Q.Find $\frac{dy}{dx}$ in the following: $\frac{\sin (ax + b)}{\cos (cx + d)}$

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 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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3. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $f(x) = \begin{cases} \dfrac{a \sin x - b x + c x^2 + x^3}{2 \log(1+x) - 2x^3 + x^4}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ is continuous at $x = 0$, then (A) $a = 2b$ (B) $a = b$ (C) $a = b = c$ (D) $b = c$

   ›Reveal solutionSolution

   Continuity at $x=0$ needs $\lim_{x\to0}f(x)=0$; since the denominator is of order $x$, the numerator's $x$-term must vanish, forcing $a=b$ — option (B).

   Continuity requirement. Because $f(0)=0$, we need

   $\lim_{x\to0}\frac{a\sin x-bx+cx^2+x^3}{2\log(1+x)-2x^3+x^4}=0.$

   Series expansions. Using $\sin x=x-\tfrac{x^3}{6}+\cdots$ and $\log(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}-\cdots$:

   $\text{Numerator}=(a-b)x+cx^2+\left(1-\tfrac{a}{6}\right)x^3+\cdots,$

   $\text{Denominator}=2\log(1+x)-2x^3+x^4=2x-x^2-\tfrac{4}{3}x^3+\cdots.$ …

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4. 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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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 $\int e^x \cos x \, dx = \dfrac{e^x}{2} (\cos x + \sin x)$ 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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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 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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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-2024-05-09-FN1 markMCQ
    Q.If the function $f(x) = \begin{cases} \dfrac{\cos ax - \cos 9x}{x^2}, & \text{if } x \neq 0 \\ 16, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$, then $a =$ (A) $\pm 8$ (B) $\pm 7$ (C) $\pm 6$ (D) $\pm 5$

    ›Reveal solutionSolution

    For continuity at $x=0$, the limit of $f(x)$ as $x\to0$ must equal $f(0)=16$. Using the cosine difference identity and the small-angle approximation $\frac{1-\cos\theta}{\theta^2}\to\frac12$, we find the limit equals $\frac{a^2-81}{2}$. Setting this equal to 16 gives $a^2=113$, which is not among the options — but careful: the identity $\cos ax - \cos 9x = -2\sin\frac{(a+9)x}{2}\sin\frac{(a-9)x}{2}$ leads to a limit of $\frac{81-a^2}{2}$. Equating to 16 yields $a^2=49$, so $a=\pm7$. The correct option is (B).

    ---

    Concept and intuition:

    Continuity at a point means the function’s value there equals the limit of the function as we approach that point. Here, $f(0)=16$ is given, so we need $\lim_{x\to0}f(x)=16$. The expression for $x\neq0$ is a fraction that looks like $0/0$ at $x=0$, so we must simplify it. The key trick: rewrite the numerator using the sum-to-product identity for cosines, then use the famous limit $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ (or equivalently $\lim_{\theta\to0}\frac{1-\cos\theta}{\theta^2}=\frac12$). This turns the problem into a simple algebraic equation.

    ---

    Step-by-step solution:

    1. Write the continuity condition For $f$ to be continuous at $x=0$:

    $\lim_{x\to 0} f(x) = f(0) = 16.$

    So we need to compute

    $L = \lim_{x\to 0} \frac{\cos ax - \cos 9x}{x^2}.$

    2. Apply the cosine difference identity Recall: $\cos P - \cos Q = -2\sin\frac{P+Q}{2}\sin\frac{P-Q}{2}$. Here $P = ax$, $Q = 9x$, so

    $\cos ax - \cos 9x = -2\sin\frac{(a+9)x}{2}\sin\frac{(a-9)x}{2}.$

    Thus

    $L = \lim_{x\to 0} \frac{-2\sin\frac{(a+9)x}{2}\sin\frac{(a-9)x}{2}}{x^2}.$

    3. Rewrite to use the standard sine limit Multiply and divide each sine by its argument:

    $L = \lim_{x\to 0} -2 \cdot \frac{\sin\frac{(a+9)x}{2}}{\frac{(a+9)x}{2}} \cdot \frac{\sin\frac{(a-9)x}{2}}{\frac{(a-9)x}{2}} \cdot \frac{\frac{(a+9)x}{2} \cdot \frac{(a-9)x}{2}}{x^2}.$

    The factors $\frac{\sin\theta}{\theta}$ each tend to 1 as $\theta\to0$ (since $x\to0$ makes both arguments go to 0). So

    $$$$

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11. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.Let ‘a’ be a positive real number. If a real valued function

    [!FORMULA] $f(x) = \begin{cases} \frac{6^{x} - 3^{x} - 2^{x} + 1}{1 - \cos\left(\frac{x}{a}\right)} & \text{if } x \neq 0 \\ \log 3 \log 4 & \text{if } x = 0 \end{cases}$

    is continuous at $x = 0$, then $a =$ (A) 1 (B) 2 (C) 3 (D) 4

    ›Reveal solutionSolution

    Factoring the numerator as $(2^x-1)(3^x-1)$ and matching the limit to $f(0)$ gives $a=1$ — option (A).

    Continuity condition. $\displaystyle\lim_{x\to0}f(x)=f(0)=\log 3\,\log 4$.

    Numerator.

    $6^x-3^x-2^x+1=3^x(2^x-1)-(2^x-1)=(2^x-1)(3^x-1).$

    As $x\to0$, $2^x-1\sim x\ln2$ and $3^x-1\sim x\ln3$, so the numerator $\sim x^2\ln2\,\ln3$.

    Denominator. $1-\cos\dfrac{x}{a}\sim\dfrac12\Big(\dfrac{x}{a}\Big)^2=\dfrac{x^2}{2a^2}$.

    Limit. …

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12. 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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