Q.Integrate the following function: $\frac{\sin^2 x}{1 + \cos x}$

Showing the 12 most recent of 34 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $\theta$ is an acute angle through which the coordinate axes are to be rotated about the origin in anti-clockwise direction to remove $xy$-term from the equation $4x^2 + 3xy + y^2 + 1 = 0$, then $(1 + \tan \theta)^2 =$ (A) $2$ (B) $4$ (C) $1$ (D) $3$

   ›Reveal solutionSolution

   To eliminate the $xy$-term in a rotated conic, we use $\cot 2\theta = \frac{A-C}{B}$. Here $A=4$, $B=3$, $C=1$, so $\cot 2\theta = 1$, giving $\tan\theta = \sqrt{2}-1$ and $(1+\tan\theta)^2 = 2$. The correct option is (A).

   ---

   Concept & Intuition

   When we rotate the coordinate axes by an angle $\theta$, the $xy$-term in a general quadratic $Ax^2 + Bxy + Cy^2 + \dots = 0$ disappears if $\theta$ satisfies

   $\cot 2\theta = \frac{A - C}{B}.$

   This formula comes from the transformation of the quadratic form under rotation: the coefficient of $x'y'$ becomes $B\cos 2\theta - (A-C)\sin 2\theta$, and setting it to zero yields the condition above. For an acute $\theta$, we take the positive root for $\tan\theta$ from the double-angle identity.

   ---

   Step-by-step solution

   1. Identify coefficients

      The given equation is $4x^2 + 3xy + y^2 + 1 = 0$.

      Here $A = 4$, $B = 3$, $C = 1$. The constant term $1$ does not affect the rotation condition.

   2. Apply the rotation condition

      To remove the $xy$-term, we need

   $\cot 2\theta = \frac{A - C}{B} = \frac{4 - 1}{3} = \frac{3}{3} = 1.$

   So $\cot 2\theta = 1$, meaning $2\theta = 45^\circ$ (since $\theta$ is acute, $2\theta$ is acute as well). Thus $\theta = 22.5^\circ$.

   3. Find $\tan\theta$ We know $\cot 2\theta = 1 \implies \tan 2\theta = 1$. Use the double-angle identity:

   $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = 1.$

   Let $t = \tan\theta$. Then

   $$$$

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $x = \sin^{-1}(\cos t)$ and $y = \tan^{-1}(\cos t)$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{\cos x}{1 + \sin^2 x}$ (B) $\dfrac{\cos t}{1 + \sin^2 t}$ (C) $\dfrac{\sin t}{1 + \sin^2 t}$ (D) $\dfrac{\sin x}{1 + \cos^2 x}$

   ›Reveal solutionSolution

   The key is to express both $x$ and $y$ in terms of $t$, then use the chain rule to find $\frac{dy}{dx}$ as a function of $x$, matching option (A).

   We are given:

   $x = \sin^{-1}(\cos t), \quad y = \tan^{-1}(\cos t).$

   We need $\frac{dy}{dx}$.

   Concept and intuition:

   Both $x$ and $y$ are functions of $t$, so we can compute $\frac{dy}{dt}$ and $\frac{dx}{dt}$ separately, then use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. The trick is to simplify $\sin^{-1}(\cos t)$ into a nicer algebraic form in terms of $t$, which will make differentiation cleaner. Then we express the final answer in terms of $x$ (since the options involve $x$ or $t$).

   ---

   1. Simplify $x$ Recall: $\sin^{-1}(\cos t) = \frac{\pi}{2} - t$ for $t$ in $[0, \pi]$, but more generally we can use the identity $\sin^{-1}(\cos t) = \frac{\pi}{2} - |t|$? Actually, a cleaner approach: Let $u = \cos t$. Then $x = \sin^{-1}(u)$. But we can also write $\cos t = \sin\left(\frac{\pi}{2} - t\right)$, so

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

   For the principal range of $\sin^{-1}$, we need $\frac{\pi}{2} - t \in [-\pi/2, \pi/2]$, i.e., $t \in [0, \pi]$. Assuming this, we get

   $x = \frac{\pi}{2} - t.$

   Hence, $t = \frac{\pi}{2} - x$.

   2. Differentiate $x$ and $y$ with respect to $t$ From $x = \frac{\pi}{2} - t$, we have

   $\frac{dx}{dt} = -1.$

   For $y = \tan^{-1}(\cos t)$, differentiate:

   $\frac{dy}{dt} = \frac{1}{1 + (\cos t)^2} \cdot (-\sin t) = -\frac{\sin t}{1 + \cos^2 t}.$

   3. Find $\frac{dy}{dx}$ Using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-\frac{\sin t}{1 + \cos^2 t}}{-1} = \frac{\sin t}{1 + \cos^2 t}.$ …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $y = \sin^{-1} \left( \frac{1 - \cos 2x}{1 + \sin^4 x} \right)$ then $\frac{dy}{dx} =$ (A) $\frac{2 \cos 2x}{1 + \sin^4 x}$ (B) $\frac{2 \sin 2x}{1 + \sin^4 x}$ (C) $\frac{2 \cos 2x}{1 + \sin^8 x}$ (D) $\frac{2 \sin 2x}{1 + \sin^8 x}$

   ›Reveal solutionSolution

   The key is to simplify the argument of the inverse sine using trigonometric identities before differentiating. After simplification, the derivative becomes $\frac{2 \sin 2x}{1 + \sin^4 x}$, which matches option (B).

   ---

   We start with

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

   Concept and intuition

   The expression inside the inverse sine looks messy, but the numerator $1 - \cos 2x$ is a classic form that simplifies to $2\sin^2 x$. That’s a huge clue: the whole fraction might simplify to something like $\frac{2\sin^2 x}{1 + \sin^4 x}$, which is a perfect candidate for the identity $\sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2\tan^{-1} t$ (for $|t|\le 1$). That substitution makes differentiation straightforward.

   ---

   Step-by-step

   1. Simplify the numerator Using the double-angle identity:

   $1 - \cos 2x = 2\sin^2 x.$

   So the argument becomes

   $\frac{2\sin^2 x}{1 + \sin^4 x}.$

   2. Recognize a standard form Let $t = \sin^2 x$. Then the argument is

   $\frac{2t}{1 + t^2}.$

   This is exactly the form $\frac{2t}{1+t^2}$, which appears in the identity

   $\sin^{-1}\left( \frac{2t}{1+t^2} \right) = 2\tan^{-1} t \quad \text{for } |t| \le 1.$

   Since $t = \sin^2 x \in [0,1]$, the condition holds. Therefore

   $y = 2\tan^{-1}(\sin^2 x).$

   3. Differentiate Differentiate $y = 2\tan^{-1}(\sin^2 x)$:

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

   The derivative of $\sin^2 x$ is $2\sin x \cos x = \sin 2x$. So …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\tan^{-1} \dfrac{1}{2} + \tan^{-1} \dfrac{1}{3} + \tan^{-1} \dfrac{2}{3} + \tan^{-1} \dfrac{1}{5} =$ (A) $\dfrac{\pi}{4}$ (B) $\tan^{-1} \left( \dfrac{7}{11} \right)$ (C) $\dfrac{\pi}{2}$ (D) $\tan^{-1} \left( \dfrac{23}{24} \right)$

   ›Reveal solutionSolution

   The sum of the four arctangents simplifies to $\pi/4$ by repeatedly applying the tangent addition formula and noticing that the combined angle lies in the first quadrant. The correct option is (A).

   We are asked to evaluate

   $\tan^{-1}\frac12 + \tan^{-1}\frac13 + \tan^{-1}\frac23 + \tan^{-1}\frac15.$

   The key idea is to combine arctangents two at a time using the formula

   $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab},$

   but we must always check the quadrant of the resulting angle. Since all given fractions are positive and less than 1, each arctangent lies in $(0,\pi/4)$. Their sum will be less than $\pi$, so we can safely use the formula without worrying about adding $\pi$ corrections.

   1. Combine the first two terms Let $\alpha = \tan^{-1}\frac12$ and $\beta = \tan^{-1}\frac13$. Then

   $\tan(\alpha+\beta) = \frac{\frac12+\frac13}{1-\frac12\cdot\frac13} = \frac{\frac{5}{6}}{1-\frac16} = \frac{5/6}{5/6} = 1.$

   Since $\alpha,\beta < \pi/4$, their sum is less than $\pi/2$ and positive, so

   $\alpha+\beta = \tan^{-1}1 = \frac{\pi}{4}.$

   2. Combine the next two terms Let $\gamma = \tan^{-1}\frac23$ and $\delta = \tan^{-1}\frac15$. Then

   $\tan(\gamma+\delta) = \frac{\frac23+\frac15}{1-\frac23\cdot\frac15} = \frac{\frac{10}{15}+\frac{3}{15}}{1-\frac{2}{15}} = \frac{13/15}{13/15} = 1.$

   Again, both angles are less than $\pi/4$, so their sum is also $\pi/4$.

   3. Add the two results Now we have

   $\left(\tan^{-1}\frac12+\tan^{-1}\frac13\right) + \left(\tan^{-1}\frac23+\tan^{-1}\frac15\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.$

   That would suggest the answer is $\pi/2$, but wait — we must check if the sum of all four original angles really equals $\pi/2$ or if there is a subtlety.

   Watch out

   The sum of two arctangents each equal to $\pi/4$ is $\pi/2$, but the original four angles are all less than $\pi/4$, so their total is less than $\pi$. However, $\pi/2$ is a valid possibility. But let’s verify by combining all four at once to be safe.

   4. Combine all four directly Let $S = \tan^{-1}\frac12 + \tan^{-1}\frac13 + \tan^{-1}\frac23 + \tan^{-1}\frac15$. …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\tan^{-1} \dfrac{3}{5} + \tan^{-1} \dfrac{6}{41} + \tan^{-1} \dfrac{9}{191} =$ (A) $\tan^{-1} \dfrac{9}{10}$ (B) $\tan^{-1} \dfrac{18}{19}$ (C) $\tan^{-1} \dfrac{3}{191}$ (D) $\tan^{-1} \dfrac{6}{205}$

   ›Reveal solutionSolution

   Adding the arctangents two at a time gives $\tan^{-1}\dfrac{9}{10}$.

   Solution

   Use $\tan^{-1}p+\tan^{-1}q=\tan^{-1}\dfrac{p+q}{1-pq}$.

   First two terms:

   $\tan^{-1}\frac{3}{5}+\tan^{-1}\frac{6}{41}=\tan^{-1}\frac{\frac{3}{5}+\frac{6}{41}}{1-\frac{3}{5}\cdot\frac{6}{41}}=\tan^{-1}\frac{123+30}{205-18}=\tan^{-1}\frac{153}{187}=\tan^{-1}\frac{9}{11}.$

   Add the third term: …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $(h, k)$ is the new origin to be chosen to eliminate first degree terms from the equation $S \equiv 2x^2 - xy - y^2 - 3x + 3y = 0$ by translation and if $\theta$ is the angle with which the axes are to be rotated about the origin in anticlockwise direction to eliminate $xy$-term from $S = 0$, then $\tan 2\theta =$ (A) $h + k$ (B) $h - k$ (C) $hk$ (D) $-\dfrac{h}{3k}$

   ›Reveal solutionSolution

   The new origin is $(h,k)=(1,1)$ and $\tan2\theta=-\dfrac13=-\dfrac{h}{3k}$, so the correct option is (D).

   Solution

   For $S\equiv 2x^2-xy-y^2-3x+3y=0$, compare with $ax^2+2h_{xy}xy+by^2+2gx+2fy+c$:

   $a=2,\quad 2h_{xy}=-1,\quad b=-1,\quad 2g=-3,\quad 2f=3,\quad c=0.$

   Translation — new origin $(h,k)$ removing the first-degree terms solves $S_x=S_y=0$:

   $S_x=4x-y-3=0,\qquad S_y=-x-2y+3=0.$

   At $(h,k)$: $4h-k=3$ and $h+2k=3$. Solving gives $h=1,\ k=1$.

   Rotation removing the $xy$-term: …

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7. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $\sin A = -\dfrac{24}{25}$, $\cos B = \dfrac{15}{17}$, $A$ does not belong to $4^{\text{th}}$ quadrant and $B$ does not belong to $1^{\text{st}}$ quadrant then $(A+B)$ lies in the quadrant (A) $1^{\text{st}}$ quadrant (B) $2^{\text{nd}}$ quadrant (C) $3^{\text{rd}}$ quadrant (D) $4^{\text{th}}$ quadrant

   ›Reveal solutionSolution

   $A+B$ lies in the third quadrant, so the answer is (C).

   Fix the quadrants.

   • $\sin A=-\dfrac{24}{25}<0$ (Q3 or Q4); $A\notin$ Q4 $\Rightarrow$ $A$ in Q3, so $\cos A=-\dfrac{7}{25}$.
   • $\cos B=\dfrac{15}{17}>0$ (Q1 or Q4); $B\notin$ Q1 $\Rightarrow$ $B$ in Q4, so $\sin B=-\dfrac{8}{17}$.

   Signs of $A+B$.

   $\sin(A+B)=\sin A\cos B+\cos A\sin B=\left(-\tfrac{24}{25}\right)\!\left(\tfrac{15}{17}\right)+\left(-\tfrac{7}{25}\right)\!\left(-\tfrac{8}{17}\right)=\frac{-360+56}{425}=-\frac{304}{425}<0.$ …

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8. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $x \in (-\pi, \pi)$ then the number of solutions of the equation $2 \sin x \sin 3x \sin 5x + \sin 5x \cos 4x = 0$ is (A) 9 (B) 13 (C) 12 (D) 14

   ›Reveal solutionSolution

   The equation simplifies to $\sin 5x\,(2\sin x\sin 3x + \cos 4x)=0$, which factors further using product‑to‑sum identities into $\sin 5x \cos 2x \cos x = 0$. Counting distinct roots in $(-\pi,\pi)$ gives 13 solutions, so the correct option is (B).

   We start with

   $2\sin x \sin 3x \sin 5x + \sin 5x \cos 4x = 0.$

   Factor out $\sin 5x$:

   $\sin 5x \bigl(2\sin x \sin 3x + \cos 4x\bigr) = 0.$

   Now we have two independent factors to solve.

   ---

   1. Solve $\sin 5x = 0$

   We know $\sin\theta = 0$ when $\theta = n\pi$, $n\in\mathbb{Z}$.

   So $5x = n\pi \implies x = \frac{n\pi}{5}$.

   We need $x\in(-\pi,\pi)$. The integers $n$ that keep $x$ in this open interval are

   $n = -4, -3, -2, -1, 0, 1, 2, 3, 4$ — that’s 9 values.

   But we must later check for duplicates with the other factor.

   ---

   2. Solve $2\sin x \sin 3x + \cos 4x = 0$

   Use the product‑to‑sum identity:

   $2\sin A \sin B = \cos(A-B) - \cos(A+B).$

   Here $A=x$, $B=3x$:

   $2\sin x \sin 3x = \cos(2x) - \cos(4x).$

   Substitute into the equation:

   $\bigl(\cos 2x - \cos 4x\bigr) + \cos 4x = 0 \implies \cos 2x = 0.$

   So the second factor reduces to $\cos 2x = 0$.

   ---

   3. Solve $\cos 2x = 0$

   $\cos\theta = 0$ when $\theta = \frac{\pi}{2} + k\pi$, $k\in\mathbb{Z}$.

   Thus $2x = \frac{\pi}{2} + k\pi \implies x = \frac{\pi}{4} + \frac{k\pi}{2}$.

   We need $x\in(-\pi,\pi)$. Let’s list values for small $k$:

   • $k=-3$: $x = \frac{\pi}{4} - \frac{3\pi}{2} = -\frac{5\pi}{4} \approx -3.93$ (outside $(-\pi,\pi)$)
   • $k=-2$: $x = \frac{\pi}{4} - \pi = -\frac{3\pi}{4}$ ✓
   • $k=-1$: $x = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4}$ ✓
   • $k=0$: $x = \frac{\pi}{4}$ ✓
   • $k=1$: $x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$ ✓
   • $k=2$: $x = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \approx 3.93$ (outside)

   So we get 4 distinct values: $-\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}$.

   ---

   4. Combine and check for duplicates

   From $\sin 5x = 0$ we had 9 values: …

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $\sin A = -\dfrac{24}{25}$, $\cos B = \dfrac{15}{17}$, A does not belong to $4^{\text{th}}$ quadrant and B does not belong to $1^{\text{st}}$ quadrant then $(A+B)$ lies in the quadrant (A) $2^{\text{nd}}$ quadrant (B) $3^{\text{rd}}$ quadrant (C) $4^{\text{th}}$ quadrant (D) $1^{\text{st}}$ quadrant

   ›Reveal solutionSolution

   We determine the quadrants of A and B from the given signs and restrictions, compute the sign of $\sin(A+B)$ and $\cos(A+B)$, and conclude that $(A+B)$ lies in the third quadrant — so the correct option is (B).

   ---

   Concept and Intuition

   When we know the sine of one angle and the cosine of another, but not the angles themselves, we can still find the quadrant of their sum by determining the signs of $\sin(A+B)$ and $\cos(A+B)$.

   The quadrant of an angle is uniquely determined by the signs of its sine and cosine:

   Quadrant	$\sin$	$\cos$
   I	+	+
   II	+	–
   III	–	–
   IV	–	+

   So our job is to find the signs of $\sin(A+B)$ and $\cos(A+B)$ using the given data.

   ---

   Step-by-step reasoning

   1. Determine the quadrant of $A$

      We are given $\sin A = -\dfrac{24}{25}$ (negative) and that $A$ is not in the 4th quadrant.

      Sine is negative in quadrants III and IV. Since IV is excluded, $A$ must be in quadrant III.

      In quadrant III, cosine is also negative. So $\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - \frac{576}{625}} = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.

   2. Determine the quadrant of $B$

      We are given $\cos B = \dfrac{15}{17}$ (positive) and that $B$ is not in the 1st quadrant.

      Cosine is positive in quadrants I and IV. Since I is excluded, $B$ must be in quadrant IV.

      In quadrant IV, sine is negative. So $\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - \frac{225}{289}} = -\sqrt{\frac{64}{289}} = -\frac{8}{17}$.

   3. Compute $\sin(A+B)$

      Using the formula:

   $\sin(A+B) = \sin A \cos B + \cos A \sin B$

   Substitute:

   $\sin(A+B) = \left(-\frac{24}{25}\right)\left(\frac{15}{17}\right) + \left(-\frac{7}{25}\right)\left(-\frac{8}{17}\right)$

   $= -\frac{360}{425} + \frac{56}{425} = -\frac{304}{425}$

   So $\sin(A+B) < 0$.

   4. Compute $\cos(A+B)$ Using the formula: $$$$

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.$\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} =$ (A) $\frac{1}{2}$ (B) $-\frac{1}{12}$ (C) $-\frac{1}{6}$ (D) $\frac{2}{3}$

    ›Reveal solutionSolution

    The limit simplifies by factoring out $\sqrt[3]{\cos x}$ and using series expansions for $\cos x$ and the binomial expansion; the result is $-\frac{1}{12}$, so the correct option is (B).

    Concept & Intuition

    When $x \to 0$, both numerator and denominator approach $0$, giving a $\frac{0}{0}$ form. Direct substitution fails. The trick: rewrite the numerator as a common power of $\cos x$, then expand $\cos x$ as a series near $0$: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots$. The square and cube roots become binomial expansions $(1+u)^p \approx 1 + pu + \frac{p(p-1)}{2}u^2 + \cdots$, which lets us isolate the leading-order cancellation. The denominator $\sin^2 x \approx x^2$ sets the scale.

    Step-by-step solution

    1. Rewrite the numerator with a common factor Let $t = \cos x$. Then the numerator is $t^{1/2} - t^{1/3}$. Factor out $t^{1/3}$:

    $\sqrt{\cos x} - \sqrt[3]{\cos x} = (\cos x)^{1/3}\left[(\cos x)^{1/6} - 1\right].$

    As $x \to 0$, $\cos x \to 1$, so $(\cos x)^{1/3} \to 1$ and won't affect the limit's value. The key is the bracket $(\cos x)^{1/6} - 1$.

    2. Expand $\cos x$ near $0$

    $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6).$

    Let $u = -\frac{x^2}{2} + \frac{x^4}{24} + \cdots$, so $\cos x = 1 + u$ with $u \to 0$.

    3. Binomial expansion for $(1+u)^{1/6}$

    $(1+u)^{1/6} = 1 + \frac{1}{6}u + \frac{(1/6)(1/6 - 1)}{2}u^2 + O(u^3).$

    Compute $u$ and $u^2$ to order $x^4$:

    • $u = -\frac{x^2}{2} + \frac{x^4}{24} + \cdots$
    • $u^2 = \left(-\frac{x^2}{2}\right)^2 + \cdots = \frac{x^4}{4} + \cdots$ (higher terms are $O(x^6)$).

    Then

    $(1+u)^{1/6} = 1 + \frac{1}{6}\left(-\frac{x^2}{2} + \frac{x^4}{24}\right) + \frac{(1/6)(-5/6)}{2}\cdot\frac{x^4}{4} + O(x^6).$

    Simplify:

    • Linear term: $-\frac{x^2}{12} + \frac{x^4}{144}$
    • Quadratic term: $\frac{-5/36}{2} \cdot \frac{x^4}{4} = -\frac{5}{72} \cdot \frac{x^4}{4} = -\frac{5x^4}{288}$.

    So

    $(\cos x)^{1/6} = 1 - \frac{x^2}{12} + \left(\frac{1}{144} - \frac{5}{288}\right)x^4 + O(x^6) = 1 - \frac{x^2}{12} - \frac{3}{288}x^4 + \cdots = 1 - \frac{x^2}{12} - \frac{x^4}{96} + \cdots.$

    4. Thus the bracket

    $(\cos x)^{1/6} - 1 = -\frac{x^2}{12} - \frac{x^4}{96} + O(x^6).$

    5. Denominator expansion $$\sin^2 x = \left(x - \frac{x^3}{6} + \cdots\right)^2 = x^2 - \frac{x^4}{3} + O(x^6). …$$

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11. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.$\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} =$ (A) $\frac{2}{3}$ (B) $-\frac{1}{6}$ (C) $-\frac{1}{12}$ (D) $\frac{1}{2}$

    ›Reveal solutionSolution

    The limit simplifies by factoring out a common power of $\cos x$ and using the series expansions for $\cos x$ and $(1+u)^\alpha$; the final value is $-\frac{1}{12}$, which corresponds to option (C).

    We want

    $L = \lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}.$

    Both numerator and denominator vanish as $x \to 0$, so this is a $\frac{0}{0}$ form. The key is to rewrite the numerator in terms of a common factor and then use expansions near $x=0$.

    ---

    1. Factor out the smallest power of $\cos x$.

    Write $\sqrt{\cos x} = (\cos x)^{1/2}$ and $\sqrt[3]{\cos x} = (\cos x)^{1/3}$. The smaller exponent is $\frac{1}{3}$, so factor $(\cos x)^{1/3}$ out:

    $(\cos x)^{1/2} - (\cos x)^{1/3} = (\cos x)^{1/3}\left[(\cos x)^{1/6} - 1\right].$

    Thus

    $L = \lim_{x \to 0} \frac{(\cos x)^{1/3}\left[(\cos x)^{1/6} - 1\right]}{\sin^2 x}.$

    As $x \to 0$, $\cos x \to 1$, so $(\cos x)^{1/3} \to 1$. That factor is harmless; the interesting part is the bracket.

    ---

    2. Expand $\cos x$ near $0$.

    We know

    $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6).$

    Also $\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$.

    ---

    3. Expand $(\cos x)^{1/6}$ using $(1+u)^\alpha$.

    Let $u = \cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} + \cdots$. Then

    $(\cos x)^{1/6} = (1+u)^{1/6} = 1 + \frac{1}{6}u + \frac{(1/6)(1/6 - 1)}{2}u^2 + \cdots$

    $= 1 + \frac{1}{6}\left(-\frac{x^2}{2} + \frac{x^4}{24}\right) + \frac{1}{6}\left(-\frac{5}{6}\right)\frac{1}{2}\left(-\frac{x^2}{2}\right)^2 + O(x^6).$

    Compute term by term:

    • Linear in $u$: $\frac{1}{6}\left(-\frac{x^2}{2}\right) = -\frac{x^2}{12}$, and the $\frac{x^4}{24}$ part gives $+\frac{x^4}{144}$.
    • Quadratic in $u$: $\frac{1}{6}\cdot\left(-\frac{5}{6}\right)\cdot\frac{1}{2} = -\frac{5}{72}$, times $u^2 = \left(-\frac{x^2}{2}\right)^2 = \frac{x^4}{4}$, gives $-\frac{5}{72}\cdot\frac{x^4}{4} = -\frac{5x^4}{288}$.

    So

    $(\cos x)^{1/6} - 1 = -\frac{x^2}{12} + \left(\frac{1}{144} - \frac{5}{288}\right)x^4 + O(x^6).$

    The $x^4$ coefficient: $\frac{1}{144} = \frac{2}{288}$, so $\frac{2}{288} - \frac{5}{288} = -\frac{3}{288} = -\frac{1}{96}$.

    Thus

    $(\cos x)^{1/6} - 1 = -\frac{x^2}{12} - \frac{x^4}{96} + O(x^6).$

    ---

    4. Assemble the limit. …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.If $\cos x + \cos y = \dfrac{2}{3}$ and $\sin x - \sin y = \dfrac{3}{4}$, then $\sin(x-y) + \cos(x-y) =$ (A) $\dfrac{161}{145}$ (B) $\dfrac{127}{145}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{8}{9}$

    ›Reveal solutionSolution

    We use sum‑to‑product identities to find $\cos\frac{x+y}{2}$ and $\sin\frac{x+y}{2}$, then compute $\sin(x-y)$ and $\cos(x-y)$ via double‑angle formulas, obtaining $\frac{161}{145}$.

    Concept & Intuition

    We are given two equations mixing sums and differences of sines and cosines. The classic trick is to rewrite each as a product using sum‑to‑product identities. That isolates the half‑sum and half‑difference angles. Then we can find $\sin\frac{x-y}{2}$ and $\cos\frac{x-y}{2}$ from the given numbers, and finally use double‑angle formulas to get $\sin(x-y)$ and $\cos(x-y)$.

    ---

    1. Apply sum‑to‑product identities

    $\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2} = \frac{2}{3}$

    $\sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2} = \frac{3}{4}$

    2. Divide the two equations to eliminate $\cos\frac{x+y}{2}$ (provided it is nonzero):

    $\frac{2\cos\frac{x+y}{2}\sin\frac{x-y}{2}}{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}} = \frac{3/4}{2/3}$

    $\tan\frac{x-y}{2} = \frac{3}{4} \cdot \frac{3}{2} = \frac{9}{8}$

    3. Find $\sin\frac{x-y}{2}$ and $\cos\frac{x-y}{2}$ from the tangent. Let $t = \frac{x-y}{2}$. Then $\tan t = \frac{9}{8}$. Construct a right triangle: opposite = 9, adjacent = 8, hypotenuse = $\sqrt{9^2+8^2} = \sqrt{145}$. Hence

    $\sin t = \frac{9}{\sqrt{145}}, \qquad \cos t = \frac{8}{\sqrt{145}}.$

    4. Use double‑angle formulas to get $\sin(x-y)$ and $\cos(x-y)$: $$$$

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