Q.Find the rate of change of the area of a circle per second with respect to its radius $r$ when $r = 5$ cm.

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.By the application of derivatives, the approximate value of $\sqrt{242}$ is (A) 2.9085 (B) 2.9975 (C) 2.9527 (D) 2.8529

   ›Reveal solutionSolution

   The answer choices (all near 3) show this is the fifth root $\sqrt[5]{242}$, not the square root. Linearising $f(x)=x^{1/5}$ about $x=243$ (where $3^5=243$) gives $\sqrt[5]{242}\approx 2.9975$, option (B).

   This is a linear (tangent-line) approximation: near a point $a$ where the value is known exactly,

   $f(a+\Delta x)\approx f(a)+f'(a)\,\Delta x.$

   Because the options cluster around 3 (whereas $\sqrt{242}\approx 15.6$), the quantity being approximated is $\sqrt[5]{242}$, and the natural reference point is $243=3^5$.

   1. Choose the point. Let $f(x)=x^{1/5}$. Then $f(243)=243^{1/5}=3$ exactly. Take $a=243$ and $\Delta x = 242-243 = -1$.

   2. Derivative. $f'(x)=\tfrac15 x^{-4/5}$. Since $243^{1/5}=3$, we have $243^{4/5}=3^4=81$, so

   $f'(243)=\frac15\cdot\frac{1}{81}=\frac{1}{405}.$

   3. Apply the approximation.

   $\sqrt[5]{242}=f(242)\approx f(243)+f'(243)\,(-1)=3-\frac{1}{405}\approx 3-0.00247 = 2.9975.$

   Tip

   Check the size of the options against the quantity. If a "root of 242" is expected to be about 3, it must be the fifth root, since $3^5=243$ sits right next to 242.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $y = (e^{2x} - 4)(6e^{2x} - 5e^x + 1)$, then $\left(\frac{dy}{dx}\right)_{x=0} - \left(\frac{d^2y}{dx^2}\right)_{x=0} =$ (A) $0$ (B) $-5$ (C) $4$ (D) $6$

   ›Reveal solutionSolution

   Differentiate the product, evaluate the first and second derivatives at $x=0$, and subtract: the result is $4$.

   Let $u = e^{2x}-4$ and $v = 6e^{2x}-5e^{x}+1$, so $y = uv$.

   Derivatives of the factors:

   $u' = 2e^{2x},\quad u'' = 4e^{2x}$

   $v' = 12e^{2x}-5e^{x},\quad v'' = 24e^{2x}-5e^{x}$

   Values at $x=0$ (where $e^{2x}=1,\ e^{x}=1$):

   $u = -3,\ v = 2,\quad u' = 2,\ v' = 7,\quad u'' = 4,\ v'' = 19$

   First derivative:

   $\left(\frac{dy}{dx}\right)_{0} = u'v + uv' = (2)(2) + (-3)(7) = 4 - 21 = -17$

   Second derivative:

   $\left(\frac{d^{2}y}{dx^{2}}\right)_{0} = u''v + 2u'v' + uv'' = (4)(2) + 2(2)(7) + (-3)(19) = 8 + 28 - 57 = -21$

   Required difference:

   $\left(\frac{dy}{dx}\right)_{0} - \left(\frac{d^{2}y}{dx^{2}}\right)_{0} = -17 - (-21) = 4$

   ✓Final answer

   The value is $4$, so the correct option is (C).

3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.

   [!FORMULA] $\left[ \frac{d}{dx} \left( (\sin x)^{\cos x} \right) \right]_{x = \frac{\pi}{4}} =$

   (A) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{\sqrt{2}+1}{\sqrt{2}}} (1 + \log \sqrt{2})$ (B) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{1}{\sqrt{2}}} (1 + \log \sqrt{2})$ (C) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{1}{\sqrt{2}}} (1 - \log \sqrt{2})$ (D) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{\sqrt{2}+1}{\sqrt{2}}} (1 - \log \sqrt{2})$

   ›Reveal solutionSolution

   Logarithmic differentiation gives $\dfrac{dy}{dx} = (\sin x)^{\cos x}\left(\dfrac{\cos^2 x}{\sin x} - \sin x\,\log(\sin x)\right)$. Evaluating at $x = \pi/4$ yields $\left(\dfrac{1}{\sqrt 2}\right)^{\frac{\sqrt 2 + 1}{\sqrt 2}}(1 + \log\sqrt 2)$, option (A).

   ---

   Concept & intuition

   Because the exponent $\cos x$ is itself a function of $x$, the ordinary power rule does not apply. Use logarithmic differentiation: take $\log$ of both sides, differentiate implicitly, then multiply back by $y$.

   ---

   1. Take logarithms Let $y = (\sin x)^{\cos x}$. Then

   $\log y = \cos x \,\log(\sin x).$

   2. Differentiate implicitly (product rule on the right):

   $\frac{1}{y}\frac{dy}{dx} = -\sin x\,\log(\sin x) + \cos x\cdot\frac{\cos x}{\sin x} = -\sin x\,\log(\sin x) + \frac{\cos^2 x}{\sin x}.$

   3. Solve for the derivative

   $\frac{dy}{dx} = (\sin x)^{\cos x}\left(\frac{\cos^2 x}{\sin x} - \sin x\,\log(\sin x)\right).$

   4. Evaluate at $x = \pi/4$ where $\sin x = \cos x = \dfrac{1}{\sqrt 2}$:

      • Base: $(\sin x)^{\cos x} = \left(\dfrac{1}{\sqrt 2}\right)^{1/\sqrt 2}$.
      • $\dfrac{\cos^2 x}{\sin x} = \dfrac{1/2}{1/\sqrt 2} = \dfrac{1}{\sqrt 2}$.
      • $-\sin x\,\log(\sin x) = -\dfrac{1}{\sqrt 2}\log\!\left(\dfrac{1}{\sqrt 2}\right) = -\dfrac{1}{\sqrt 2}(-\log\sqrt 2) = \dfrac{1}{\sqrt 2}\log\sqrt 2.$

      The bracket becomes

   $\frac{1}{\sqrt 2} + \frac{1}{\sqrt 2}\log\sqrt 2 = \frac{1}{\sqrt 2}\,(1 + \log\sqrt 2).$

   5. Combine

   $\left.\frac{dy}{dx}\right|_{\pi/4} = \left(\frac{1}{\sqrt 2}\right)^{1/\sqrt 2}\cdot\frac{1}{\sqrt 2}\,(1 + \log\sqrt 2).$

   Since $\dfrac{1}{\sqrt 2} = \left(\dfrac{1}{\sqrt 2}\right)^1$, the two base factors add exponents:

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

   Tip

   The stray factor $\tfrac{1}{\sqrt 2}$ is absorbed into the base's exponent ($\tfrac{1}{\sqrt 2} + 1 = \tfrac{\sqrt 2 + 1}{\sqrt 2}$) — a standard way the options disguise the answer.

   Watch out

   Watch the sign: $\log(1/\sqrt 2) = -\log\sqrt 2$, so $-\sin x\,\log(\sin x)$ is positive. This makes the bracket $1 + \log\sqrt 2$, not $1 - \log\sqrt 2$.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $y = \tan^{-1} \left( \frac{\sqrt{1+9x^2}-1}{3x} \right)$ then $\left( \frac{dy}{dx} \right)_{x = \frac{1}{3}} =$ (A) $\frac{2}{3}$ (B) $\frac{1}{3}$ (C) $\frac{3}{4}$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The key is to simplify the inverse tangent expression using a trigonometric substitution, turning it into a linear function of $x$, which makes differentiation trivial. The derivative at $x = \frac13$ is $\frac{3}{4}$, so option (C) is correct.

   ---

   The expression inside the arctan looks messy, but the presence of $\sqrt{1+9x^2}$ suggests a substitution like $3x = \tan\theta$ or $3x = \sinh t$. Since we have a square root of a sum of squares, the hyperbolic substitution is cleaner: let $3x = \sinh t$. Then $\sqrt{1+9x^2} = \sqrt{1+\sinh^2 t} = \cosh t$. This turns the fraction into something like $\frac{\cosh t - 1}{\sinh t}$, which simplifies using hyperbolic half-angle identities. The result is that the arctan becomes just $t/2$, i.e., a linear function of $t$, and hence a simple function of $x$.

   ---

   1. Substitute to simplify Let $3x = \sinh t$, so $x = \frac13 \sinh t$. Then

   $\sqrt{1+9x^2} = \sqrt{1+\sinh^2 t} = \cosh t.$

   The argument of the arctan becomes

   $\frac{\sqrt{1+9x^2} - 1}{3x} = \frac{\cosh t - 1}{\sinh t}.$

   2. Use the hyperbolic half-angle identity Recall: $\cosh t - 1 = 2\sinh^2(t/2)$ and $\sinh t = 2\sinh(t/2)\cosh(t/2)$. Hence

   $\frac{\cosh t - 1}{\sinh t} = \frac{2\sinh^2(t/2)}{2\sinh(t/2)\cosh(t/2)} = \frac{\sinh(t/2)}{\cosh(t/2)} = \tanh(t/2).$

   3. Simplify the inverse tangent So

   $y = \tan^{-1}\bigl(\tanh(t/2)\bigr).$

   But $\tanh(t/2)$ is always between $-1$ and $1$, and $\tan^{-1}$ of a hyperbolic tangent is not simply $t/2$ — wait, we need a different identity. Actually, there is a known relation:

   $\tan^{-1}\left(\frac{\cosh t - 1}{\sinh t}\right) = \frac{t}{2}.$

   Let’s verify: $\tan(t/2) = \frac{\sin t}{1+\cos t}$, but here we have hyperbolic functions. The correct identity is:

   $\tanh^{-1} u = \frac12 \log\frac{1+u}{1-u},$

   but we have $\tan^{-1}$, not $\tanh^{-1}$. So let’s check numerically: if $t=1$, $\frac{\cosh 1 - 1}{\sinh 1} \approx 0.462$, $\tan^{-1}(0.462) \approx 0.433$, and $t/2 = 0.5$. Not equal. So the direct half-angle idea needs adjustment.

   Better approach: Use the identity

   $\tan^{-1}\left(\frac{\sqrt{1+a^2}-1}{a}\right) = \frac12 \tan^{-1} a,$

   for $a>0$. Let’s prove it: set $a = \tan\theta$, then $\sqrt{1+a^2} = \sec\theta$, so

   $\frac{\sec\theta - 1}{\tan\theta} = \frac{1/\cos\theta - 1}{\sin\theta/\cos\theta} = \frac{1 - \cos\theta}{\sin\theta} = \tan(\theta/2).$

   Hence $\tan^{-1}(\tan(\theta/2)) = \theta/2 = \frac12 \tan^{-1} a$. Perfect.

   4. Apply the identity Here $a = 3x$. So for $x>0$ (and $x=1/3>0$), we have

   $y = \tan^{-1}\left(\frac{\sqrt{1+9x^2}-1}{3x}\right) = \frac12 \tan^{-1}(3x).$

   5. Differentiate

   $\frac{dy}{dx} = \frac12 \cdot \frac{1}{1+(3x)^2} \cdot 3 = \frac{3}{2(1+9x^2)}.$

   6. Evaluate at $x = \frac13$

   $\left.\frac{dy}{dx}\right|_{x=1/3} = \frac{3}{2\left(1+9\cdot\frac19\right)} = \frac{3}{2(1+1)} = \frac{3}{4}.$

   Watch out

   A common mistake is to forget the domain condition: the identity $\tan^{-1}\bigl(\frac{\sqrt{1+a^2}-1}{a}\bigr) = \frac12 \tan^{-1} a$ holds for $a>0$. At $x=1/3$, $a=1>0$, so it’s valid. For negative $x$, the sign flips.

   Tip

   The identity used here is a neat trick: whenever you see $\sqrt{1+u^2}$ in an arctan, try substituting $u = \tan\theta$ to simplify.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\lim_{x \to 0} \frac{\log{(4+x)^x} - \log{4^x}}{\sin^2{x}} =$ (A) $4$ (B) $\frac{1}{4}$ (C) $2$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The numerator reduces to $x\log\!\left(1+\tfrac{x}{4}\right)\sim\tfrac{x^2}{4}$, and $\sin^2 x\sim x^2$, so the limit is $\tfrac14$.

   Simplify the numerator. Using $\log(a^b)=b\log a$:

   $\log\!\big((4+x)^x\big)-\log\!\big(4^x\big)=x\log(4+x)-x\log 4=x\log\!\frac{4+x}{4}=x\log\!\left(1+\frac{x}{4}\right).$

   Take the limit. As $x\to0$, $\log\!\left(1+\tfrac{x}{4}\right)\sim\tfrac{x}{4}$ and $\sin^2 x\sim x^2$:

   $\lim_{x\to0}\frac{x\log\!\left(1+\tfrac{x}{4}\right)}{\sin^2 x}=\lim_{x\to0}\frac{x\cdot\tfrac{x}{4}}{x^2}=\frac{x^2/4}{x^2}=\frac14.$

   ✓Final answer

   The limit is $\dfrac{1}{4}$ — option (B).

6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the rate of change of the slope of the tangent drawn to the curve $y = x^3 - 2x^2 + 3x - 2$ at the point $(2, 4)$ is $k$ times the rate of change of its abscissa, then $k=$ (A) $2$ (B) $4$ (C) $6$ (D) $8$

   ›Reveal solutionSolution

   The problem asks for the constant $k$ such that the rate of change of the slope of the tangent equals $k$ times the rate of change of the abscissa. This means we need the second derivative of $y$ with respect to $x$, evaluated at $x=2$, which gives $k=8$.

   We are told that the rate of change of the slope of the tangent is $k$ times the rate of change of the abscissa.

   The “slope of the tangent” is just the first derivative $\frac{dy}{dx}$.

   The “rate of change” of that slope with respect to time (or any parameter) is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$.

   The “rate of change of its abscissa” is $\frac{dx}{dt}$.

   The condition is:

   $\frac{d}{dt}\left(\frac{dy}{dx}\right) = k \cdot \frac{dx}{dt}$

   By the chain rule, $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2} \cdot \frac{dx}{dt}$.

   So the equation becomes:

   $\frac{d^2y}{dx^2} \cdot \frac{dx}{dt} = k \cdot \frac{dx}{dt}$

   Assuming $\frac{dx}{dt} \neq 0$ (the abscissa is changing), we cancel it and get:

   $\frac{d^2y}{dx^2} = k$

   Thus, $k$ is simply the second derivative of $y$ with respect to $x$ at the given point.

   Now we compute:

   1. First derivative:

      $y = x^3 - 2x^2 + 3x - 2$

      $\frac{dy}{dx} = 3x^2 - 4x + 3$

   2. Second derivative:

      $\frac{d^2y}{dx^2} = 6x - 4$

   3. Evaluate at $x=2$:

      $\frac{d^2y}{dx^2}\big|_{x=2} = 6(2) - 4 = 12 - 4 = 8$

   Therefore $k = 8$.

   Watch out

   A common mistake is to confuse “rate of change of slope” with the slope itself, or to try using the point $(2,4)$ to find something about the curve’s value — but the $y$-coordinate is irrelevant here; only $x=2$ matters.

   Tip

   The phrase “rate of change of … is $k$ times the rate of change of its abscissa” is a disguised way of saying “the second derivative equals $k$”. Always translate such language into derivatives with respect to time, then use the chain rule.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The number of all common roots of the equation $x^4 - 10x^3 + 37x^2 - 60x + 36 = 0$ and the transformed equation of it obtained by increasing any two distinct roots of it by 1, keeping the other two roots fixed, is (A) 1 (B) 3 (C) 4 (D) 2

   ›Reveal solutionSolution

   We first find the roots of the given quartic equation. Then, we identify the roots that remain unchanged when two distinct roots are increased by 1. The number of such common roots is $\boxed{3}$.

   Concept and Intuition

   The problem asks for the number of common roots between an original polynomial equation and a new equation formed by a specific transformation of its roots.

   1. Understanding "Common Roots": When we talk about common roots of two polynomial equations, we are looking for the values of $x$ that satisfy both equations. If a root has a certain multiplicity in both equations, its contribution to the "number of common roots" is the minimum of its multiplicities in the two equations. For example, if $x=a$ is a root of multiplicity 3 in the first equation and multiplicity 2 in the second, then $x=a$ contributes 2 to the count of common roots.

   2. The Transformation: The original equation has four roots, say $r_1, r_2, r_3, r_4$. The transformation states that we "increase any two distinct roots of it by 1, keeping the other two roots fixed".

      • "Distinct roots" here refers to distinct values of the roots. For example, if the roots are $2, 2, 3, 3$, the distinct values are 2 and 3. So, we must choose one root with value 2 and one root with value 3 to be increased.
      • Let's say we choose $r_a$ and $r_b$ (where $r_a \neq r_b$) to be increased by 1. The new roots will be $r_a+1$ and $r_b+1$.
      • The "other two roots fixed" means the remaining two roots, say $r_c$ and $r_d$, are unchanged.
      • So, if the original set of roots is $R = \{r_a, r_b, r_c, r_d\}$, the transformed set of roots is $R' = \{r_a+1, r_b+1, r_c, r_d\}$.
      • The common roots are the elements in the intersection of these two multisets, $R \cap R'$.

   The strategy is to first find all roots of the original equation, then apply the transformation to form the new set of roots, and finally count the common roots.

   Step-by-Step Solution

   1. Find the roots of the original equation:

      The given equation is $P(x) = x^4 - 10x^3 + 37x^2 - 60x + 36 = 0$.

      We can try to find rational roots using the Rational Root Theorem (divisors of 36).

      • Let's test $x=1$: $1 - 10 + 37 - 60 + 36 = 4 \neq 0$.
      • Let's test $x=2$: $2^4 - 10(2^3) + 37(2^2) - 60(2) + 36$ $= 16 - 10(8) + 37(4) - 120 + 36$ $= 16 - 80 + 148 - 120 + 36$ $= (16 + 148 + 36) - (80 + 120)$ $= 200 - 200 = 0$. So, $x=2$ is a root. This means $(x-2)$ is a factor.
      • We can perform polynomial division or synthetic division:

        2 | 1  -10   37  -60   36
          |    2   -16   42  -36
          --------------------
            1  -8    21  -18    0
        

        The quotient is $x^3 - 8x^2 + 21x - 18$. Let's call this $Q(x)$.
      • Let's test $x=2$ again for $Q(x)$: $2^3 - 8(2^2) + 21(2) - 18$ $= 8 - 8(4) + 42 - 18$ $= 8 - 32 + 42 - 18$ $= (8 + 42) - (32 + 18)$ $= 50 - 50 = 0$. So, $x=2$ is a root of $Q(x)$ as well, meaning it's a root of multiplicity at least 2 for $P(x)$.
      • Divide $Q(x)$ by $(x-2)$:

        2 | 1  -8   21  -18
          |    2  -12   18
          -----------------
            1  -6    9    0
        

        The quotient is $x^2 - 6x + 9$. This is a perfect square: $(x-3)^2$.
      • Therefore, the original equation can be factored as: $P(x) = (x-2)(x-2)(x-3)^2 = (x-2)^2 (x-3)^2 = 0$.
      • The roots of the original equation are $2, 2, 3, 3$.

   2. Apply the transformation to the roots:

      The original roots are $R = \{2, 2, 3, 3\}$.

      The distinct values of the roots are 2 and 3.

      The problem states we increase "any two distinct roots" by 1. This means we must choose one root with value 2 and one root with value 3 to be increased.

      Let's pick one instance of '2' (say $2_A$) and one instance of '3' (say $3_A$) to be increased.

      The other two roots, the remaining '2' (say $2_B$) and the remaining '3' (say $3_B$), are kept fixed.

      • Roots to be increased: $2_A$ and $3_A$.
      • Roots to be fixed: $2_B$ and $3_B$.
      • The transformed roots are: $2_A+1 = 2+1 = 3$ $3_A+1 = 3+1 = 4$ $2_B = 2$ $3_B = 3$
      • So, the set of roots for the transformed equation is $R' = \{3, 4, 2, 3\}$. (As a multiset, this is $\{2, 3, 3, 4\}$).

   3. Find the common roots:

      We need to find the common roots between the original set $R = \{2, 2, 3, 3\}$ and the transformed set $R' = \{2, 3, 3, 4\}$.

      To find the intersection of multisets, for each element, its multiplicity in the intersection is the minimum of its multiplicities in the individual multisets.

      • For the root 2: Multiplicity of 2 in $R$ is 2. Multiplicity of 2 in $R'$ is 1. The minimum is 1. So, 2 is a common root with multiplicity 1.
      • For the root 3: Multiplicity of 3 in $R$ is 2. Multiplicity of 3 in $R'$ is 2. The minimum is 2. So, 3 is a common root with multiplicity 2.
      • For the root 4: Multiplicity of 4 in $R$ is 0. Multiplicity of 4 in $R'$ is 1. The minimum is 0. So, 4 is not a common root.

      The common roots are $\{2, 3, 3\}$.

      The total number of common roots is $1 + 2 = 3$.

   ✓Final answer

   The number of all common roots is $\boxed{3}$.

8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The radius of a cone of height 9 units is changed from 2 units to 2.12 units. The exact change and approximate change in the volume of the cone are respectively (A) $(1.4437)\pi, (1.44)\pi$ (B) $(1.4832)\pi, (1.479)\pi$ (C) $(1.4842)\pi, (1.48)\pi$ (D) $(1.4832)\pi, (1.44)\pi$

   ›Reveal solutionSolution

   The exact change in volume is found by subtracting the original volume from the new volume; the approximate change uses the derivative (differential) of the volume with respect to the radius. The exact change is $(1.4832)\pi$ and the approximate change is $(1.44)\pi$, so the correct option is (D).

   Concept & Intuition

   The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. Here height $h$ is fixed at 9, so $V = 3\pi r^2$. When the radius changes from $r=2$ to $r=2.12$, the exact change is simply $V(2.12) - V(2)$. The approximate change uses the differential $dV = V'(r)\, dr$, which gives a linear approximation valid for small changes. This is a classic application of calculus: the derivative tells us the instantaneous rate of change, and multiplying by the small change in radius gives a quick estimate.

   1. Write the volume formula

      $V = \frac{1}{3}\pi r^2 h$. With $h=9$, this becomes $V = 3\pi r^2$.

   2. Compute the exact change

      Original volume: $V(2) = 3\pi (2)^2 = 12\pi$.

      New volume: $V(2.12) = 3\pi (2.12)^2 = 3\pi \times 4.4944 = 13.4832\pi$.

      Exact change: $\Delta V = 13.4832\pi - 12\pi = (1.4832)\pi$.

   3. Compute the approximate change using differentials

      Derivative: $\frac{dV}{dr} = 6\pi r$. At $r=2$, $\frac{dV}{dr} = 12\pi$.

      Change in radius: $dr = 2.12 - 2 = 0.12$.

      Approximate change: $dV = 12\pi \times 0.12 = 1.44\pi$.

   4. Compare with options

      Exact: $(1.4832)\pi$, Approximate: $(1.44)\pi$. This matches option (D).

   Watch out

   A common mistake is to forget that the height is constant and treat the cone’s volume as if both radius and height change. Here height is fixed, so the formula simplifies nicely.

   Tip

   The differential approximation $dV = V'(r)\, dr$ is essentially the tangent line approximation. It works well when $dr$ is small relative to $r$ — here $0.12$ is 6% of 2, so the approximation is decent but not perfect, which is why the exact value is slightly larger.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $f : [a, b] \to [c, d]$ is a continuous and strictly increasing function, then $\dfrac{d-c}{b-a}$ is (A) Value of the function at a point $t \in (a, b)$ (B) Value of the function at $t \in (a, b)$ such that $f'(t) = 0$ (C) Slope of the tangent drawn to the curve $y = f(t)$ at a point $t \in (c, d)$ (D) Slope of the tangent drawn to the curve $y = f(t)$ at a point $t \in (a, b)$

   ›Reveal solutionSolution

   The expression $\frac{d-c}{b-a}$ is the average rate of change of $f$ over $[a,b]$, and by the Mean Value Theorem it equals $f'(t)$ for some $t \in (a,b)$, which is the slope of the tangent at that point. So the correct option is (D).

   The key idea is to recognize that $\frac{d-c}{b-a}$ is not just any number — it’s the slope of the secant line connecting the endpoints $(a, c)$ and $(b, d)$ of the graph of $f$. Since $f$ is continuous on $[a,b]$ and differentiable (strictly increasing implies differentiable almost everywhere, but we need the Mean Value Theorem which requires differentiability on $(a,b)$ and continuity on $[a,b]$; here $f$ is continuous and strictly increasing, but not necessarily differentiable everywhere — however, the problem likely assumes $f$ is differentiable, or we interpret the slope of the tangent as the derivative where it exists). The Mean Value Theorem guarantees that for a differentiable function, there is some $t$ in $(a,b)$ where the instantaneous slope $f'(t)$ equals that average slope. That instantaneous slope is precisely the slope of the tangent line to $y=f(t)$ at $t$.

   Now let’s walk through the reasoning step by step.

   1. Interpret the expression geometrically. The points $(a, f(a)) = (a, c)$ and $(b, f(b)) = (b, d)$ lie on the curve $y=f(x)$. The slope of the line joining them is

   $\frac{f(b)-f(a)}{b-a} = \frac{d-c}{b-a}.$

   This is the average rate of change of $f$ over $[a,b]$.

   2. Apply the Mean Value Theorem (MVT). Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ (strictly increasing functions on an interval are differentiable almost everywhere, but we assume differentiability here as typical for such problems), the MVT says there exists some $t \in (a,b)$ such that

   $f'(t) = \frac{f(b)-f(a)}{b-a} = \frac{d-c}{b-a}.$

   3. Interpret $f'(t)$ geometrically.

      The derivative $f'(t)$ is the slope of the tangent line to the curve $y=f(x)$ at the point $(t, f(t))$. So $\frac{d-c}{b-a}$ equals the slope of that tangent line at some $t$ in $(a,b)$.

   4. Match with the options.

      • (A) says it’s the value of the function at some $t$ — no, it’s a slope, not a function value.
      • (B) says it’s the value at a point where $f'(t)=0$ — but the average slope is generally not zero, so this is false.
      • (C) says it’s the slope of the tangent at a point $t \in (c,d)$ — but $t$ is in the domain $(a,b)$, not the range $(c,d)$.
      • (D) says it’s the slope of the tangent at a point $t \in (a,b)$ — exactly what we derived.

   Watch out

   A common mistake is to confuse the domain and range: $t$ must be in $(a,b)$ because the MVT gives a point in the interval where the function is defined, not in the codomain.

   Tip

   The MVT is the bridge between average and instantaneous rates of change. Whenever you see $\frac{f(b)-f(a)}{b-a}$, think “secant slope” and then “MVT → tangent slope somewhere inside.”

   ✓Final answer

   The correct option is (D).

   ANSWER: D

10. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.If $y(\cos x)^{\sin x} = (\sin x)^{\sin x}$ then the value of $\dfrac{dy}{dx}$ at $x = \dfrac{\pi}{4}$ is (A) $0$ (B) $1$ (C) $\sqrt{2}$ (D) $\dfrac{\sqrt{3}}{2}$

    ›Reveal solutionSolution

    The relation simplifies to $y=(\tan x)^{\sin x}$; logarithmic differentiation gives $\dfrac{dy}{dx}=\sqrt{2}$ at $x=\dfrac{\pi}{4}$.

    Simplify. From $y(\cos x)^{\sin x}=(\sin x)^{\sin x}$,

    $y=\frac{(\sin x)^{\sin x}}{(\cos x)^{\sin x}}=(\tan x)^{\sin x}.$

    Logarithmic differentiation.

    $\log y=\sin x\,\log(\tan x),$

    $\frac{1}{y}\frac{dy}{dx}=\cos x\,\log(\tan x)+\sin x\cdot\frac{\sec^2x}{\tan x}.$

    Since $\dfrac{\sec^2x}{\tan x}=\dfrac{1}{\sin x\cos x}$, the second term is $\dfrac{1}{\cos x}=\sec x$:

    $\frac{dy}{dx}=y\big(\cos x\,\log(\tan x)+\sec x\big).$

    At $x=\dfrac{\pi}{4}$. Here $\tan\tfrac{\pi}{4}=1\Rightarrow\log 1=0$, and $y=1^{\sin(\pi/4)}=1$, $\sec\tfrac{\pi}{4}=\sqrt2$:

    $\frac{dy}{dx}=1\cdot\big(0+\sqrt2\big)=\sqrt2.$

    ✓Final answer

    $\dfrac{dy}{dx}\Big|_{x=\pi/4}=\sqrt{2}$ — option (C).

11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The approximate value of $\sqrt[3]{730}$ obtained by the application of derivatives is (A) 9.0041 (B) 9.01 (C) 9.006 (D) 9.05

    ›Reveal solutionSolution

    Using the linear approximation formula $f(x+\Delta x) \approx f(x) + f'(x)\,\Delta x$ with $f(x)=\sqrt[3]{x}$, we set $x=729$ and $\Delta x=1$ to get $\sqrt[3]{730} \approx 9 + \frac{1}{3\cdot 9^2} = 9.004115...$, which matches option (A) 9.0041.

    We want an approximate value of $\sqrt[3]{730}$ without a calculator, using derivatives. The key idea is linear approximation (also called the tangent‑line approximation): for a small change $\Delta x$ near a point $x$, the function’s value changes at roughly the rate given by its derivative. So we pick a perfect cube close to 730 — namely $729 = 9^3$ — and use the derivative to estimate the small step from 729 to 730.

    1. Choose the function and the known point.

       Let $f(x) = \sqrt[3]{x} = x^{1/3}$. We know exactly that $f(729) = \sqrt[3]{729} = 9$. We want $f(730) = f(729 + 1)$. Here $\Delta x = 1$, which is small relative to 729, so the linear approximation should be accurate to several decimal places.

    2. Recall the linear approximation formula.

       For a differentiable function $f$,

    $f(x + \Delta x) \approx f(x) + f'(x)\,\Delta x.$

    This is just the first‑order Taylor expansion; geometrically, we follow the tangent line at $x$ instead of the curve.

    3. Compute the derivative at $x=729$.

    $f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}.$

    At $x = 729$, $\sqrt[3]{729^2} = (\sqrt[3]{729})^2 = 9^2 = 81$, so

    $f'(729) = \frac{1}{3 \cdot 81} = \frac{1}{243}.$

    4. Apply the approximation.

    $\sqrt[3]{730} = f(729+1) \approx f(729) + f'(729) \cdot 1 = 9 + \frac{1}{243}.$

    Now $\frac{1}{243}$ is about $0.004115226...$ (since $243 \times 4 = 972$, so $1/243 \approx 0.004115$). Adding to 9 gives

    $\sqrt[3]{730} \approx 9.004115...$

    5. Compare with the options.

       • (A) 9.0041 — matches our result to four decimal places.
       • (B) 9.01 — too large.
       • (C) 9.006 — slightly larger.
       • (D) 9.05 — far too large.

       The approximation $9.004115...$ rounds to $9.0041$ when truncated to four decimal places, so option (A) is correct.

    Watch out

    A common mistake is to use $x=730$ and $\Delta x = -1$ with $f(729) \approx f(730) - f'(730)$, but that requires knowing $f'(730)$ exactly — which we don’t. Always center the approximation at a point where you know both the function value and the derivative exactly.

    Tip

    Notice that $\frac{1}{243} = \frac{1}{3^5}$; you can compute it mentally: $1/243 \approx 0.004115$. This is a neat fraction to remember for cube‑root approximations near perfect cubes.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The vertical angle of a right circular cone is $60^\circ$. If water is being poured into the cone at the rate of $\dfrac{1}{\sqrt{3}}$ m$^3$/min, then the rate (m/min) at which the radius of the water level is increasing when the height of the water level is 3 m is (A) $\dfrac{1}{3\sqrt{3}\pi}$ (B) $\dfrac{1}{9\sqrt{3}\pi}$ (C) $\dfrac{1}{9\pi}$ (D) $\dfrac{1}{3\pi}$

    ›Reveal solutionSolution

    The problem is a classic related‑rates cone problem. Using the vertical angle to relate radius and height, we differentiate the volume formula and substitute the given rate to find $\frac{dr}{dt} = \frac{1}{9\pi}$ when $h=3$. The correct option is (C).

    ---

    Concept & Intuition

    When water is poured into a cone, the volume, radius, and height all change with time. The vertical angle gives a fixed ratio between radius and height: for a right circular cone with vertical angle $60^\circ$, the radius $r$ and height $h$ satisfy $\tan(30^\circ) = r/h$, so $r = h/\sqrt{3}$. This lets us express volume solely in terms of $h$ (or $r$). Then we differentiate with respect to time and plug in the known rate $dV/dt$ to find $dr/dt$ at the instant $h=3$.

    ---

    Step‑by‑step solution

    1. Relate radius and height using the vertical angle The vertical angle is $60^\circ$, meaning the angle at the apex between two opposite generatrices is $60^\circ$. The half‑angle at the apex is $30^\circ$. In the cross‑section (a triangle), $\tan 30^\circ = \frac{r}{h}$. Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we have

    $\frac{r}{h} = \frac{1}{\sqrt{3}} \quad\Rightarrow\quad r = \frac{h}{\sqrt{3}}.$

    2. Write the volume of water in terms of $h$ only Volume of a cone: $V = \frac{1}{3}\pi r^2 h$. Substitute $r = h/\sqrt{3}$:

    $V = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{3} \cdot h = \frac{\pi}{9} h^3.$

    3. Differentiate with respect to time $t$

    $\frac{dV}{dt} = \frac{\pi}{9} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{3} h^2 \frac{dh}{dt}.$

    We are given $\frac{dV}{dt} = \frac{1}{\sqrt{3}}$ m³/min.

    4. Find $\frac{dh}{dt}$ when $h = 3$

    $\frac{1}{\sqrt{3}} = \frac{\pi}{3} (3)^2 \frac{dh}{dt} = \frac{\pi}{3} \cdot 9 \frac{dh}{dt} = 3\pi \frac{dh}{dt}.$

    So

    $\frac{dh}{dt} = \frac{1}{3\pi\sqrt{3}}.$

    5. Find $\frac{dr}{dt}$ using $r = h/\sqrt{3}$ Differentiate: $\frac{dr}{dt} = \frac{1}{\sqrt{3}} \frac{dh}{dt}$. Substitute $\frac{dh}{dt}$:

    $\frac{dr}{dt} = \frac{1}{\sqrt{3}} \cdot \frac{1}{3\pi\sqrt{3}} = \frac{1}{3\pi \cdot 3} = \frac{1}{9\pi}.$

    Tip

    Notice we never needed to solve for $h$ as a function of $t$ — related rates work directly with the instantaneous relationship.

    Watch out

    A common mistake is to forget that $r$ and $h$ are both changing, so you must use the cone’s geometry to eliminate one variable before differentiating. Differentiating $V = \frac13\pi r^2 h$ without substituting the fixed ratio leads to extra terms.

    ---

    ✓Final answer

    The correct option is (C).

    ANSWER: C