Q.If 𝑓(𝑥) = 𝑥 tan−1 𝑥 , then 𝑓′(1)is equal to
(A) 𝜋 4 − 1 2
(B) 𝜋 4 + 1 2
(C) − 𝜋 4 − 1 2
(D) − 𝜋 4 + 1 2

Showing the 12 most recent of 50 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $f(x) = \sqrt{\cos^{-1} \sqrt{1-x^2}}$, then $f'\left(\dfrac{1}{2}\right) =$ (A) $\sqrt{\dfrac{2}{\pi}}$ (B) $\sqrt{\dfrac{\pi}{2}}$ (C) $-\sqrt{\dfrac{2}{\pi}}$ (D) $-\sqrt{\dfrac{\pi}{2}}$

   ›Reveal solutionSolution

   For $x\ge0$, $\cos^{-1}\sqrt{1-x^2}=\sin^{-1}x$, so $f(x)=\sqrt{\sin^{-1}x}$ and $f'\!\left(\tfrac12\right)=\sqrt{\tfrac{2}{\pi}}$ — option (A).

   Simplify. For $0\le x\le 1$, let $\theta=\cos^{-1}\sqrt{1-x^2}\in\big[0,\tfrac{\pi}{2}\big]$. Then $\cos^2\theta=1-x^2$, so $\sin\theta=x$ and $\theta=\sin^{-1}x$. Hence

   $f(x)=\sqrt{\sin^{-1}x}.$

   Differentiate.

   $f'(x)=\frac{1}{2\sqrt{\sin^{-1}x}}\cdot\frac{1}{\sqrt{1-x^2}}.$

   Evaluate at $x=\tfrac12$. $\sin^{-1}\tfrac12=\dfrac{\pi}{6}$ and $\sqrt{1-\tfrac14}=\dfrac{\sqrt3}{2}$: …

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2. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $\int (\sqrt{\csc x + 1})\, dx = k \tan^{-1}(f(x)) + c$, then $\frac{1}{k} f\left(\frac{\pi}{6}\right) =$ (A) $\frac{1}{2}$ (B) $\frac{1}{4}$ (C) $-\frac{1}{4}$ (D) $-\frac{1}{2}$

   ›Reveal solutionSolution

   $\displaystyle\int\sqrt{\csc x + 1}\,dx = -2\tan^{-1}\!\left(\sqrt{\csc x - 1}\right) + c$, so $k=-2$, $f(x)=\sqrt{\csc x-1}$ and $\tfrac{1}{k}f\!\left(\tfrac{\pi}{6}\right) = -\tfrac{1}{2}$.

   Evaluating the integral.

   Write $\sqrt{\csc x + 1} = \sqrt{\dfrac{1+\sin x}{\sin x}}$. We claim

   $\int \sqrt{\csc x + 1}\,dx = -2\tan^{-1}\!\left(\sqrt{\csc x - 1}\right) + c.$

   Verification by differentiation.

   Let $g = \sqrt{\csc x - 1}$, so $g^2 = \csc x - 1$ and $1 + g^2 = \csc x$. Then

   $2g\,g' = -\csc x\cot x \quad\Rightarrow\quad g' = \frac{-\csc x\cot x}{2g}.$

   $\frac{d}{dx}\left[-2\tan^{-1} g\right] = \frac{-2g'}{1+g^2} = \frac{-2g'}{\csc x} = \frac{\cot x}{g} = \frac{\cos x}{\sin x\,\sqrt{\dfrac{1-\sin x}{\sin x}}} = \frac{\cos x}{\sqrt{\sin x}\,\sqrt{1-\sin x}}.$

   Since $\cos x = \sqrt{1-\sin^2 x} = \sqrt{1-\sin x}\,\sqrt{1+\sin x}$, this reduces to …

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3. 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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4. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $x = \log_e \left( \cot \left( \frac{\pi}{4} + \theta \right) \right)$, then $\lim_{\theta \to 0} \frac{\theta}{(\sinh x)(\cosh x)} =$ (A) $0$ (B) $-\frac{1}{2}$ (C) $-2$ (D) $1$

   ›Reveal solutionSolution

   The key idea is to simplify $x$ using log properties and trig identities, then rewrite the limit in terms of $\theta$ using hyperbolic function definitions. The limit evaluates to $-\frac{1}{2}$, so option (B) is correct.

   We start with the given expression:

   $x = \log_e \left( \cot \left( \frac{\pi}{4} + \theta \right) \right)$.

   The goal is to find $\lim_{\theta \to 0} \frac{\theta}{(\sinh x)(\cosh x)}$.

   The presence of $\sinh x$ and $\cosh x$ suggests we should first simplify $x$ itself. The argument of the log involves a cotangent of a sum, which often simplifies using trigonometric identities. Once $x$ is expressed in terms of $\theta$, the hyperbolic functions become manageable, and the limit reduces to a standard form.

   1. Simplify $x$ using a trig identity.

      Recall that $\cot\left(\frac{\pi}{4} + \theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta}$.

      This comes from $\cot(A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$, but a quicker route:

      $\cot\left(\frac{\pi}{4} + \theta\right) = \frac{\cos(\frac{\pi}{4}+\theta)}{\sin(\frac{\pi}{4}+\theta)}$, and using $\cos(\frac{\pi}{4}+\theta) = \frac{1}{\sqrt{2}}(\cos\theta - \sin\theta)$, $\sin(\frac{\pi}{4}+\theta) = \frac{1}{\sqrt{2}}(\cos\theta + \sin\theta)$, the ratio gives $\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \frac{1 - \tan\theta}{1 + \tan\theta}$.

      So $x = \log_e \left( \frac{1 - \tan\theta}{1 + \tan\theta} \right)$.

   2. Rewrite $x$ using log properties.

      $x = \log_e(1 - \tan\theta) - \log_e(1 + \tan\theta)$.

      For small $\theta$, $\tan\theta \approx \theta$, so we can expand:

      $\log_e(1 - \theta) \approx -\theta - \frac{\theta^2}{2} - \cdots$ and $\log_e(1 + \theta) \approx \theta - \frac{\theta^2}{2} + \cdots$.

      Thus $x \approx (-\theta - \frac{\theta^2}{2}) - (\theta - \frac{\theta^2}{2}) = -2\theta$, ignoring higher-order terms.

      More precisely, $x = -2\theta + O(\theta^3)$ as $\theta \to 0$.

   3. Express $\sinh x$ and $\cosh x$ in terms of $\theta$.

      For small $x$, $\sinh x \approx x$ and $\cosh x \approx 1 + \frac{x^2}{2}$.

      But here $x \approx -2\theta$, so $\sinh x \approx -2\theta$ and $\cosh x \approx 1 + \frac{(-2\theta)^2}{2} = 1 + 2\theta^2$.

      Therefore $(\sinh x)(\cosh x) \approx (-2\theta)(1 + 2\theta^2) = -2\theta - 4\theta^3$.

   4. Form the limit expression.

      $\frac{\theta}{(\sinh x)(\cosh x)} \approx \frac{\theta}{-2\theta - 4\theta^3} = \frac{1}{-2 - 4\theta^2}$. …

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5. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.$\lim_{n \to \infty} \left[ \frac{1}{n^2} \sec^2 \frac{1}{n} + \frac{2}{n^2} \sec^2 \frac{2}{n} + \frac{4}{n^2} \sec^2 \frac{4}{n} + \ldots + \frac{1}{n} \sec^2 1 \right] =$ (A) $\frac{1}{2} \sec(1)$ (B) $\frac{1}{2} \cos ec(1)$ (C) $\tan(1)$ (D) $\frac{1}{2} \tan(1)$

   ›Reveal solutionSolution

   The sum is a Riemann sum for $\displaystyle\int_0^1 x\sec^2(x^2)\,dx=\tfrac12\tan 1$.

   Identify the Riemann sum. The general term is

   $t_k=\frac{k}{n^2}\sec^2\!\frac{k^2}{n^2}=\frac{1}{n}\cdot\frac{k}{n}\,\sec^2\!\left(\frac{k}{n}\right)^{2},$

   and the last term ($k=n$) is $\dfrac{n}{n^2}\sec^2\dfrac{n^2}{n^2}=\dfrac1n\sec^2 1$, matching the final entry. Writing $x_k=\dfrac{k}{n}$ with spacing $\Delta x=\dfrac1n$,

   $\sum_{k=1}^{n} t_k=\sum_{k=1}^{n}\frac{1}{n}\,f\!\left(\frac{k}{n}\right),\qquad f(x)=x\sec^2(x^2).$

   Take the limit as an integral. …

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6. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $\tan h\,x = \dfrac{1}{2}$ then $\sin h\,2x - \sec h\,2x =$ (A) $\dfrac{29}{15}$ (B) $\dfrac{11}{15}$ (C) $3$ (D) $\dfrac{-13}{15}$

   ›Reveal solutionSolution

   Use the given $\tanh x = \frac12$ to find $\sinh 2x$ and $\sech 2x$ via hyperbolic identities, then subtract to get $\frac{29}{15}$.

   The core idea here is that hyperbolic functions obey identities very similar to trigonometric ones, but with sign differences. Given $\tanh x$, you can find $\sinh x$ and $\cosh x$ using the fundamental relation $\cosh^2 x - \sinh^2 x = 1$, then compute double-angle forms. The trick is to avoid solving for $x$ directly — work algebraically with the ratios.

   1. Find $\cosh x$ and $\sinh x$ from $\tanh x$. Let $\tanh x = \frac{\sinh x}{\cosh x} = \frac12$. So $\sinh x = \frac12 \cosh x$. Use $\cosh^2 x - \sinh^2 x = 1$:

   $\cosh^2 x - \left(\frac12 \cosh x\right)^2 = 1 \quad\Rightarrow\quad \cosh^2 x - \frac14 \cosh^2 x = \frac34 \cosh^2 x = 1.$

   Hence $\cosh^2 x = \frac43$, so $\cosh x = \frac{2}{\sqrt{3}}$ (positive, since $\cosh x > 0$ for all real $x$).

   Then $\sinh x = \frac12 \cdot \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.

   2. Compute $\sinh 2x$. Using $\sinh 2x = 2 \sinh x \cosh x$:

   $\sinh 2x = 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}} = \frac{4}{3}.$

   3. Compute $\cosh 2x$ and then $\sech 2x$. Use $\cosh 2x = \cosh^2 x + \sinh^2 x$ (note: plus sign, unlike $\cos 2x$): $$$$

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7. 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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8. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.

   [!FORMULA] $\lim_{x \to 0^-} \frac{\sqrt{\frac{1}{2}(1 - \cos^2 x)}}{x} =$

   (A) $\frac{1}{\sqrt{2}}$ (B) $\frac{-1}{\sqrt{2}}$ (C) $-1$ (D) does not exist

   ›Reveal solutionSolution

   The key is to simplify the numerator using $\sqrt{\sin^2 x} = |\sin x|$, then consider the left-hand limit where $\sin x$ is negative. The limit equals $-\frac{1}{\sqrt{2}}$.

   The expression involves a square root of a trigonometric square, which always gives a non-negative result. That absolute-value behaviour is the entire crux — the sign of $\sin x$ near $x = 0^-$ decides the answer.

   1. Simplify the numerator. Recall $\cos^2 x = 1 - \sin^2 x$, so

   $\frac{1}{2}(1 - \cos^2 x) = \frac{1}{2}\bigl(1 - (1 - \sin^2 x)\bigr) = \frac{1}{2}\sin^2 x.$

   Therefore

   $\sqrt{\frac{1}{2}(1 - \cos^2 x)} = \sqrt{\frac{1}{2}\sin^2 x} = \frac{1}{\sqrt{2}} \, |\sin x|.$

   2. Rewrite the limit. The given limit becomes

   $\lim_{x \to 0^-} \frac{\frac{1}{\sqrt{2}} |\sin x|}{x} = \frac{1}{\sqrt{2}} \lim_{x \to 0^-} \frac{|\sin x|}{x}.$

   3. Handle the sign for $x \to 0^-$. For $x$ just less than $0$, $\sin x$ is negative (since $\sin$ is odd and positive for small positive $x$). Hence $|\sin x| = -\sin x$. So

   $\frac{|\sin x|}{x} = \frac{-\sin x}{x} = -\frac{\sin x}{x}.$

   4. Use the standard limit. …

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9. 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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10. 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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11. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.If Rolle's Theorem is applicable for the function $f(x) = \begin{cases} x^p \log x, & x \neq 0 \\ 0, & x = 0 \end{cases}$ on the interval $[0, 1]$, then a possible value of $p$ is (A) $-2$ (B) $-1$ (C) $0$ (D) $1$

    ›Reveal solutionSolution

    Rolle’s Theorem requires continuity on $[0,1]$ and differentiability on $(0,1)$ with $f(0)=f(1)$. For $f(x)=x^p\log x$ (with $f(0)=0$), continuity at $0$ forces $p>0$, so only $p=1$ works among the options.

    ---

    The key idea is that Rolle’s Theorem is a three‑part contract: the function must be continuous on the closed interval, differentiable on the open interval, and take equal values at the endpoints. Here the only tricky part is the behavior at $x=0$, because $\log x$ blows up as $x\to 0^+$. The exponent $p$ must be strong enough to tame that blow‑up.

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    1. Check endpoint equality

       We have $f(1) = 1^p \log 1 = 0$, and $f(0)=0$ by definition. So $f(0)=f(1)$ holds for any $p$ — no restriction yet.

    2. Continuity on $[0,1]$

       The only potential trouble is at $x=0$. For $x>0$, $f(x)=x^p\log x$. We need

    $\lim_{x\to 0^+} x^p \log x = f(0)=0.$

    Recall that $\log x \to -\infty$ as $x\to 0^+$, so $x^p$ must go to $0$ fast enough to overpower it.

    • If $p>0$, then $x^p \to 0$ and $x^p \log x \to 0$ (a standard limit: $x^\epsilon \log x \to 0$ for any $\epsilon>0$).
    • If $p=0$, then $f(x)=\log x \to -\infty$, so the limit is not $0$.
    • If $p<0$, then $x^p \to +\infty$ and the product $x^p \log x \to -\infty$ (or $+\infty$ depending on sign), so the limit is not $0$. Hence continuity at $0$ forces $p>0$.

    3. Differentiability on $(0,1)$ …

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12. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.Approximate volume of a cone whose semi vertical angle is $\tan^{-1}(3)$ and base radius is $15.001$ is (A) $(375.025)\pi$ (B) $(325.025)\pi$ (C) $(375.075)\pi$ (D) $(325.075)\pi$

    ›Reveal solutionSolution

    With $\tan\alpha=r/h=3$, $V=\dfrac{\pi r^3}{9}$; using $dV=\dfrac{\pi r^2}{3}\,dr$ at $r=15,\ dr=0.001$ gives $V\approx(375.075)\pi$.

    Semi-vertical angle $\alpha=\tan^{-1}(3)$, so $\tan\alpha=\dfrac{r}{h}=3\Rightarrow h=\dfrac{r}{3}$.

    $V=\frac13\pi r^2 h=\frac13\pi r^2\cdot\frac{r}{3}=\frac{\pi r^3}{9}.$

    At $r=15$: $V=\dfrac{\pi(3375)}{9}=375\pi$.

    Differential for a small change $dr=0.001$ about $r=15$: …

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