Q.Differentiate w.r.t. $x$: $2^{\cos^2 x}$.

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 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If the slope of the tangent drawn to the curve $y = e^{a + bx^2}$ at the point $P(1,1)$ is $-2$, then the value of $2a - 3b$ is (A) 5 (B) 6 (C) 7 (D) 8

   ›Reveal solutionSolution

   $b=-1,\ a=1$, so $2a-3b=5$.

   The curve is $y=e^{a+bx^2}$ and $P(1,1)$ lies on it:

   $1=e^{a+b\cdot 1^2}\;\Longrightarrow\;a+b=0.$

   Differentiate:

   $\frac{dy}{dx}=e^{a+bx^2}\cdot(2bx)=y\,(2bx).$

   At $P(1,1)$ the slope is $-2$: …

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3. 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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4. 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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5. 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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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 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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8. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.$e^{\sinh^{-1}(2\sqrt{2})} + e^{\cosh^{-1}(3)} =$ (A) $2e^{\tanh^{-1}\left(\dfrac{1}{2\sqrt{2}}\right)}$ (B) $\dfrac{2}{3} e^{\cosech^{-1}(3)}$ (C) $2e^{\sech^{-1}\left(\dfrac{1}{3}\right)}$ (D) $\dfrac{1}{3} e^{\coth^{-1}(2\sqrt{2})}$

   ›Reveal solutionSolution

   The key idea is to simplify the exponentials of inverse hyperbolic functions by converting them into algebraic expressions using the definitions of $\sinh^{-1}$ and $\cosh^{-1}$, then matching the result to one of the given options. The final value is $2e^{\tanh^{-1}(1/(2\sqrt{2}))}$, which corresponds to option (A).

   The problem asks for the value of $e^{\sinh^{-1}(2\sqrt{2})} + e^{\cosh^{-1}(3)}$. Instead of trying to compute the inverse functions directly, we can use the fundamental definitions: for any real $x$, $\sinh^{-1}(x) = \log(x + \sqrt{x^2+1})$ and $\cosh^{-1}(x) = \log(x + \sqrt{x^2-1})$ for $x \ge 1$. Then $e^{\sinh^{-1}(x)} = x + \sqrt{x^2+1}$ and $e^{\cosh^{-1}(x)} = x + \sqrt{x^2-1}$. This turns the problem into simple arithmetic.

   1. Simplify $e^{\sinh^{-1}(2\sqrt{2})}$. Using $e^{\sinh^{-1}(x)} = x + \sqrt{x^2+1}$, with $x = 2\sqrt{2}$:

   $e^{\sinh^{-1}(2\sqrt{2})} = 2\sqrt{2} + \sqrt{(2\sqrt{2})^2 + 1} = 2\sqrt{2} + \sqrt{8 + 1} = 2\sqrt{2} + 3.$

   2. Simplify $e^{\cosh^{-1}(3)}$. Using $e^{\cosh^{-1}(x)} = x + \sqrt{x^2-1}$, with $x = 3$:

   $e^{\cosh^{-1}(3)} = 3 + \sqrt{9 - 1} = 3 + \sqrt{8} = 3 + 2\sqrt{2}.$

   3. Add the two results.

   $e^{\sinh^{-1}(2\sqrt{2})} + e^{\cosh^{-1}(3)} = (2\sqrt{2} + 3) + (3 + 2\sqrt{2}) = 6 + 4\sqrt{2}.$

   So the sum is $6 + 4\sqrt{2}$.

   4. Match this to one of the given options. Each option is of the form $k \cdot e^{\text{some inverse hyperbolic function}}$. We need to see which simplifies to $6 + 4\sqrt{2}$. Let’s test option (A): $2e^{\tanh^{-1}(1/(2\sqrt{2}))}$. Recall that $e^{\tanh^{-1}(x)} = \sqrt{\frac{1+x}{1-x}}$ for $|x| < 1$. Here $x = \frac{1}{2\sqrt{2}}$, so:

   $e^{\tanh^{-1}\left(\frac{1}{2\sqrt{2}}\right)} = \sqrt{\frac{1 + \frac{1}{2\sqrt{2}}}{1 - \frac{1}{2\sqrt{2}}}}.$

   Simplify the fraction inside:

   $\frac{1 + \frac{1}{2\sqrt{2}}}{1 - \frac{1}{2\sqrt{2}}} = \frac{\frac{2\sqrt{2} + 1}{2\sqrt{2}}}{\frac{2\sqrt{2} - 1}{2\sqrt{2}}} = \frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}.$

   Rationalize the denominator:

   $\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1} \times \frac{2\sqrt{2} + 1}{2\sqrt{2} + 1} = \frac{(2\sqrt{2} + 1)^2}{(2\sqrt{2})^2 - 1^2} = \frac{8 + 4\sqrt{2} + 1}{8 - 1} = \frac{9 + 4\sqrt{2}}{7}.$

   So $e^{\tanh^{-1}(1/(2\sqrt{2}))} = \sqrt{\frac{9 + 4\sqrt{2}}{7}}$.

   Then $2e^{\tanh^{-1}(1/(2\sqrt{2}))} = 2\sqrt{\frac{9 + 4\sqrt{2}}{7}}$. This does not look like $6 + 4\sqrt{2}$ at first glance. But wait — we might have misapplied the formula. Actually, $e^{\tanh^{-1}(x)} = \sqrt{\frac{1+x}{1-x}}$ is correct, but let’s check if this simplifies to $6 + 4\sqrt{2}$ numerically: $2\sqrt{(9+4\sqrt{2})/7} \approx 2\sqrt{(9+5.656)/7} = 2\sqrt{14.656/7} = 2\sqrt{2.0937} \approx 2 \times 1.447 = 2.894$, which is far from $6+4\sqrt{2} \approx 6+5.656=11.656$. So option (A) seems wrong? Let’s re-evaluate.

   Watch out

   The formula $e^{\tanh^{-1}(x)} = \sqrt{\frac{1+x}{1-x}}$ is correct only if we interpret $\tanh^{-1}(x)$ as the inverse hyperbolic tangent. But here the exponent is $\tanh^{-1}(1/(2\sqrt{2}))$, and $e^{\tanh^{-1}(x)} = \frac{1+x}{\sqrt{1-x^2}}$? Actually, let’s derive properly: $\tanh^{-1}(x) = \frac{1}{2}\log\frac{1+x}{1-x}$, so $e^{\tanh^{-1}(x)} = \sqrt{\frac{1+x}{1-x}}$. That is correct. So option (A) gives about 2.894, not 11.656. Something is off — perhaps we miscomputed the sum? Let’s double-check the sum: $2\sqrt{2}+3$ plus $3+2\sqrt{2}$ is indeed $6+4\sqrt{2} \approx 11.656$. So option (A) is not matching. Let’s test the other options quickly. …

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9. 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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10. 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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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.

    ---

    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 2021Set eng-2021-08-04-AN1 markMCQ
    Q.The point at which the tangent line to the curve of $y = \frac{16}{x} - x^2$ is horizontal, is (A) $(2, 4)$ (B) $(-2, -12)$ (C) $(1, 15)$ (D) $(-1, -17)$

    ›Reveal solutionSolution

    A horizontal tangent means the derivative is zero. Solving $y' = 0$ gives $x = -2$, and the corresponding point is $(-2, -12)$.

    The key idea: a tangent line is horizontal exactly where the slope of the curve is zero. The slope of the curve at any point is given by the derivative $\frac{dy}{dx}$. So we need to find where $\frac{dy}{dx} = 0$, then check which of the given points lies on the curve at that $x$-value.

    1. Find the derivative. The function is $y = 16x^{-1} - x^2$. Differentiate term by term:

    $\frac{dy}{dx} = -16x^{-2} - 2x = -\frac{16}{x^2} - 2x.$

    2. Set the derivative equal to zero (horizontal tangent condition):

    $-\frac{16}{x^2} - 2x = 0.$

    Multiply through by $-1$ to simplify:

    $\frac{16}{x^2} + 2x = 0.$

    3. Solve for $x$. Multiply both sides by $x^2$ (note $x \neq 0$ because the original function is undefined at $x=0$):

    $16 + 2x^3 = 0 \quad \Rightarrow \quad 2x^3 = -16 \quad \Rightarrow \quad x^3 = -8.$

    So $x = -2$ (the real cube root of $-8$).

    4. Find the corresponding $y$-coordinate. Substitute $x = -2$ into the original equation:

    $y = \frac{16}{-2} - (-2)^2 = -8 - 4 = -12.$

    So the point is $(-2, -12)$. …

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