Q.Find $\frac{dy}{dx}$ in the following: $x = 2at^2, y = at^4$

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $x^2 + y^2 = t - \dfrac{1}{t}$ and $x^4 + y^4 = t^2 + \dfrac{1}{t^2}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{y}{x}$ (B) $\dfrac{y^2}{x^2}$ (C) $\sqrt{\dfrac{y}{x}}$ (D) $-\dfrac{y}{x}$

   ›Reveal solutionSolution

   Eliminating the parameter $t$ between the two given equations produces the direct relation $x^2y^2=-1$ between $x$ and $y$, whose implicit derivative is $-y/x$.

   Concept and Intuition

   When $x$ and $y$ are both linked to a parameter $t$ through two equations, differentiating each with respect to $t$ separately (and dividing) works, but it is often faster — and here it is exact — to first eliminate $t$ algebraically to get a direct $x$–$y$ relation, then differentiate that implicitly in the ordinary way.

   Step-by-Step Solution

   1. Square the first equation: $(x^2+y^2)^2=\left(t-\dfrac1t\right)^2=t^2-2+\dfrac1{t^2}$, i.e. $x^4+2x^2y^2+y^4=t^2+\dfrac1{t^2}-2$.
   2. The second equation says $x^4+y^4=t^2+\dfrac1{t^2}$. Substitute this in: $\left(t^2+\dfrac1{t^2}\right)+2x^2y^2=t^2+\dfrac1{t^2}-2$.
   3. This forces $2x^2y^2=-2\Rightarrow x^2y^2=-1$ — a $t$-free relation directly linking $x$ and $y$.
   4. Differentiate $x^2y^2=-1$ implicitly w.r.t. $x$: $2xy^2+x^2\cdot2y\dfrac{dy}{dx}=0$.
   5. Divide through by $2xy$ (nonzero): $y+x\dfrac{dy}{dx}=0$.
   6. So $\dfrac{dy}{dx}=-\dfrac{y}{x}$.

   Common Mistakes

   • Trying to differentiate both original equations w.r.t. $t$ and eliminate $dt$ directly — doable but far more error-prone than first eliminating $t$ algebraically.
   • Sign slip in the final division step, giving $+y/x$ instead of $-y/x$.

   ✓Final answer

   The correct option is (D) — $-\dfrac{y}{x}$.

   ANSWER: D

2. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $(a + \sqrt{2}\,b\cos x)(a - \sqrt{2}\,b\cos y) = a^2 - b^2$ where $a > b > 0$, then at $\left(\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)$, $\dfrac{dy}{dx} =$ (A) $\dfrac{a + b}{a - b}$ (B) $\dfrac{a - b}{a + b}$ (C) $\dfrac{a - 2b}{a + 2b}$ (D) $\dfrac{2a + b}{2a - b}$

   ›Reveal solutionSolution

   Expand the product, simplify by dividing by the common factor $b$, then implicitly differentiate and evaluate at $x=y=\pi/4$. Answer: $\dfrac{a-b}{a+b}$.

   Concept and Intuition

   The given relation looks intimidating as a product, but expanding it cancels the $a^2$ on both sides (since the RHS is $a^2-b^2$) and leaves a much simpler equation relating $\cos x,\cos y$, and $\cos x\cos y$. From there it's routine implicit differentiation; the special evaluation point $x=y=\pi/4$ is chosen because $\sin$ and $\cos$ coincide there, which cancels neatly.

   Step-by-Step Solution

   1. Expand: $a^2 - a\sqrt2 b\cos y + a\sqrt2 b\cos x - 2b^2\cos x\cos y = a^2-b^2$.
   2. Cancel $a^2$ from both sides: $\sqrt2 ab(\cos x-\cos y) - 2b^2\cos x\cos y = -b^2$.
   3. Divide through by $b$ (nonzero): $\sqrt2 a(\cos x-\cos y) - 2b\cos x\cos y + b = 0$.
   4. Differentiate implicitly w.r.t. $x$ (treat $y=y(x)$):

   $\sqrt2 a\left(-\sin x + \sin y\cdot y'\right) - 2b\left(-\sin x\cos y - \cos x\sin y\cdot y'\right) = 0.$

   5. Group $y'$ terms: $y'\left(\sqrt2 a\sin y + 2b\cos x\sin y\right) = \sqrt2 a\sin x - 2b\sin x\cos y$.
   6. So $y' = \dfrac{\sin x(\sqrt2 a - 2b\cos y)}{\sin y(\sqrt2 a + 2b\cos x)}$.
   7. At $x=y=\dfrac\pi4$: $\sin x=\sin y=\cos x=\cos y=\dfrac1{\sqrt2}$. Substitute:

   $y'=\frac{\frac1{\sqrt2}\left(\sqrt2 a - \frac{2b}{\sqrt2}\right)}{\frac1{\sqrt2}\left(\sqrt2 a + \frac{2b}{\sqrt2}\right)}=\frac{\sqrt2 a-\sqrt2 b}{\sqrt2 a+\sqrt2 b}=\frac{a-b}{a+b}.$

   Common Mistakes

   • Forgetting to expand the product first and instead trying to implicitly differentiate the product form directly, which is far more error-prone.
   • Sign slips when differentiating $\cos x\cos y$ as a product (needs the product rule with $y'$ attached only to the $\cos y$ factor).

   ✓Final answer

   The correct option is (B) — $\dfrac{a - b}{a + b}$.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $x^2+y^2=t+\dfrac{1}{t}$ and $x^4+y^4=t^2+\dfrac{1}{t^2}$, then $x^3 y\dfrac{dy}{dx}=$ (A) -1 (B) 0 (C) 1 (D) 2

   ›Reveal solutionSolution

   The two given relations force $x^2y^2=1$ (i.e. $xy$ is constant), from which $x^3y\,dy/dx=-1$.

   Concept and Intuition

   Rather than solving for $x,y$ in terms of $t$ explicitly, combine the two given equations algebraically (square the first, subtract the second) to eliminate $t$ entirely and land on a simple constant-product relation between $x$ and $y$.

   Step-by-Step Solution

   1. Square the first relation: $(x^2+y^2)^2=\left(t+\dfrac1t\right)^2=t^2+2+\dfrac{1}{t^2}$, i.e.

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

   2. The second given relation is $x^4+y^4=t^2+\dfrac{1}{t^2}$.
   3. Subtract: $2x^2y^2 = \left(t^2+2+\dfrac1{t^2}\right)-\left(t^2+\dfrac1{t^2}\right)=2$, so $x^2y^2=1$.
   4. This means $xy=\pm1$, a constant independent of $t$. Differentiate $xy=\text{const}$ implicitly:

   $x\frac{dy}{dx}+y=0 \Rightarrow \frac{dy}{dx}=-\frac{y}{x}$

   5. Then $x^3y\dfrac{dy}{dx}=x^3y\left(-\dfrac{y}{x}\right)=-x^2y^2=-1$ (using step 3).

   Common Mistakes

   • Trying to solve for $x,y$ individually in terms of $t$ (unnecessarily complicated) instead of eliminating $t$ algebraically.
   • Sign error when substituting $dy/dx=-y/x$.

   ✓Final answer

   The correct option is (A) — -1.

   ANSWER: A

4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $\dfrac{y}{x}\cos^4\alpha + \dfrac{x}{y}\sin^4\alpha = 2\sin^2\alpha \cdot \cos^2\alpha$, then $\dfrac{dy}{dx} =$ (A) $\sin^3\alpha \cos\alpha$ (B) $\sin^2\alpha \cos^2\alpha$ (C) $\dfrac{\sin^2\alpha}{\cos^2\alpha}$ (D) $\sin\alpha \cos^3\alpha$

   ›Reveal solutionSolution

   The given relation is secretly a perfect square in disguise; it forces $y=x\tan^2\alpha$, so $dy/dx=\tan^2\alpha$.

   Concept and Intuition

   Rather than differentiating implicitly right away, it pays to recognise the algebraic structure first. Multiplying by $xy$ converts the equation into a quadratic in $x$ and $y$ that factors as a perfect square, revealing $y/x$ is actually a constant (independent of $x$), which makes the derivative trivial.

   Step-by-Step Solution

   1. Start from $\dfrac{y}{x}\cos^4\alpha+\dfrac{x}{y}\sin^4\alpha = 2\sin^2\alpha\cos^2\alpha$.
   2. Multiply both sides by $xy$: $y^2\cos^4\alpha + x^2\sin^4\alpha = 2xy\sin^2\alpha\cos^2\alpha$.
   3. Rearrange: $y^2\cos^4\alpha - 2xy\sin^2\alpha\cos^2\alpha + x^2\sin^4\alpha = 0$.
   4. This is $(y\cos^2\alpha - x\sin^2\alpha)^2 = 0$, so $y\cos^2\alpha = x\sin^2\alpha$, i.e. $y = x\tan^2\alpha$.
   5. Since $\tan^2\alpha$ is a constant (does not depend on $x$), $\dfrac{dy}{dx}=\tan^2\alpha = \dfrac{\sin^2\alpha}{\cos^2\alpha}$.

   Common Mistakes

   • Jumping straight into implicit differentiation of the original messy relation instead of spotting the perfect square — much harder and error-prone.
   • Forgetting that $y=x\tan^2\alpha$ makes this literally a line through the origin, so the derivative is just its slope.

   ✓Final answer

   The correct option is (C) — $\dfrac{\sin^2\alpha}{\cos^2\alpha}$.

   ANSWER: C

5. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $3\sin xy + 4\cos xy = 5$, then $\dfrac{dy}{dx}$ is equal to ____ (A) $\dfrac{3\sin xy + 4\cos xy}{3\cos xy - 4\sin xy}$ (B) $\dfrac{3\cos xy + 4\sin xy}{4\cos xy - 3\sin xy}$ (C) $\dfrac{-y}{x}$ (D) $\dfrac{x}{y}$

   ›Reveal solutionSolution

   Since $3\sin\theta+4\cos\theta$ has maximum value exactly $5$ (as $\sqrt{3^2+4^2}=5$), equating it to $5$ forces $xy$ to be a fixed constant, so implicit differentiation of $xy=c$ gives $dy/dx=-y/x$.

   Concept and Intuition

   $a\sin\theta+b\cos\theta$ always has amplitude $\sqrt{a^2+b^2}$ — here $\sqrt{9+16}=5$. So the equation $3\sin(xy)+4\cos(xy)=5$ isn't a "generic" implicit curve; it can only be satisfied when the expression sits exactly at its maximum, which happens at one specific angle. That pins $xy$ to a single constant value, turning a trigonometric-looking implicit relation into the much simpler $xy=\text{const}$.

   Step-by-Step Solution

   1. Note $3^2+4^2=25=5^2$, so $3\sin\theta+4\cos\theta$ has maximum value $5$, attained only when $\theta$ equals the specific angle $\phi=\tan^{-1}(3/4)$ (mod $2\pi$).
   2. The given equation demands $3\sin(xy)+4\cos(xy)=5$, i.e. the maximum — so $xy=\phi$ is fixed, a constant independent of which point on the curve we pick.
   3. Differentiate $xy=\text{constant}$ implicitly: $\dfrac{d}{dx}(xy)=0 \Rightarrow y+x\dfrac{dy}{dx}=0$.
   4. Solve: $\dfrac{dy}{dx}=-\dfrac{y}{x}$.

   Common Mistakes

   • Differentiating the trig terms directly (product/chain rule on $\sin(xy),\cos(xy)$) without noticing the amplitude equals the RHS, missing the much simpler $xy=$const shortcut and getting stuck in messy algebra.
   • Sign error in the implicit derivative of $xy$.

   ✓Final answer

   The correct option is (C) — $\dfrac{-y}{x}$.

   ANSWER: C

6. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $x^2 \tan^{-1}\dfrac{y}{x} - y^2 \tan^{-1}\dfrac{x}{y} = k$, then $\left(\dfrac{dy}{dx}\right)_{(1,1)} =$ (A) $0$ (B) $\pi/4$ (C) $1$ (D) $\pi/2$

   ›Reveal solutionSolution

   Implicit differentiation of $x^2\tan^{-1}(y/x)-y^2\tan^{-1}(x/y)=k$ evaluated at $(1,1)$ gives $dy/dx=1$.

   Concept and Intuition

   Differentiate both terms using product and chain rules, being careful with the derivatives of $\tan^{-1}(y/x)$ and $\tan^{-1}(x/y)$ with respect to $x$ (treating $y$ as a function of $x$).

   Step-by-Step Solution

   1. $\dfrac{d}{dx}\left[x^2\tan^{-1}\dfrac{y}{x}\right] = 2x\tan^{-1}\dfrac{y}{x} + x^2\cdot\dfrac{y'x-y}{x^2+y^2}$.
   2. $\dfrac{d}{dx}\left[y^2\tan^{-1}\dfrac{x}{y}\right] = 2yy'\tan^{-1}\dfrac{x}{y} + y^2\cdot\dfrac{y-xy'}{x^2+y^2}$.
   3. Setting derivative of LHS $=0$ and evaluating at $(1,1)$, where $\tan^{-1}(1)=\pi/4$ for both terms: $2\cdot\dfrac\pi4 + \dfrac{y'-1}{2} - 2y'\cdot\dfrac\pi4 - \dfrac{1-y'}{2} = 0$.
   4. Combine the two $/2$ terms: $\dfrac{y'-1}{2}-\dfrac{1-y'}{2}=y'-1$.
   5. So: $\dfrac\pi2 + (y'-1) - \dfrac\pi2 y' = 0 \Rightarrow y'\left(1-\dfrac\pi2\right) = 1-\dfrac\pi2 \Rightarrow y'=1$.

   Common Mistakes

   • Sign slips differentiating $\tan^{-1}(x/y)$ with respect to $x$ (the $y'$ appears with a minus sign inside).

   ✓Final answer

   The correct option is (C) — $1$.

   ANSWER: C

7. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The length of the tangent drawn at the point $p\left(\dfrac{\pi}{4}\right)$ on the curve $x^{2/3}+y^{2/3}=2^{2/3}$ is (A) $\dfrac{2}{3}$ (B) 1 (C) $\dfrac{4}{3}$ (D) 2

   ›Reveal solutionSolution

   A length-of-tangent computation on the astroid at its symmetric point $\theta=\pi/4$; the length comes out to exactly $1$.

   Concept and Intuition

   The astroid $x^{2/3}+y^{2/3}=a^{2/3}$ can be parametrized as $x=a\cos^3\theta,\ y=a\sin^3\theta$, which automatically satisfies the equation. "The point $P(\pi/4)$" refers to this parameter value. The length of the tangent line segment (from the point of tangency to where it meets the $x$-axis) is given by the standard formula $\left|\dfrac{y\sqrt{1+y'^2}}{y'}\right|$.

   Step-by-Step Solution

   1. Here $a=2$ (since $2^{2/3}=a^{2/3}$), so $x=2\cos^3\theta$, $y=2\sin^3\theta$.
   2. At $\theta=\pi/4$: $\cos(\pi/4)=\sin(\pi/4)=1/\sqrt2$, so $x=y=2\left(\dfrac{1}{\sqrt2}\right)^3=2\cdot\dfrac{1}{2\sqrt2}=\dfrac{1}{\sqrt2}$.
   3. Differentiate $x^{2/3}+y^{2/3}=a^{2/3}$ implicitly: $\dfrac23x^{-1/3}+\dfrac23y^{-1/3}y'=0\Rightarrow y'=-\left(\dfrac{y}{x}\right)^{1/3}$.
   4. Since $x=y$ here, $y'=-1$.
   5. Length of tangent $=\left|\dfrac{y\sqrt{1+y'^2}}{y'}\right|=\dfrac{(1/\sqrt2)\sqrt{1+1}}{1}=\dfrac{(1/\sqrt2)\sqrt2}{1}=1$.

   Common Mistakes

   • Confusing length of tangent with length of normal (the normal formula has no division by $y'$).
   • Sign errors treating $y'$ as $+1$ instead of $-1$ (though this specific length formula uses $|y'|$, so it doesn't affect the final magnitude here).

   ✓Final answer

   The correct option is (B) — 1.

   ANSWER: B

8. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If the tangent to the curve $xy + ax + by = 0$ at (1,1) makes an angle $\tan^{-1}2$ with X-axis, then $\dfrac{ab}{a+b} =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Use the point lying on the curve plus the given slope at that point to get two linear equations in $a,b$. Answer: $\dfrac{ab}{a+b}=2$.

   Concept and Intuition

   A point on an implicitly-defined curve gives one equation relating the curve's parameters; the slope condition (via implicit differentiation) gives a second. Two equations, two unknowns.

   Step-by-Step Solution

   1. Since $(1,1)$ lies on $xy+ax+by=0$: $1+a+b=0 \Rightarrow a+b=-1$.
   2. Differentiate implicitly: $y+xy'+a+by'=0 \Rightarrow y'(x+b) = -(y+a) \Rightarrow y'=-\dfrac{y+a}{x+b}$.
   3. At $(1,1)$, slope $=\tan^{-1}2 \Rightarrow y'=2$: $-\dfrac{1+a}{1+b}=2 \Rightarrow -(1+a)=2(1+b) \Rightarrow -1-a=2+2b \Rightarrow a+2b=-3$.
   4. Subtract $a+b=-1$ from $a+2b=-3$: $b=-2$. Then $a=-1-b=-1-(-2)=1$.
   5. Verify: curve $xy+x-2y=0$; at $(1,1)$: $1+1-2=0$ ✓; slope $y'=-\dfrac{1+1}{1-2}=-\dfrac{2}{-1}=2$ ✓.
   6. $\dfrac{ab}{a+b}=\dfrac{(1)(-2)}{-1}=\dfrac{-2}{-1}=2$.

   Common Mistakes

   • Sign error differentiating $by$ term (giving $by'$, not $b'y$ — $b$ is a constant here).
   • Forgetting to verify the solution satisfies both original conditions.

   ✓Final answer

   The correct option is (B) — $2$.

   ANSWER: B

9. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\mathrm{Tan}^{-1}x^2 + \mathrm{Tan}^{-1}y^2 = \dfrac{\pi}{2}$, then $\left(\dfrac{dy}{dx}\right)_{(-1,2)} =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $-\dfrac{1}{2}$

   ›Reveal solutionSolution

   Reducing to $y^2=x^{-2}$ gives $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$, which at $(-1,2)$ equals $\dfrac12$.

   Concept and Intuition

   If $\mathrm{Tan}^{-1}a+\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}$ with $a,b>0$, then $\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}-\mathrm{Tan}^{-1}a=\mathrm{Cot}^{-1}a$, so $b=\dfrac{1}{a}$. Applying this to $a=x^2,\ b=y^2$ collapses the relation into an algebraic one.

   Step-by-Step Solution

   1. From $\mathrm{Tan}^{-1}x^2+\mathrm{Tan}^{-1}y^2=\dfrac{\pi}{2}$ we get $y^2=\dfrac{1}{x^2}=x^{-2}$.
   2. Differentiate: $2y\,\dfrac{dy}{dx}=-2x^{-3}$.
   3. Hence $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$.
   4. At $(-1,2)$: $x^3=(-1)^3=-1$, so $\dfrac{dy}{dx}=-\dfrac{1}{(-1)(2)}=\dfrac{1}{2}$.

   Common Mistakes

   • Getting the sign wrong by mishandling $x^3=-1$ at $x=-1$.
   • Trying to differentiate the arctangents directly without first simplifying, which is far messier.

   ✓Final answer

   The correct option is (C) — $\dfrac{dy}{dx}=\dfrac{1}{2}$.

   ANSWER: C

10. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If $\sqrt{x-xy}+\sqrt{y-xy}=1$, then $\dfrac{dy}{dx}=$ (A) $-\sqrt{\dfrac{y-y^2}{x-x^2}}$ (B) $-\sqrt{\dfrac{1-y^2}{1-x^2}}$ (C) $-\sqrt{\dfrac{1-y}{1-x}}$ (D) $-\sqrt{\dfrac{x-y}{x+y}}$

    ›Reveal solutionSolution

    Squaring the constraint reveals that it forces $x+y=1$ identically, so $\dfrac{dy}{dx}=-1$ throughout; matching this against the options singles out $-\sqrt{\dfrac{y-y^2}{x-x^2}}$, since it equals $-1$ for every point satisfying $y=1-x$.

    Concept and Intuition

    Rather than blindly grinding through implicit differentiation of two square roots, it pays to first understand the curve itself. Squaring $\sqrt{x-xy}+\sqrt{y-xy}=1$ carefully (using $s=x+y,\,p=xy$) collapses to a perfect square equalling zero, revealing that the relation is nothing but the straight line $x+y=1$. Once we know that, $\dfrac{dy}{dx}=-1$ is immediate, and we just need to find which option reduces to $-1$ on this line.

    Step-by-Step Solution

    1. Square the given equation:

    $x(1-y)+y(1-x)+2\sqrt{xy(1-x)(1-y)} = 1$

    $x+y-2xy+2\sqrt{xy(1-x)(1-y)} = 1$

    2. Let $s=x+y,\ p=xy$. Then:

    $2\sqrt{p(1-x)(1-y)} = 1-s+2p$

    Note $(1-x)(1-y)=1-s+p$. Squaring again:

    $4p(1-s+p) = (1-s+2p)^2$

    Let $q=1-s$. Expanding both sides: LHS $=4pq+4p^2$; RHS $=q^2+4pq+4p^2$. So $0=q^2$, i.e. $q=0$, i.e. $s=1$.

    3. Hence $x+y=1$ is forced — the given relation is the line $y=1-x$ (restricted to the domain where the square roots are real, $0\le x,y\le1$).

    4. Differentiating $y=1-x$ directly: $\dfrac{dy}{dx}=-1$.

    5. Check which option gives $-1$ identically along $y=1-x$:

    • (A): $y-y^2=y(1-y)$. Substituting $y=1-x$: $y(1-y)=(1-x)\cdot x=x-x^2$. So the ratio is exactly $1$, and $-\sqrt1=-1$ for every $x$. ✓
    • (B): at $x=0.3,y=0.7$: $\dfrac{1-y^2}{1-x^2}=\dfrac{0.51}{0.91}\ne1$. ✗
    • (C): at $x=0.3,y=0.7$: $\dfrac{1-y}{1-x}=\dfrac{0.3}{0.7}\ne1$. ✗
    • (D): at $x=0.3,y=0.7$: $\dfrac{x-y}{x+y}=\dfrac{-0.4}{1}<0$, square root undefined. ✗

    Common Mistakes

    • Grinding through raw implicit differentiation without noticing the constraint simplifies to a straight line — leads to messy, error-prone algebra.
    • Checking the options only at the symmetric point $x=y=0.5$, where several options coincidentally also give $-1$ — a second, asymmetric test point is needed to discriminate.
    • Sign error in expanding $(1-s+2p)^2$ vs $4p(1-s+p)$.

    ✓Final answer

    The correct option is (A) — $-\sqrt{\dfrac{y-y^2}{x-x^2}}$.

    ANSWER: A

11. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.If $y = 4x - 6$ is a tangent to the curve $y^2 = ax^4 + b$ at $(3, 6)$, then the values of $a$ and $b$ are (A) $a = \dfrac{4}{9}$ & $b = \dfrac{-4}{9}$ (B) $a = 0$ & $b = \dfrac{4}{9}$ (C) $a = \dfrac{-4}{9}$ & $b = \dfrac{-4}{9}$ (D) $a = \dfrac{4}{9}$ & $b = 0$

    ›Reveal solutionSolution

    This tests using both the point-on-curve condition and the matching-slope condition for a tangent line to solve for two unknown parameters. Answer: $a=\tfrac49,\ b=0$.

    Concept and Intuition

    A tangent line to a curve at a given point must (1) pass through that point (the point lies on the curve) and (2) have the same slope as the curve's derivative there. Two conditions determine the two unknowns $a,b$.

    Step-by-Step Solution

    1. Since $(3,6)$ lies on $y^2=ax^4+b$: $6^2 = a\cdot 3^4+b \Rightarrow 36 = 81a+b$. — (i)
    2. Differentiate implicitly: $2y\,y' = 4ax^3 \Rightarrow y' = \dfrac{2ax^3}{y}$.
    3. At $(3,6)$: $y'(3,6) = \dfrac{2a(27)}{6} = 9a$.
    4. The tangent line $y=4x-6$ has slope $4$, so $9a=4 \Rightarrow a=\dfrac49$.
    5. Substitute into (i): $36 = 81\left(\dfrac49\right)+b = 36+b \Rightarrow b=0$.

    Common Mistakes

    • Only using the point-on-curve condition and forgetting to also match the derivative to the tangent's slope (or vice versa) — both are needed to pin down two unknowns.

    ✓Final answer

    The correct option is (D) — $a=\dfrac49$ & $b=0$.

    ANSWER: D

12. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
    Q.If $y^{y^{y^{\cdot^{\cdot^{\cdot^{\infty}}}}}} = \log\{x + \log\{x + \cdots\}\}$, then $\dfrac{dy}{dx}$ at $x = e^2 - 2$, $y = \sqrt{2}$ equals _____ (A) $\dfrac{\log 2}{2\sqrt{2}(e^2 - 1)}$ (B) $\dfrac{1 - \log 2}{2\sqrt{2}(e^2 - 1)}$ (C) $\dfrac{\sqrt{2}(1 - \log 2)}{e^2 - 1}$ (D) $\dfrac{\log 2}{\sqrt{2}(e^2 - 1)}$

    ›Reveal solutionSolution

    Both sides define the same implicit quantity $u$ via a self-referential equation; differentiate each side's defining equation implicitly and combine using the chain rule. The answer is (B).

    Concept and Intuition

    The infinite power tower $y^{y^{y^{\cdots}}}=u$ satisfies the self-consistency equation $u=y^u$ (the tower "regenerates" itself). Likewise the infinite nested logarithm $\log\{x+\log\{x+\cdots\}\}=u$ satisfies $u=\log(x+u)$. Since the problem states these two quantities are equal (both equal to the same $u$), $u$ is implicitly a common function linking $x$ and $y$; differentiating each defining relation gives $du/dy$ and $du/dx$, and the chain rule combines them into $dy/dx$.

    Step-by-Step Solution

    1. Verify $u=2$ at the given point. Nested log: $u=\log(x+u)$ at $x=e^2-2$: try $u=2$: $\log(e^2-2+2)=\log(e^2)=2$ ✓. Tower: $u=y^u$ at $y=\sqrt2$: try $u=2$: $(\sqrt2)^2=2$ ✓. Both consistent with $u=2$.
    2. Differentiate the tower relation $u=y^u$ w.r.t. $y$. Take $\log$: $\log u = u\log y$. Differentiate: $\dfrac1u\dfrac{du}{dy} = \dfrac{du}{dy}\log y + \dfrac uy$ $\Rightarrow \left(\dfrac1u-\log y\right)\dfrac{du}{dy}=\dfrac uy \Rightarrow \dfrac{du}{dy}=\dfrac{u^2}{y(1-u\log y)}$. At $y=\sqrt2,\ u=2$: $\dfrac{du}{dy}=\dfrac{4}{\sqrt2(1-2\log\sqrt2)}=\dfrac{4}{\sqrt2(1-\ln2)}=\dfrac{2\sqrt2}{1-\ln2}$.
    3. Differentiate the nested-log relation $u=\log(x+u)$ w.r.t. $x$: $\dfrac{du}{dx}=\dfrac{1+\frac{du}{dx}}{x+u} \Rightarrow \dfrac{du}{dx}(x+u-1)=1 \Rightarrow \dfrac{du}{dx}=\dfrac1{x+u-1}$. At $x=e^2-2,\ u=2$: $x+u-1=e^2-1$, so $\dfrac{du}{dx}=\dfrac1{e^2-1}$.
    4. Combine via chain rule: since both equal the same $u$, $\dfrac{dy}{dx}=\dfrac{du/dx}{du/dy}=\dfrac{\;\dfrac1{e^2-1}\;}{\dfrac{2\sqrt2}{1-\ln2}}=\dfrac{1-\ln2}{2\sqrt2(e^2-1)}$.

    Common Mistakes

    • Forgetting to verify $u=2$ actually satisfies both self-consistency equations before differentiating.
    • Sign/algebra slips in the implicit differentiation of $u=y^u$ (logarithmic differentiation) — a very common source of error in infinite-tower problems.

    ✓Final answer

    The correct option is (B) — $\dfrac{1 - \log 2}{2\sqrt{2}(e^2 - 1)}$.

    ANSWER: B