Q.Differentiate with respect to $x$ using first principle: $x\cos x$.

Showing the 12 most recent of 25 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If the function $f(x) = \begin{cases} \dfrac{p(1+\sin 3x)}{(\pi+6x)^2}, & \text{if } -\dfrac{\pi}{2} < x < -\dfrac{\pi}{6} \\ \cos 2x, & \text{if } x = -\dfrac{\pi}{6} \\ \dfrac{q(\sin 12x + 2\sin 6x)}{\cos^3 \left( \dfrac{\pi + 12x}{2} \right)}, & \text{if } -\dfrac{\pi}{6} < x < 0 \end{cases}$ is continuous at $x = -\dfrac{\pi}{6}$, then $p + 2q =$ (A) 3 (B) 2 (C) 1 (D) 0

   ›Reveal solutionSolution

   For continuity at $x = -\frac{\pi}{6}$, the left-hand limit, right-hand limit, and function value must all be equal. Using standard trigonometric limits and expansions, we find $p = \frac{1}{2}$ and $q = \frac{1}{4}$, so $p + 2q = 1$.

   The core idea is that continuity at a point means the function’s value there equals both one-sided limits. Here, the function is defined piecewise around $x = -\frac{\pi}{6}$, so we compute the left-hand limit (from the first piece), the right-hand limit (from the third piece), and set them equal to the given function value at that point, which is $\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}$.

   The trick is in handling the limits — each involves a trigonometric expression that simplifies nicely when you substitute $x = -\frac{\pi}{6} + h$ and let $h \to 0$. This substitution shifts the point to zero, making standard limits like $\lim_{t\to 0} \frac{\sin t}{t} = 1$ directly applicable.

   1. Function value at $x = -\frac{\pi}{6}$

      From the second piece: $f\left(-\frac{\pi}{6}\right) = \cos\left(2 \cdot -\frac{\pi}{6}\right) = \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}$.

   2. Left-hand limit (LHL): $x \to -\frac{\pi}{6}^-$

      For $x < -\frac{\pi}{6}$, we use the first piece:

   $f(x) = \frac{p(1 + \sin 3x)}{(\pi + 6x)^2}.$

   Substitute $x = -\frac{\pi}{6} + h$, where $h \to 0^-$ (so $h < 0$). Then:

   $3x = 3\left(-\frac{\pi}{6} + h\right) = -\frac{\pi}{2} + 3h,$

   $\sin 3x = \sin\left(-\frac{\pi}{2} + 3h\right) = -\cos(3h).$

   So $1 + \sin 3x = 1 - \cos(3h)$.

   Also, $\pi + 6x = \pi + 6\left(-\frac{\pi}{6} + h\right) = \pi - \pi + 6h = 6h$.

   Hence the limit becomes:

   $\lim_{h\to 0^-} \frac{p(1 - \cos 3h)}{(6h)^2} = \lim_{h\to 0^-} \frac{p(1 - \cos 3h)}{36h^2}.$

   Using $1 - \cos t \approx \frac{t^2}{2}$ for small $t$, we have $1 - \cos 3h \approx \frac{(3h)^2}{2} = \frac{9h^2}{2}$.

   So:

   $\text{LHL} = \lim_{h\to 0^-} \frac{p \cdot \frac{9h^2}{2}}{36h^2} = \frac{9p}{72} = \frac{p}{8}.$

   3. Right-hand limit (RHL): $x \to -\frac{\pi}{6}^+$ For $x > -\frac{\pi}{6}$, use the third piece:

   $f(x) = \frac{q(\sin 12x + 2\sin 6x)}{\cos^3\left(\frac{\pi + 12x}{2}\right)}.$

   Again substitute $x = -\frac{\pi}{6} + h$, now $h \to 0^+$.

   Compute each part:

   $12x = 12\left(-\frac{\pi}{6} + h\right) = -2\pi + 12h$, so $\sin 12x = \sin(-2\pi + 12h) = \sin(12h)$ (since $\sin(\theta - 2\pi) = \sin\theta$).

   $6x = 6\left(-\frac{\pi}{6} + h\right) = -\pi + 6h$, so $\sin 6x = \sin(-\pi + 6h) = -\sin(6h)$ (since $\sin(\theta - \pi) = -\sin\theta$).

   Thus $\sin 12x + 2\sin 6x = \sin(12h) - 2\sin(6h)$.

   For the denominator: $\frac{\pi + 12x}{2} = \frac{\pi + 12\left(-\frac{\pi}{6} + h\right)}{2} = \frac{\pi - 2\pi + 12h}{2} = \frac{-\pi + 12h}{2} = -\frac{\pi}{2} + 6h$.

   So $\cos\left(\frac{\pi + 12x}{2}\right) = \cos\left(-\frac{\pi}{2} + 6h\right) = \sin(6h)$ (since $\cos(\theta - \frac{\pi}{2}) = \sin\theta$).

   Hence the denominator is $\cos^3(\cdots) = \sin^3(6h)$.

   The RHL is: …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $f(x) = \pi - \cos^{-1} \left( \dfrac{x^2 + 4x + 3}{x^2 + 4x + 5} \right)$, then $f'(1) =$ (A) $\dfrac{4}{5}$ (B) $2$ (C) $\dfrac{1}{5}$ (D) $-2$

   ›Reveal solutionSolution

   Differentiate $f(x)=\pi-\cos^{-1}(t)$ with $t=\dfrac{x^2+4x+3}{x^2+4x+5}$. This gives $f'(1)=\dfrac{1}{5}$ — option (C).

   The constant $\pi$ contributes nothing to the derivative, so $f'(x)=-\dfrac{d}{dx}\cos^{-1}(t)=\dfrac{t'}{\sqrt{1-t^2}}$, where the extra minus sign comes from the derivative of $\cos^{-1}$.

   1. Differentiate the inner ratio. With $t=\dfrac{x^2+4x+3}{x^2+4x+5}$, both numerator and denominator share the derivative $2x+4$, so by the quotient rule

   $t'=\frac{(2x+4)(x^2+4x+5)-(x^2+4x+3)(2x+4)}{(x^2+4x+5)^2}=\frac{(2x+4)\cdot 2}{(x^2+4x+5)^2}=\frac{4(x+2)}{(x^2+4x+5)^2}.$

   2. Evaluate the pieces at $x=1$. …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The local minimum value of the function $f(x) = 2x - 3 \tan^{-1} x$ is (A) $3 \tan^{-1} \left( \dfrac{1}{\sqrt{2}} \right) - \sqrt{2}$ (B) $\sqrt{2} - 3 \tan^{-1} \left( \dfrac{1}{\sqrt{2}} \right)$ (C) $3 \tan^{-1} (\sqrt{2}) - \dfrac{1}{\sqrt{2}}$ (D) $\dfrac{1}{\sqrt{2}} - 3 \tan^{-1} (\sqrt{2})$

   ›Reveal solutionSolution

   The function has a local minimum at $x = \frac{1}{\sqrt{2}}$, and the minimum value is $\sqrt{2} - 3\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$, which matches option (B).

   The key to this problem is recognizing that a local minimum of a differentiable function occurs where the derivative changes sign from negative to positive. So we first find the critical points by setting $f'(x) = 0$, then classify them using the second derivative test or sign analysis, and finally evaluate the function at the minimum point.

   Let’s go step by step.

   1. Find the derivative. $f(x) = 2x - 3\tan^{-1}x$ Differentiate term by term: $\frac{d}{dx}(2x) = 2$ $\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}$, so

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

   2. Set the derivative to zero to find critical points.

   $2 - \frac{3}{1+x^2} = 0 \quad\Rightarrow\quad \frac{3}{1+x^2} = 2 \quad\Rightarrow\quad 1+x^2 = \frac{3}{2} \quad\Rightarrow\quad x^2 = \frac{1}{2}.$

   So $x = \pm \frac{1}{\sqrt{2}}$ are the two critical points.

   3. Classify each critical point. Compute the second derivative:

   $f''(x) = \frac{d}{dx}\left(2 - \frac{3}{1+x^2}\right) = 0 - 3\cdot\frac{d}{dx}(1+x^2)^{-1} = -3\cdot(-1)(1+x^2)^{-2}\cdot 2x = \frac{6x}{(1+x^2)^2}.$

   • At $x = \frac{1}{\sqrt{2}}$: $f''\left(\frac{1}{\sqrt{2}}\right) = \frac{6\cdot\frac{1}{\sqrt{2}}}{(1+\frac{1}{2})^2} = \frac{6/\sqrt{2}}{(3/2)^2} = \frac{6/\sqrt{2}}{9/4} = \frac{24}{9\sqrt{2}} > 0$. So this is a local minimum. …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $(x^2 - 3x + 2) e^{\frac{y}{x-1}} = x + 2$ then $\left(\frac{dy}{dx}\right)_{x=0} =$ (A) $2$ (B) $-2$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   Find $y(0)=0$ from the equation, then differentiate implicitly: $\left(\dfrac{dy}{dx}\right)_{x=0}=-2$ — option (B).

   Given $(x^2-3x+2)\,e^{\frac{y}{x-1}}=x+2$.

   Value of $y$ at $x=0$. Substituting $x=0$:

   $2\,e^{-y}=2\ \Rightarrow\ e^{-y}=1\ \Rightarrow\ y=0.$

   So we work at the point $(0,0)$.

   Implicit differentiation. Differentiating both sides w.r.t. $x$:

   $(2x-3)\,e^{\frac{y}{x-1}}+(x^2-3x+2)\,e^{\frac{y}{x-1}}\cdot\frac{y'(x-1)-y}{(x-1)^2}=1.$ …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $y = (\sin^{-1} x)^2$, then $(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} =$ (A) $\dfrac{1}{2}$ (B) $2$ (C) $-\dfrac{1}{2}$ (D) $4$

   ›Reveal solutionSolution

   The key idea is to differentiate $y = (\sin^{-1} x)^2$ twice and substitute into the given expression; the result simplifies to the constant $2$, so the correct option is (B).

   We start with $y = (\sin^{-1} x)^2$. The expression $(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx}$ looks like it might simplify nicely because derivatives of inverse trigonometric functions often produce algebraic terms that cancel. The plan: compute $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$, then substitute.

   1. First derivative Let $u = \sin^{-1} x$, so $y = u^2$. Then

   $\frac{dy}{dx} = 2u \cdot \frac{du}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}}.$

   Hence

   $\frac{dy}{dx} = \frac{2\sin^{-1} x}{\sqrt{1-x^2}}.$

   2. Second derivative Differentiate $\frac{dy}{dx}$ using the quotient rule (or product rule). Write

   $\frac{dy}{dx} = 2\sin^{-1} x \cdot (1-x^2)^{-1/2}.$

   Differentiate:

   $\frac{d^2y}{dx^2} = 2\left[ \frac{1}{\sqrt{1-x^2}} \cdot (1-x^2)^{-1/2} + \sin^{-1} x \cdot \left(-\frac12\right)(1-x^2)^{-3/2} \cdot (-2x) \right].$

   Simplify the second term: $(-1/2)(-2x) = x$, so

   $\frac{d^2y}{dx^2} = 2\left[ \frac{1}{1-x^2} + \frac{x \sin^{-1} x}{(1-x^2)^{3/2}} \right].$

   Thus

   $\frac{d^2y}{dx^2} = \frac{2}{1-x^2} + \frac{2x \sin^{-1} x}{(1-x^2)^{3/2}}.$

   3. Plug into the expression We need $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx}$. First term: $(1-x^2)\frac{d^2y}{dx^2} = (1-x^2)\left( \frac{2}{1-x^2} + \frac{2x \sin^{-1} x}{(1-x^2)^{3/2}} \right) = 2 + \frac{2x \sin^{-1} x}{\sqrt{1-x^2}}.$ …

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6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If the range of the real valued function $f(x) = \dfrac{x^2 + x + k}{x^2 - x + k}$ is $\left[\dfrac{1}{3}, 3\right]$, then $k =$ (A) $-2$ (B) $-1$ (C) $1$ (D) $2$

   ›Reveal solutionSolution

   The key idea is to treat the range condition as a quadratic in $x$ having real solutions, leading to inequalities in $k$; solving them gives $k = 1$, which is option (C).

   We are given

   $f(x) = \frac{x^2 + x + k}{x^2 - x + k}$

   and told its range is exactly $\left[\frac13, 3\right]$. That means for every $y$ in that interval, there is some real $x$ such that $f(x)=y$, and no $y$ outside the interval is attained.

   ---

   Concept and Intuition

   Instead of analyzing the function directly, we set $y = f(x)$ and cross-multiply to get a quadratic in $x$:

   $y = \frac{x^2 + x + k}{x^2 - x + k} \quad\Longrightarrow\quad y(x^2 - x + k) = x^2 + x + k.$

   Rearranging gives

   $(y-1)x^2 - (y+1)x + k(y-1) = 0.$

   For a given $y$, this quadratic in $x$ has real solutions exactly when its discriminant is non‑negative. The range of $f$ is precisely the set of $y$ for which this discriminant condition holds (and the denominator is not zero, but that will be automatically satisfied for the correct $k$). So the problem reduces to: find $k$ such that the inequality $\Delta(y) \ge 0$ yields exactly the interval $[\frac13, 3]$.

   ---

   Step‑by‑step solution

   1. Set up the discriminant condition The quadratic in $x$ is

   $(y-1)x^2 - (y+1)x + k(y-1) = 0.$

   Its discriminant is

   $\Delta(y) = (y+1)^2 - 4(y-1)\cdot k(y-1) = (y+1)^2 - 4k(y-1)^2.$

   For real $x$, we need $\Delta(y) \ge 0$.

   2. Interpret the range condition

      The problem says the range is exactly $[\frac13, 3]$. That means:

      • For $y$ in $[\frac13, 3]$, we have $\Delta(y) \ge 0$.
      • For $y$ outside that interval, $\Delta(y) < 0$.
      • The endpoints $y = \frac13$ and $y = 3$ should give $\Delta(y) = 0$ (since the range is closed).

   3. Use the endpoints to find $k$

      At $y = 3$:

   $\Delta(3) = (3+1)^2 - 4k(3-1)^2 = 16 - 4k \cdot 4 = 16 - 16k.$

   Setting $\Delta(3)=0$ gives $16 - 16k = 0 \Rightarrow k = 1$.

   At $y = \frac13$:

   $\Delta\!\left(\frac13\right) = \left(\frac13+1\right)^2 - 4k\left(\frac13-1\right)^2 = \left(\frac43\right)^2 - 4k\left(-\frac23\right)^2 = \frac{16}{9} - 4k \cdot \frac{4}{9} = \frac{16}{9} - \frac{16k}{9}.$

   Setting this to zero also gives $\frac{16}{9}(1 - k) = 0 \Rightarrow k = 1$.

   So $k=1$ makes both endpoints satisfy $\Delta=0$. …

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7. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.Define $f:\mathbb{R} \rightarrow \mathbb{R}$ by $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x} - 4}}, & x > 0 \end{cases}$ Then the value of ‘$a$’ so that $f$ is continuous at $x = 0$ is (A) 8 (B) 4 (C) 2 (D) 1

   ›Reveal solutionSolution

   To make $f$ continuous at $x=0$, the left-hand limit, right-hand limit, and $f(0)$ must all be equal. Computing both limits gives $8$, so $a = 8$, which corresponds to option (A).

   We need $f$ to be continuous at $x = 0$. That means

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

   We already know $f(0) = a$, so we just need to find the two one-sided limits and set them equal to $a$.

   ---

   1. Left-hand limit ($x \to 0^-$)

   For $x < 0$,

   $f(x) = \frac{1 - \cos 4x}{x^2}.$

   As $x \to 0$, $\cos 4x \approx 1 - \frac{(4x)^2}{2} + \cdots$, so $1 - \cos 4x \approx \frac{16x^2}{2} = 8x^2$.

   More rigorously, use the standard limit

   $\lim_{u \to 0} \frac{1 - \cos u}{u^2} = \frac12.$

   Let $u = 4x$. Then

   $\frac{1 - \cos 4x}{x^2} = \frac{1 - \cos 4x}{(4x)^2} \cdot 16 = 16 \cdot \frac{1 - \cos 4x}{(4x)^2}.$

   As $x \to 0$, $4x \to 0$, so

   $\lim_{x \to 0^-} f(x) = 16 \cdot \frac12 = 8.$

   Tip

   The identity $1 - \cos \theta = 2\sin^2(\theta/2)$ also works:

   $\frac{1 - \cos 4x}{x^2} = \frac{2\sin^2 2x}{x^2} = 8 \cdot \frac{\sin^2 2x}{(2x)^2} \to 8$.

   ---

   2. Right-hand limit ($x \to 0^+$)

   For $x > 0$,

   $f(x) = \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}.$

   Direct substitution gives $0/0$, so we rationalize the denominator:

   Multiply numerator and denominator by $\sqrt{16 + \sqrt{x}} + 4$:

   $$f(x) = \frac{\sqrt{x} \left( \sqrt{16 + \sqrt{x}} + 4 \right)}{(16 + \sqrt{x}) - 16} …$$

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8. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $\int (1 + x - x^2) e^{x + x^2} dx = f(x) + c$, then $f(1) - f(-1) =$ (A) $e^2 - \dfrac{1}{e^2}$ (B) $e^2 + \dfrac{1}{e^2}$ (C) $e + \dfrac{1}{e}$ (D) $e - \dfrac{1}{e}$

   ›Reveal solutionSolution

   The integrand is the exact derivative of $x\,e^{x+\frac1x}$, so $f(x)=x\,e^{x+1/x}$ and $f(1)-f(-1)=e^{2}+e^{-2}$ — option (B).

   The concept first

   Whenever an exponential $e^{g(x)}$ is multiplied by a bracket, try the product rule backwards:

   $\frac{d}{dx}\Big[h(x)e^{g(x)}\Big]=\Big[h'(x)+h(x)g'(x)\Big]e^{g(x)} .$

   If the bracket can be split as $h'+h\,g'$, the antiderivative is simply $h\,e^{g}$ — no integration by parts, no substitution.

   Step 1 — Identify $g$, guess $h$

   The exponent is $g(x)=x+\dfrac1x$, so $g'(x)=1-\dfrac{1}{x^{2}}$. Try the simplest candidate $h(x)=x$ (so $h'=1$):

   $h'+hg'=1+x\left(1-\frac{1}{x^{2}}\right)=1+x-\frac{1}{x},$

   which is precisely the bracket multiplying the exponential.

   Step 2 — Write the antiderivative

   $\int\left(1+x-\frac1x\right)e^{\,x+\frac1x}\,dx=x\,e^{\,x+\frac1x}+c\qquad\Rightarrow\qquad f(x)=x\,e^{\,x+\frac1x}.$

   Step 3 — Evaluate at the two points …

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9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots$ to $n$ terms $= n(n+1)f(n)$, then $f(2) =$ (A) $12$ (B) $42$ (C) $18$ (D) $20$

   ›Reveal solutionSolution

   Sum of the first $2$ terms is $120$; setting $2\cdot 3\cdot f(2)=120$ gives $f(2)=20$ (option D).

   The series is $\displaystyle\sum (2r-1)(2r+1)(2r+3)$, and its sum to $n$ terms is written as $n(n+1)f(n)$.

   Evaluate directly for $n=2$ by adding the first two terms:

   $1\cdot 3\cdot 5 = 15, \qquad 3\cdot 5\cdot 7 = 105$ …

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10. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.If $\int e^x (x^3 + x^2 - x + 4) dx = e^x f(x) + c$, then $f(1) =$ (A) $0$ (B) $1$ (C) $2$ (D) $3$

    ›Reveal solutionSolution

    The key idea is that integrals of the form $\int e^x P(x)\,dx$ equal $e^x Q(x)+c$ where $Q(x)$ is a polynomial found by comparing derivatives. Here $f(x)=x^3-2x^2+3x+1$, so $f(1)=3$, making the answer (D).

    We are given:

    $\int e^x (x^3 + x^2 - x + 4) \, dx = e^x f(x) + c$

    and need $f(1)$. The structure suggests that $f(x)$ is a polynomial, because differentiating $e^x f(x)$ gives $e^x (f(x) + f'(x))$, which must match the integrand $e^x (x^3 + x^2 - x + 4)$. This is a classic method: when the integrand is $e^x$ times a polynomial, the antiderivative is $e^x$ times another polynomial of the same degree.

    1. Set up the derivative condition Differentiate the right-hand side:

    $\frac{d}{dx}\left(e^x f(x)\right) = e^x f(x) + e^x f'(x) = e^x \bigl(f(x) + f'(x)\bigr)$

    This must equal the integrand $e^x (x^3 + x^2 - x + 4)$. Cancel $e^x$ (nonzero) to get:

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

    2. Assume a polynomial form for $f(x)$ Since the right side is degree 3, $f(x)$ must also be degree 3 (because $f'(x)$ is one degree lower, so the sum’s leading term comes from $f(x)$). Write:

    $f(x) = ax^3 + bx^2 + cx + d$

    Then

    $f'(x) = 3ax^2 + 2bx + c$

    3. Match coefficients Substitute into $f(x) + f'(x)$:

    $(ax^3 + bx^2 + cx + d) + (3ax^2 + 2bx + c) = ax^3 + (b+3a)x^2 + (c+2b)x + (d+c)$

    Equate to $x^3 + x^2 - x + 4$:

    • $x^3$: $a = 1$ …

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11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If $f(x) = x^{\tan x} + (\tan x)^x$, then $f'\left(\dfrac{\pi}{4}\right) =$ (A) $1 + \dfrac{\pi}{2} \log \left(\dfrac{e\pi}{4}\right)$ (B) $\dfrac{\pi}{2} \left(\log \dfrac{\pi}{4} + 1\right)$ (C) $1$ (D) $0$

    ›Reveal solutionSolution

    The derivative of a sum of two variable-exponent functions is found by logarithmic differentiation on each term separately; evaluating at $x = \pi/4$ gives $f'(\pi/4) = 1 + \frac{\pi}{2}\log\frac{e\pi}{4}$, which matches option (A).

    We have $f(x) = x^{\tan x} + (\tan x)^x$.

    To differentiate terms where both base and exponent are functions of $x$, we use logarithmic differentiation: take the natural log, differentiate implicitly, then solve for the derivative. This works because the exponent is not constant, so the power rule alone fails.

    ---

    1. Differentiate $u(x) = x^{\tan x}$ Let $u = x^{\tan x}$. Take $\log u = \tan x \cdot \log x$. Differentiate:

    $\frac{u'}{u} = \sec^2 x \cdot \log x + \tan x \cdot \frac{1}{x}$

    So

    $u' = x^{\tan x} \left( \sec^2 x \log x + \frac{\tan x}{x} \right)$

    2. Differentiate $v(x) = (\tan x)^x$ Let $v = (\tan x)^x$. Take $\log v = x \cdot \log(\tan x)$. Differentiate:

    $\frac{v'}{v} = \log(\tan x) + x \cdot \frac{\sec^2 x}{\tan x}$

    Since $\frac{\sec^2 x}{\tan x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$, we have

    $v' = (\tan x)^x \left( \log(\tan x) + \frac{2x}{\sin 2x} \right)$

    3. Combine

    $f'(x) = x^{\tan x} \left( \sec^2 x \log x + \frac{\tan x}{x} \right) + (\tan x)^x \left( \log(\tan x) + \frac{2x}{\sin 2x} \right)$

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

       At $x = \pi/4$:

       • $\tan(\pi/4) = 1$, so $x^{\tan x} = (\pi/4)^1 = \pi/4$ and $(\tan x)^x = 1^{\pi/4} = 1$.
       • $\sec^2(\pi/4) = 2$, $\log(\pi/4)$ stays as is, $\tan(\pi/4)=1$, $\sin(2\cdot\pi/4)=\sin(\pi/2)=1$.

       Plug in:

    $f'(\pi/4) = \frac{\pi}{4} \left( 2 \log\frac{\pi}{4} + \frac{1}{\pi/4} \right) + 1 \left( \log 1 + \frac{2\cdot\pi/4}{1} \right)$

    Simplify:

    • $\frac{\pi}{4} \cdot 2 \log\frac{\pi}{4} = \frac{\pi}{2} \log\frac{\pi}{4}$
    • $\frac{\pi}{4} \cdot \frac{1}{\pi/4} = \frac{\pi}{4} \cdot \frac{4}{\pi} = 1$ …

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12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.If the slope of the tangent drawn at any point $(x, y)$ to the curve $y = f(x)$ is $3x^2 - 5$ and $f(1) = 2$, then the tangent at $(1, 2)$ to the curve $y = f(x)$ intersects the curve at the point (A) $(2, 0)$ (B) $(-2, 8)$ (C) $(3, -2)$ (D) $(-1, 6)$

    ›Reveal solutionSolution

    The slope of the tangent is given by the derivative, so we integrate to find $f(x)$, then find the tangent line at $(1,2)$, and solve for its second intersection with the curve. The required point is $(-2,8)$, which corresponds to option (B).

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    We are told that the slope of the tangent at any point $(x, y)$ on the curve $y = f(x)$ is $3x^2 - 5$. That means

    $\frac{dy}{dx} = 3x^2 - 5.$

    Since the derivative is given, we can recover $f(x)$ by integration. Then we will have the full equation of the curve. The tangent line at $(1,2)$ can be written using the slope at that point, and we find where else this line meets the curve — that is the second intersection point.

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    1. Find $f(x)$ by integrating the derivative

    $f(x) = \int (3x^2 - 5)\,dx = x^3 - 5x + C.$

    We use the condition $f(1) = 2$:

    $1^3 - 5(1) + C = 2 \quad\Rightarrow\quad 1 - 5 + C = 2 \quad\Rightarrow\quad C = 6.$

    So the curve is

    $y = x^3 - 5x + 6.$

    2. Find the slope of the tangent at $(1,2)$ At $x = 1$,

    $m = 3(1)^2 - 5 = -2.$

    So the tangent line has slope $-2$ and passes through $(1,2)$. Its equation:

    $y - 2 = -2(x - 1) \quad\Rightarrow\quad y = -2x + 4.$

    3. Find intersection points of the tangent and the curve Set the curve equal to the line:

    $x^3 - 5x + 6 = -2x + 4.$

    Simplify:

    $x^3 - 5x + 6 + 2x - 4 = 0 \quad\Rightarrow\quad x^3 - 3x + 2 = 0.$

    We already know $x = 1$ is a root (since the tangent touches at $(1,2)$). Factor out $(x-1)$:

    $$$$

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