Q.Find the derivative of the function given by $f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)$ and hence find $f'(1)$.

Showing the 12 most recent of 26 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $y = \tan^{-1} \left[ \left( \frac{1 - \cos 2\sqrt{x}}{1 + \cos 2\sqrt{x}} \right)^{1/2} \right],\ 0 < x < \frac{\pi^2}{4}$, then $y(2y' + y) =$ (A) $1$ (B) $x + 1$ (C) $\sqrt{x}$ (D) $\sqrt{x} + 1$

   ›Reveal solutionSolution

   Simplify the argument with half-angle identities to get $y=\sqrt{x}$, so $y'=\dfrac{1}{2\sqrt{x}}$ and $y(2y'+y)=x+1$ — option (B).

   Simplify the inside first. Using $1-\cos 2\theta = 2\sin^2\theta$ and $1+\cos 2\theta = 2\cos^2\theta$ with $\theta=\sqrt{x}$:

   $\frac{1-\cos 2\sqrt{x}}{1+\cos 2\sqrt{x}} = \frac{2\sin^2\sqrt{x}}{2\cos^2\sqrt{x}} = \tan^2\sqrt{x}.$

   Taking the square root gives $\left(\tan^2\sqrt{x}\right)^{1/2}=|\tan\sqrt{x}|$. For $0<x<\tfrac{\pi^2}{4}$ we have $0<\sqrt{x}<\tfrac{\pi}{2}$, so $\tan\sqrt{x}>0$ and

   $y=\tan^{-1}\!\left(\tan\sqrt{x}\right)=\sqrt{x},$

   since $\sqrt{x}$ lies in the principal range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ of $\tan^{-1}$. …

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8. 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 $y$.

   ---

   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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9. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $x^2 + y^2 = t - \dfrac{1}{t},\ x^4 + y^4 = t^2 + \dfrac{1}{t^2}$ then $\dfrac{dy}{dx} =$ (A) $\dfrac{x}{y}$ (B) $-\dfrac{x}{y}$ (C) $\dfrac{y}{x}$ (D) $-\dfrac{y}{x}$

   ›Reveal solutionSolution

   The key is to relate the given equations by squaring the first and comparing with the second, which reveals that $x^2 y^2 = -1$, leading to $\frac{dy}{dx} = -\frac{x}{y}$.

   We have two equations linking $x$, $y$, and a parameter $t$:

   $x^2 + y^2 = t - \frac{1}{t}, \quad x^4 + y^4 = t^2 + \frac{1}{t^2}.$

   The goal is to find $\frac{dy}{dx}$ without explicitly solving for $t$. The trick is to notice that squaring the first equation will produce $x^4 + y^4 + 2x^2 y^2$, which we can compare with the second equation to eliminate $t$.

   1. Square the first equation:

   $(x^2 + y^2)^2 = \left(t - \frac{1}{t}\right)^2.$

   Expanding both sides:

   $x^4 + y^4 + 2x^2 y^2 = t^2 + \frac{1}{t^2} - 2.$

   2. Substitute the second equation $x^4 + y^4 = t^2 + \frac{1}{t^2}$ into the left side:

   $\left(t^2 + \frac{1}{t^2}\right) + 2x^2 y^2 = t^2 + \frac{1}{t^2} - 2.$

   3. Cancel $t^2 + \frac{1}{t^2}$ from both sides, leaving:

   $2x^2 y^2 = -2 \quad \Rightarrow \quad x^2 y^2 = -1.$

   Watch out

   $x^2 y^2 = -1$ means $x$ and $y$ cannot both be real numbers — but the problem is algebraic, so we proceed with the relation as given. In implicit differentiation, this relation is enough.

   4. Differentiate $x^2 y^2 = -1$ implicitly with respect to $x$. Using the product rule: …

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10. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.If $x = \sqrt{1 - \tan y}$, then $\dfrac{dy}{dx} =$ (A) $-\dfrac{2x}{x^4 + 2x^2 + 2}$ (B) $-\dfrac{2x}{x^4 - 2x^2 + 2}$ (C) $\dfrac{2x}{x^4 - 2x^2 + 2}$ (D) $\dfrac{2x}{x^4 + 2x^2 + 2}$

    ›Reveal solutionSolution

    Implicit differentiation gives $\dfrac{dy}{dx}=-\dfrac{2x}{x^{4}-2x^{2}+2}$. Option (B).

    Solution

    From $x=\sqrt{1-\tan y}$, square both sides:

    $x^{2}=1-\tan y\ \Longrightarrow\ \tan y=1-x^{2}.$

    Differentiate $x^{2}=1-\tan y$ with respect to $x$:

    $2x=-\sec^{2}y\,\frac{dy}{dx}\ \Longrightarrow\ \frac{dy}{dx}=-\frac{2x}{\sec^{2}y}.$

    Express $\sec^{2}y$ in terms of $x$ using $\sec^{2}y=1+\tan^{2}y$: …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.If $x = \sqrt{1 - \tan y}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{2x}{x^4 + 2x^2 + 2}$ (B) $-\dfrac{2x}{x^4 - 2x^2 + 2}$ (C) $\dfrac{2x}{x^4 - 2x^2 + 2}$ (D) $-\dfrac{2x}{x^4 + 2x^2 + 2}$

    ›Reveal solutionSolution

    Square the given relation to get $\tan y=1-x^{2}$, i.e. $y=\tan^{-1}(1-x^{2})$, and differentiate: $\dfrac{dy}{dx}=-\dfrac{2x}{x^{4}-2x^{2}+2}$ — option (B).

    The concept first

    When $y$ is buried inside a trigonometric function and $x$ sits outside a radical, do not rush into implicit differentiation. It is far cleaner to tidy the relation algebraically first, so that $y$ is written explicitly in terms of $x$; then a single application of the chain rule finishes it. The tool you need is

    $\frac{d}{dx}\tan^{-1}u=\frac{1}{1+u^{2}}\cdot\frac{du}{dx}.$

    Step 1 — Make $y$ explicit

    $x=\sqrt{1-\tan y}\;\Longrightarrow\;x^{2}=1-\tan y\;\Longrightarrow\;\tan y=1-x^{2}\;\Longrightarrow\;y=\tan^{-1}\!\left(1-x^{2}\right).$

    Step 2 — Differentiate with the chain rule

    With $u=1-x^{2}$, $\dfrac{du}{dx}=-2x$:

    $\frac{dy}{dx}=\frac{1}{1+\left(1-x^{2}\right)^{2}}\times(-2x)=\frac{-2x}{1+\left(1-x^{2}\right)^{2}} .$

    Step 3 — Simplify the denominator

    $\left(1-x^{2}\right)^{2}=1-2x^{2}+x^{4},$

    $1+\left(1-x^{2}\right)^{2}=1+1-2x^{2}+x^{4}=x^{4}-2x^{2}+2 .$

    Therefore …

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12. 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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