2 + 7t)$ …

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9. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.The equation of the tangent to the curve $x^2 + y - 7 = 4x$ at the point $(1, 10)$ is (A) $y = 2x + 8$ (B) $y = x + 8$ (C) $y = -2x - 14$ (D) $y = x - 4$

   ›Reveal solutionSolution

   The equation of the tangent to a curve at a given point is found by calculating the derivative (which gives the slope of the tangent) at that point and then using the point-slope form of a line. The equation of the tangent is $\boxed{y = 2x + 8}$.

   To find the equation of the tangent to a curve at a specific point, we need two pieces of information: the slope of the tangent line and a point it passes through. The given point $(1, 10)$ provides the latter. The crucial concept here is that the derivative of a function at a particular point gives the slope of the tangent line to the curve at that very point.

   Here's how we approach this problem:

   1. Rearrange the curve's equation:

      The given equation of the curve is $x^2 + y - 7 = 4x$. It's often helpful to express $y$ explicitly as a function of $x$ if possible, as it simplifies differentiation.

      $y = 4x - x^2 + 7$

   2. Find the derivative of the curve:

      We differentiate $y$ with respect to $x$ to find $\frac{dy}{dx}$. This derivative represents the general formula for the slope of the tangent line at any point $(x, y)$ on the curve.

   $\frac{dy}{dx} = \frac{d}{dx}(4x - x^2 + 7)$

   $\frac{dy}{dx} = 4 - 2x + 0$

   $\frac{dy}{dx} = 4 - 2x$

   3. Calculate the slope of the tangent at the given point: Now we substitute the $x$-coordinate of the given point $(1, 10)$ into the derivative to find the specific slope of the tangent at that point. Let $m$ be the slope of the tangent.

   $m = \left.\frac{dy}{dx}\right|_{(1, 10)} = 4 - 2(1)$

   $m = 4 - 2$

   $m = 2$

   So, the slope of the tangent line at $(1, 10)$ is $2$. …
   

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10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If $|x|$ is so small that $x^3$ and higher powers of $x$ can be neglected, then an approximate value of $\dfrac{1}{\sqrt{4-x(2+x)^3}}$ is (A) $\dfrac{1}{16} \left( 1 + \dfrac{13x}{8} + \dfrac{219}{128} x^2 \right)$ (B) $\dfrac{1}{16} \left( 1 + \dfrac{11x}{8} - \dfrac{165}{128} x^2 \right)$ (C) $\dfrac{1}{32} \left( 1 - \dfrac{11x}{8} + \dfrac{219}{128} x^2 \right)$ (D) $\dfrac{1}{16} \left( 1 - \dfrac{11x}{8} + \dfrac{171}{128} x^2 \right)$

    ›Reveal solutionSolution

    Expanding $\dfrac{1}{\sqrt{4-x}\,(2+x)^3}$ to order $x^2$ gives $\dfrac{1}{16}\!\left(1 - \dfrac{11x}{8} + \dfrac{171}{128}x^2\right)$ (option D).

    The expression is $\dfrac{1}{\sqrt{4-x}\,(2+x)^3}$ (value $\tfrac{1}{2\cdot 8}=\tfrac{1}{16}$ at $x=0$).

    First factor:

    $(4-x)^{-1/2} = \tfrac12\left(1-\tfrac{x}{4}\right)^{-1/2} = \tfrac12\!\left(1 + \tfrac{x}{8} + \tfrac{3x^2}{128}\right).$

    Second factor:

    $(2+x)^{-3} = \tfrac18\left(1+\tfrac{x}{2}\right)^{-3} = \tfrac18\!\left(1 - \tfrac{3x}{2} + \tfrac{3x^2}{2}\right).$

    Product ($\tfrac12\cdot\tfrac18 = \tfrac{1}{16}$):

    $\frac{1}{16}\left(1 + \tfrac{x}{8} + \tfrac{3x^2}{128}\right)\left(1 - \tfrac{3x}{2} + \tfrac{3x^2}{2}\right).$ …

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11. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.On differentiation if we get $f(x, y) \, dy - g(x, y) \, dx = 0$ from $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$ then $\dfrac{g(2,2)}{f(1,1)}$ (A) $\dfrac{11}{7}$ (B) $-3$ (C) $-\dfrac{1}{3}$ (D) $7$

    ›Reveal solutionSolution

    The problem asks for the ratio $\frac{g(2,2)}{f(1,1)}$ where $f$ and $g$ come from rewriting the differential of the given implicit equation. We find $f$ and $g$ by implicit differentiation, then evaluate and compute the ratio, obtaining $-\frac{1}{3}$.

    We start with the equation

    $2x^2 - 3xy + y^2 + x + 2y - 8 = 0.$

    The phrase "On differentiation if we get $f(x, y) \, dy - g(x, y) \, dx = 0$" means we differentiate the equation implicitly and then rearrange terms so that all $dy$ terms are grouped together and all $dx$ terms are grouped together, with the form $f\,dy - g\,dx = 0$. Here $f$ and $g$ are functions of $x$ and $y$.

    Concept & Intuition:

    When we differentiate an implicit relation $F(x,y)=0$, we get $F_x\,dx + F_y\,dy = 0$. That can be rewritten as $F_y\,dy = -F_x\,dx$, or equivalently $F_y\,dy + F_x\,dx = 0$. But the problem wants the form $f\,dy - g\,dx = 0$, so we match: $f = F_y$ and $g = -F_x$. Then the ratio $\frac{g}{f}$ is just $-\frac{F_x}{F_y}$, which is the negative of the derivative $\frac{dy}{dx}$ (since $\frac{dy}{dx} = -\frac{F_x}{F_y}$). So we are essentially computing a ratio of partial derivatives at given points.

    Let’s do it step by step.

    1. Differentiate implicitly with respect to $x$, treating $y$ as a function of $x$:

    $\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0.$

    This gives:

    $4x - 3\left( y + x\frac{dy}{dx} \right) + 2y\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0.$

    2. Simplify:

    $4x - 3y - 3x\frac{dy}{dx} + 2y\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0.$

    3. Collect terms with $\frac{dy}{dx}$:

    $(-3x + 2y + 2)\frac{dy}{dx} + (4x - 3y + 1) = 0.$

    4. Isolate $\frac{dy}{dx}$:

    $(-3x + 2y + 2)\frac{dy}{dx} = -(4x - 3y + 1).$

    So

    $\frac{dy}{dx} = -\frac{4x - 3y + 1}{-3x + 2y + 2}.$

    5. Rewrite in the form $f\,dy - g\,dx = 0$: Multiply both sides of the differentiated equation by $dx$:

    $(-3x + 2y + 2)\,dy + (4x - 3y + 1)\,dx = 0.$

    But the problem wants $f\,dy - g\,dx = 0$. So we need a minus sign before the $dx$ term. Write: …

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12. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.If m and M are respectively the absolute minimum and absolute maximum values of a function $f(x) = 2x^3 + 9x^2 + 12x + 1$ defined on $[-3,0]$, then m + M = (A) $-7$ (B) $0$ (C) $1$ (D) $5$

    ›Reveal solutionSolution

    The absolute minimum and maximum of a cubic on a closed interval occur either at critical points inside the interval or at the endpoints. For $f(x)=2x^3+9x^2+12x+1$ on $[-3,0]$, the sum is $m+M = 1$, so the answer is (C).

    We are asked for the sum of the absolute minimum $m$ and absolute maximum $M$ of a cubic function on a closed interval. The key idea: for a continuous function on a closed interval, the extreme values occur either at critical points (where the derivative is zero) or at the endpoints. So we find all candidates, evaluate $f$ at each, pick the smallest and largest, and add them.

    ---

    1. Find the derivative and critical points.

    The derivative is

    $f'(x) = 6x^2 + 18x + 12.$

    Factor out 6:

    $f'(x) = 6(x^2 + 3x + 2) = 6(x+1)(x+2).$

    Set $f'(x)=0$:

    $x = -1 \quad \text{or} \quad x = -2.$

    Both lie inside the interval $[-3,0]$, so they are candidates.

    2. Evaluate $f$ at the critical points and endpoints.

    • At $x = -3$ (left endpoint):

    $f(-3) = 2(-27) + 9(9) + 12(-3) + 1 = -54 + 81 - 36 + 1 = -8.$

    • At $x = -2$ (critical point):

    $f(-2) = 2(-8) + 9(4) + 12(-2) + 1 = -16 + 36 - 24 + 1 = -3.$

    • At $x = -1$ (critical point):

    $f(-1) = 2(-1) + 9(1) + 12(-1) + 1 = -2 + 9 - 12 + 1 = -4.$

    • At $x = 0$ (right endpoint):

    $f(0) = 1.$

    3. Identify the absolute minimum and maximum. …

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