Q.Integrate the function $f'(ax+b)\,[f(ax+b)]^n$

Showing the 12 most recent of 26 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $f(x)$ is a differentiable function and $y = e^{f(x)} + e^{f(x)} + e^{f(x)} + \ldots \infty$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{y f'(x)}{1 + y}$ (B) $\dfrac{(1 + y) f'(x)}{y}$ (C) $\dfrac{(1 - y) f'(x)}{y}$ (D) $\dfrac{y f'(x)}{1 - y}$

   ›Reveal solutionSolution

   The expression is an infinite nested exponent, so it satisfies $y = e^{\,f(x)+y}$. Implicit differentiation gives $\dfrac{dy}{dx} = \dfrac{y\,f'(x)}{1-y}$.

   Setting up the self-similar equation. The right-hand side is an infinitely nested tower $y = e^{\,f(x)+e^{\,f(x)+\cdots}}$. Because the exponent contains an exact copy of the whole expression, the tower folds into itself:

   $y = e^{\,f(x) + y}$

   Take logarithms:

   $\log y = f(x) + y$

   Differentiate both sides with respect to $x$:

   $\frac{1}{y}\frac{dy}{dx} = f'(x) + \frac{dy}{dx}$

   Collect the derivative terms:

   $\frac{1}{y}\frac{dy}{dx} - \frac{dy}{dx} = f'(x) \;\Longrightarrow\; \frac{dy}{dx}\left(\frac{1-y}{y}\right) = f'(x)$

   Solve for the derivative: …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $f:\mathbb{R}-\{0\} \to \mathbb{R}$ is a differentiable function such that $\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right) = x - \frac{10}{3}$, then $f'(3) - f'\left(\frac{1}{3}\right) =$ (A) $\frac{12}{5}$ (B) $\frac{80}{9}$ (C) $3$ (D) $5$

   ›Reveal solutionSolution

   We differentiate the given functional equation with respect to $x$ and then substitute $x=3$ to directly find the required expression. The value of $f'(3) - f'\left(\frac{1}{3}\right)$ is $\boxed{3}$.

   The problem presents a functional equation involving $f(x)$ and $f(1/x)$, and asks for an expression involving their derivatives, $f'(3)$ and $f'(1/3)$. The most direct approach to solve such problems is to differentiate the given functional equation.

   Here's why this approach works:

   When you have an equation relating $f(x)$ and $f(1/x)$, differentiating it will introduce $f'(x)$ and $f'(1/x)$. The chain rule will be crucial for the term $f(1/x)$. After differentiation, we will have a new equation involving derivatives. By carefully choosing a value for $x$ (in this case, $x=3$), we can make the arguments of the derivatives match the terms we need to find.

   Let's work through the steps.

   1. Write down the given functional equation: We are given the equation:

   $\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right) = x - \frac{10}{3}$

   This equation holds for all $x \in \mathbb{R}-\{0\}$.
   

   2. Differentiate both sides with respect to $x$:

   Since the function $f(x)$ is differentiable, we can differentiate both sides of the equation with respect to $x$.

   Recall the chain rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$.

   Here, for the term $f(1/x)$, $g(x) = 1/x$, so $g'(x) = -1/x^2$.

   Differentiating the left side:
   

   $\frac{d}{dx}\left[\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right)\right] = \frac{1}{3}f'(x) + 3f'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$

   $= \frac{1}{3}f'(x) - \frac{3}{x^2}f'\left(\frac{1}{x}\right)$

   Differentiating the right side:
   

   $\frac{d}{dx}\left[x - \frac{10}{3}\right] = 1 - 0 = 1$

   Equating the derivatives of both sides, we get:
   

   $\frac{1}{3}f'(x) - \frac{3}{x^2}f'\left(\frac{1}{x}\right) = 1$

   This is a new functional equation involving the derivatives. …
   

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3. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $f(x) = \sum_{p=1}^{7} p^2 \sin^{-1} \left( \frac{4}{5} \sin(px) - \frac{3}{5} \cos(px) \right)$ then the value of $\frac{df}{dx}$ at $x=1$ is (Given that $\sin^{-1}(\sin x) = x$) (A) 0 (B) 628 (C) 1140 (D) 784

   ›Reveal solutionSolution

   The core idea is to simplify the argument of the inverse sine function using a trigonometric identity, which then allows us to use the given property $\sin^{-1}(\sin x) = x$. After simplification, the function $f(x)$ becomes a sum of linear terms, making its derivative straightforward to calculate. The final value of $\frac{df}{dx}$ at $x=1$ is $\boxed{784}$.

   The problem asks for the derivative of a function $f(x)$ at a specific point. The function $f(x)$ involves a sum and an inverse trigonometric function whose argument is a linear combination of $\sin(px)$ and $\cos(px)$. The key to solving this problem lies in simplifying the argument of the $\sin^{-1}$ function.

   Concept and Intuition

   1. Trigonometric Transformation: An expression of the form $a \sin \theta + b \cos \theta$ can always be rewritten as a single sine or cosine function. Specifically, we can write $a \sin \theta + b \cos \theta = R \sin(\theta + \alpha)$, where $R = \sqrt{a^2+b^2}$, $\cos \alpha = \frac{a}{R}$, and $\sin \alpha = \frac{b}{R}$. This transformation is crucial because it allows us to simplify the argument of $\sin^{-1}$.
   2. Inverse Sine Property: The problem explicitly states that $\sin^{-1}(\sin x) = x$. This is a very important piece of information. Normally, $\sin^{-1}(\sin x)$ equals $x$ only for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. However, by providing this identity, the problem simplifies the situation, allowing us to directly replace $\sin^{-1}(\sin(\text{expression}))$ with the $\text{expression}$ itself, regardless of its range. This avoids complex principal value considerations.
   3. Differentiation of a Sum: The function $f(x)$ is a sum of terms. The derivative of a sum is the sum of the derivatives, which simplifies the differentiation process.

   Let's apply these concepts step-by-step.

   1. Simplify the argument of $\sin^{-1}$: The argument of the inverse sine function is $\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px)$. This is in the form $a \sin \theta + b \cos \theta$, where $a = \frac{4}{5}$, $b = -\frac{3}{5}$, and $\theta = px$. First, calculate $R = \sqrt{a^2+b^2}$:

   $R = \sqrt{\left(\frac{4}{5}\right)^2 + \left(-\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25} + \frac{9}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1$

   Now, we want to express the argument as $R \sin(\theta - \alpha)$.
   We need $\cos \alpha = \frac{a}{R} = \frac{4/5}{1} = \frac{4}{5}$ and $\sin \alpha = \frac{b}{R} = \frac{-3/5}{1} = -\frac{3}{5}$.
   Let $\alpha_0$ be an angle such that $\cos \alpha_0 = \frac{4}{5}$ and $\sin \alpha_0 = \frac{3}{5}$. (This $\alpha_0$ is a constant acute angle, specifically $\alpha_0 = \tan^{-1}(\frac{3}{4})$).
   Then, the expression becomes:
   

   $1 \cdot \left( \cos \alpha_0 \sin(px) - \sin \alpha_0 \cos(px) \right)$

   Using the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, with $A=px$ and $B=\alpha_0$:
   

   $\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px) = \sin(px - \alpha_0)$

   So, the argument simplifies to $\sin(px - \alpha_0)$.
   

   2. Substitute the simplified argument back into $f(x)$:

   Now, $f(x)$ can be written as:

   $f(x) = \sum_{p=1}^{7} p^2 \sin^{-1} \left( \sin(px - \alpha_0) \right)$

   3. Apply the given identity $\sin^{-1}(\sin x) = x$: …

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4. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $f(x) = (1 + x^3)(1 + x^6)(1 + x^{12})(1 + x^{24})$, then $f'(-1) =$ (A) 24 (B) 12 (C) 48 (D) 60

   ›Reveal solutionSolution

   At $x=-1$ the factor $(1+x^3)$ vanishes, so only the term where it is differentiated survives: $f'(-1)=24$ — option (A).

   For a product $f=f_1 f_2 f_3 f_4$, the derivative is $f'=f_1'f_2f_3f_4+f_1f_2'f_3f_4+\cdots$. Every term keeps three of the original factors undifferentiated.

   1. Note the vanishing factor. At $x=-1$, $\;1+x^3=1+(-1)^3=0$. Any product-rule term that still contains the factor $(1+x^3)$ is therefore $0$. Only the single term in which $(1+x^3)$ is the one being differentiated can be non-zero.

   2. Keep the surviving term. …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.$f(x)$ is a real valued bijective function and twice differentiable function. If $g(x)$ is inverse of $f(x)$ and $f(0) = \alpha$, then $g''(\alpha) =$ (A) $\dfrac{f''(0)}{[f'(0)]^3}$ (B) $\dfrac{f''(\alpha)}{[f'(\alpha)]^3}$ (C) $\dfrac{f''(0)}{[f'(\alpha)]^2}$ (D) $\dfrac{f''(\alpha)}{[f'(0)]^2}$

   ›Reveal solutionSolution

   The second derivative of the inverse function is found by differentiating the relation $g'(f(x)) = 1/f'(x)$ using the chain rule, yielding $g''(\alpha) = -f''(0)/[f'(0)]^3$, which matches option (A).

   We are given that $f$ is bijective (so invertible), twice differentiable, and $g$ is its inverse: $g(f(x)) = x$ and $f(g(y)) = y$. The problem asks for $g''(\alpha)$ where $\alpha = f(0)$. That means we evaluate the second derivative of the inverse at the point where the original function’s value is $\alpha$ — which corresponds to $x = 0$ in the original function.

   Concept & Intuition

   The key idea: derivatives of inverse functions are linked by the reciprocal relation for the first derivative, but for the second derivative we must differentiate that relation carefully using the chain rule. The result expresses the curvature of the inverse in terms of the curvature of the original function at the corresponding point. A common mistake is to forget that when differentiating $g'(f(x)) = 1/f'(x)$, the argument of $g'$ is $f(x)$, so the chain rule brings in $f'(x)$ again.

   Let’s work it out step by step.

   1. Start with the fundamental inverse relation Since $g$ is the inverse of $f$, we have for all $x$ in the domain:

   $g(f(x)) = x.$

   Differentiate both sides with respect to $x$. The left side uses the chain rule:

   $g'(f(x)) \cdot f'(x) = 1.$

   Hence,

   $g'(f(x)) = \frac{1}{f'(x)}. \qquad(1)$

   This is the well-known formula for the derivative of an inverse.

   2. Differentiate again to get the second derivative Differentiate both sides of (1) with respect to $x$. The left side is a composition: $g'(f(x))$. Its derivative is

   $\frac{d}{dx} \bigl[ g'(f(x)) \bigr] = g''(f(x)) \cdot f'(x).$

   The right side is $1/f'(x)$, whose derivative is

   $\frac{d}{dx} \left( \frac{1}{f'(x)} \right) = -\frac{f''(x)}{[f'(x)]^2}.$

   Equating:

   $g''(f(x)) \cdot f'(x) = -\frac{f''(x)}{[f'(x)]^2}. \qquad(2)$

   3. Solve for $g''(f(x))$ Divide both sides of (2) by $f'(x)$ (which is nonzero because $f$ is bijective and differentiable, so $f'$ cannot change sign and is never zero):

   $g''(f(x)) = -\frac{f''(x)}{[f'(x)]^3}. \qquad(3)$

   4. Evaluate at the specific point We need $g''(\alpha)$, and we know $\alpha = f(0)$. So set $x = 0$ in (3):

   $g''(f(0)) = g''(\alpha) = -\frac{f''(0)}{[f'(0)]^3}.$

   The negative sign is important — it tells us that the curvature of the inverse has the opposite sign to the curvature of the original function at corresponding points.

   5. Match with the options The expression we obtained is $-\dfrac{f''(0)}{[f'(0)]^3}$. Looking at the choices: …

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6. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $a$ and $b$ are non-negative real numbers and $\lim_{x \to 0} \frac{e^{ax} - \cos bx}{1 - \cos x} = 4$, then $\lim_{x \to a} \frac{\sin(bx - ab)}{(x - a)} =$ (A) $1$ (B) $2$ (C) $4$ (D) $8$

   ›Reveal solutionSolution

   The first limit forces $a=0,\ b=2$; then $\displaystyle\lim_{x\to a}\frac{\sin(bx-ab)}{x-a}=\lim_{x\to0}\frac{\sin 2x}{x}=2$.

   Expand the first limit near $x=0$:

   $e^{ax}-\cos bx=\Big(1+ax+\tfrac{a^2x^2}{2}+\cdots\Big)-\Big(1-\tfrac{b^2x^2}{2}+\cdots\Big)=ax+\tfrac{a^2+b^2}{2}x^2+\cdots,$

   $1-\cos x=\tfrac{x^2}{2}+\cdots.$

   For the ratio to be finite the linear term $ax$ must vanish, so $a=0$. Then …

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7. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $y = e^{2x} + \sin x$, then $2y'' - 5y' + 2y =$ (A) $4 \sin x$ (B) $-5 \cos x$ (C) $-4 \sin x$ (D) $5 \cos x$

   ›Reveal solutionSolution

   The key idea is to compute the first and second derivatives of $y = e^{2x} + \sin x$, substitute them into $2y'' - 5y' + 2y$, and simplify. The result is $-5 \cos x$, which corresponds to option (B).

   We start with the function

   $y = e^{2x} + \sin x.$

   The expression we need is $2y'' - 5y' + 2y$. Instead of solving a differential equation, we simply differentiate and substitute — this is a direct computation.

   1. First derivative Differentiate term by term:

   $y' = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(\sin x) = 2e^{2x} + \cos x.$

   2. Second derivative Differentiate $y'$:

   $y'' = \frac{d}{dx}(2e^{2x}) + \frac{d}{dx}(\cos x) = 4e^{2x} - \sin x.$

   3. Substitute into $2y'' - 5y' + 2y$

   $2y'' = 2(4e^{2x} - \sin x) = 8e^{2x} - 2\sin x,$

   $-5y' = -5(2e^{2x} + \cos x) = -10e^{2x} - 5\cos x,$

   $2y = 2(e^{2x} + \sin x) = 2e^{2x} + 2\sin x.$

   4. Add them together Combine the $e^{2x}$ terms: $8e^{2x} - 10e^{2x} + 2e^{2x} = 0$. Combine the $\sin x$ terms: $-2\sin x + 2\sin x = 0$. The only remaining term is $-5\cos x$. …

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8. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If $f(x) = \dfrac{\cos^2 x}{1 + \sin^2 x}$, then $f\left(\dfrac{\pi}{4}\right) - 3 f'\left(\dfrac{\pi}{4}\right) =$ (A) $\dfrac{5}{3}$ (B) $\dfrac{11}{3}$ (C) $\dfrac{13}{9}$ (D) $3$

   ›Reveal solutionSolution

   $f\!\left(\tfrac\pi4\right)=\tfrac13$ and $f'\!\left(\tfrac\pi4\right)=-\tfrac89$, so $f\!\left(\tfrac\pi4\right)-3f'\!\left(\tfrac\pi4\right)=\tfrac13+\tfrac{24}{9}=3$ — option (D).

   Evaluate $f(\pi/4)$. With $f(x)=\dfrac{\cos^2x}{1+\sin^2x}$ and $\cos^2\tfrac\pi4=\sin^2\tfrac\pi4=\tfrac12$:

   $f\!\left(\tfrac\pi4\right)=\frac{1/2}{1+1/2}=\frac{1/2}{3/2}=\frac13.$

   Differentiate. With $u=\cos^2x,\;v=1+\sin^2x$ (so $u'=-\sin2x,\;v'=\sin2x$):

   $f'(x)=\frac{u'v-uv'}{v^2}=\frac{-\sin2x\,(1+\sin^2x)-\cos^2x\,\sin2x}{(1+\sin^2x)^2}=\frac{-\sin2x\,(2)}{(1+\sin^2x)^2}=\frac{-2\sin2x}{(1+\sin^2x)^2}.$

   At $x=\tfrac\pi4$: $\sin2x=1$ and $1+\sin^2\tfrac\pi4=\tfrac32$, so …

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9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $\sinh x = \dfrac{12}{5}$, then $\sinh 3x + \cosh 3x =$ (A) $125$ (B) $144$ (C) $169$ (D) $216$

   ›Reveal solutionSolution

   The key is to use the identity $\sinh 3x + \cosh 3x = e^{3x}$, then find $e^x$ from $\sinh x = \frac{12}{5}$ using $\cosh x = \sqrt{1+\sinh^2 x}$ and $e^x = \sinh x + \cosh x$. The result is $125$, so the correct option is (A).

   ---

   The problem asks for $\sinh 3x + \cosh 3x$ given $\sinh x = \frac{12}{5}$. The direct approach would be to compute $\sinh 3x$ and $\cosh 3x$ using triple-angle formulas, but that’s messy. Instead, recall the elegant identity:

   For any real $x$, $\sinh x + \cosh x = e^x$.

   Similarly, $\sinh 3x + \cosh 3x = e^{3x}$.

   So the problem reduces to finding $e^{3x}$ from $\sinh x = \frac{12}{5}$. That’s much simpler.

   ---

   Step-by-step reasoning:

   1. Find $\cosh x$ from $\sinh x$. The fundamental identity for hyperbolic functions is:

   $\cosh^2 x - \sinh^2 x = 1$

   Given $\sinh x = \frac{12}{5}$, we have:

   $\cosh^2 x = 1 + \left(\frac{12}{5}\right)^2 = 1 + \frac{144}{25} = \frac{169}{25}$

   Since $\cosh x \ge 1$ for all real $x$, we take the positive root:

   $\cosh x = \sqrt{\frac{169}{25}} = \frac{13}{5}$

   2. Find $e^x$ using the sum identity. As noted:

   $e^x = \sinh x + \cosh x = \frac{12}{5} + \frac{13}{5} = \frac{25}{5} = 5$

   So $e^x = 5$.

   3. Compute $e^{3x}$. $$$$

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10. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
    Q.If $f(x) = \dfrac{e^{-x} \sin x}{\log_e x}$ and $f'(x) = f(x) \cdot g(x)$, then $g'(e) =$ (A) $e^{-2} - \csc^2(e)$ (B) $2e^{2} - \csc^2(e)$ (C) $2e^{-2} - \csc^2(e)$ (D) $2e^{-2} + \csc^2(e)$

    ›Reveal solutionSolution

    We use logarithmic differentiation to simplify $f(x)$ into a sum of terms, which directly gives us $g(x) = \frac{f'(x)}{f(x)}$. Differentiating $g(x)$ and substituting $x=e$ then yields the result. The value of $g'(e)$ is $\boxed{2e^{-2} - \csc^2(e)}$.

    The problem asks us to find $g'(e)$ given a function $f(x)$ and the relationship $f'(x) = f(x) \cdot g(x)$. The function $f(x)$ is a product and quotient of several functions, making direct differentiation quite cumbersome.

    The key insight here is to recognize that the expression $g(x) = \frac{f'(x)}{f(x)}$ is precisely the derivative of $\log_e |f(x)|$. This means we can use logarithmic differentiation to find $g(x)$ efficiently. By taking the natural logarithm of $f(x)$ first, we convert products and quotients into sums and differences, which are much simpler to differentiate.

    Here's how we approach the problem:

    1. Express $g(x)$ using logarithmic differentiation:

       Given $f'(x) = f(x) \cdot g(x)$, we can write $g(x) = \frac{f'(x)}{f(x)}$.

       This expression is the result of differentiating $\log_e f(x)$ with respect to $x$.

       So, our first step is to take the natural logarithm of $f(x)$ and then differentiate it.

       We have $f(x) = \dfrac{e^{-x} \sin x}{\log_e x}$.

       Taking the natural logarithm on both sides:

    $\log_e f(x) = \log_e \left( \dfrac{e^{-x} \sin x}{\log_e x} \right)$

    Using the properties of logarithms ($\log(AB/C) = \log A + \log B - \log C$):
    

    $\log_e f(x) = \log_e (e^{-x}) + \log_e (\sin x) - \log_e (\log_e x)$

    Simplify the first term: $\log_e (e^{-x}) = -x$.
    

    $\log_e f(x) = -x + \log_e (\sin x) - \log_e (\log_e x)$

    2. Differentiate to find $g(x)$: Now, differentiate both sides of the equation with respect to $x$:

    $\frac{d}{dx} (\log_e f(x)) = \frac{d}{dx} (-x) + \frac{d}{dx} (\log_e (\sin x)) - \frac{d}{dx} (\log_e (\log_e x))$

    We know that $\frac{d}{dx} (\log_e f(x)) = \frac{f'(x)}{f(x)}$, which is $g(x)$.
    Differentiating each term on the right side:
    *   $\frac{d}{dx} (-x) = -1$
    *   $\frac{d}{dx} (\log_e (\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$
    *   $\frac{d}{dx} (\log_e (\log_e x)) = \frac{1}{\log_e x} \cdot \frac{1}{x}$ (using the chain rule)
    
    Combining these, we get $g(x)$:
    

    $g(x) = -1 + \cot x - \frac{1}{x \log_e x}$

    3. Differentiate $g(x)$ to find $g'(x)$: Now we need to find the derivative of $g(x)$:

    $g'(x) = \frac{d}{dx} (-1) + \frac{d}{dx} (\cot x) - \frac{d}{dx} \left( \frac{1}{x \log_e x} \right)$

    *   $\frac{d}{dx} (-1) = 0$
    *   $\frac{d}{dx} (\cot x) = -\csc^2 x$ …
    

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11. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
    Q.If c is the value lying in the interval $(1,3)$ such that Lagrange's mean value theorem holds for $f(x) = x^3 - 2x^2 + x - 1$ on $[1,3]$, then $9c^2 - 12c =$ (A) $15$ (B) $18$ (C) $24$ (D) $27$

    ›Reveal solutionSolution

    Lagrange’s Mean Value Theorem guarantees a point $c$ in $(1,3)$ where the derivative equals the average rate of change. Solving $f'(c) = \frac{f(3)-f(1)}{3-1}$ gives $3c^2 - 4c + 1 = 6$, so $9c^2 - 12c = 15$. The answer is (A).

    ---

    Concept & Intuition

    Lagrange’s Mean Value Theorem says: if a function is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is some $c$ in $(a,b)$ where the instantaneous slope (the derivative) equals the average slope over the whole interval.

    Here we are given $f(x) = x^3 - 2x^2 + x - 1$ on $[1,3]$. Instead of solving for $c$ directly, we can find the combination $9c^2 - 12c$ by manipulating the equation that $c$ satisfies.

    ---

    Step-by-step solution

    1. Compute the average rate of change

    $f(1) = 1^3 - 2\cdot1^2 + 1 - 1 = 1 - 2 + 1 - 1 = -1$

    $f(3) = 27 - 2\cdot9 + 3 - 1 = 27 - 18 + 3 - 1 = 11$

    The average slope is

    $\frac{f(3)-f(1)}{3-1} = \frac{11 - (-1)}{2} = \frac{12}{2} = 6.$

    2. Find the derivative

    $f'(x) = 3x^2 - 4x + 1.$

    3. Apply Lagrange’s theorem There exists $c \in (1,3)$ such that

    $f'(c) = 6 \quad\Longrightarrow\quad 3c^2 - 4c + 1 = 6.$

    4. Simplify the equation

    $3c^2 - 4c + 1 - 6 = 0 \quad\Longrightarrow\quad 3c^2 - 4c - 5 = 0.$

    5. Find the required expression We need $9c^2 - 12c$. Notice that $$$$

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12. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.If $y = \sqrt{\sin(\log 2x)} + \sqrt{\sin(\log 2x)} + \sqrt{\sin(\log 2x)} + \ldots \infty$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{\cos(\log 2x)}{2x(2y-1)}$ (B) $\dfrac{\cos(\log 2x)}{(2y-1)}$ (C) $\dfrac{\cos(\log 2x)}{x(2y-1)}$ (D) $\dfrac{\sin(\log 2x)}{x(2y-1)}$

    ›Reveal solutionSolution

    The infinite sum collapses to a simple equation $y = \sqrt{\sin(\log 2x)} + y$, which forces us to reinterpret the expression as a self-repeating pattern. The correct interpretation is $y = \sqrt{\sin(\log 2x) + \sqrt{\sin(\log 2x) + \dots}}$, leading to $y^2 = \sin(\log 2x) + y$, and differentiating gives $\frac{dy}{dx} = \frac{\cos(\log 2x)}{x(2y-1)}$, so the answer is (C).

    ---

    The key here is to first understand what the infinite expression actually means. At first glance, it looks like a sum of identical terms: $\sqrt{\sin(\log 2x)} + \sqrt{\sin(\log 2x)} + \dots$ to infinity. But that sum would diverge (unless the term is zero), so it cannot be that. Instead, the notation is a classic trick: it means an infinite nested radical, where each radical contains the entire rest of the expression. That is:

    $y = \sqrt{\sin(\log 2x) + \sqrt{\sin(\log 2x) + \sqrt{\sin(\log 2x) + \dots}}}$

    This is a self-similar structure: the whole expression appears again inside itself. That self-reference lets us write a simple algebraic equation for $y$.

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    1. Write the self-referential equation Since the expression inside the first square root is exactly the same as the whole $y$, we have:

    $y = \sqrt{\sin(\log 2x) + y}$

    This is the crucial step — it turns an infinite process into a finite equation.

    2. Square both sides

    $y^2 = \sin(\log 2x) + y$

    Rearranging:

    $y^2 - y = \sin(\log 2x)$

    3. Differentiate implicitly with respect to $x$ Differentiate both sides:

    $2y \frac{dy}{dx} - \frac{dy}{dx} = \cos(\log 2x) \cdot \frac{1}{2x} \cdot 2$

    The derivative of $\sin(\log 2x)$ uses the chain rule: derivative of $\sin$ is $\cos$, derivative of $\log 2x$ is $\frac{1}{2x} \cdot 2 = \frac{1}{x}$. So:

    $$$$

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