Q.Integrate the following function: $\int x^2 \left(1 - \frac{1}{x^2}\right) dx$

Showing the 12 most recent of 18 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\int \frac{3x \sec^2 \sqrt{9x^2 - 12x + 1} - 2 \sec^2 \sqrt{(3x-2)^2 - 3}}{\sqrt{9x^2 - 12x + 1}} dx =$ (A) $\sqrt{9x^2 - 12x + 1} + c$ (B) $\frac{1}{3} \cos \sqrt{9x^2 - 12x + 1} + c$ (C) $\frac{1}{2\sqrt{9x^2 - 12x + 1}} + c$ (D) $\frac{1}{3} \tan \sqrt{9x^2 - 12x + 1} + c$

   ›Reveal solutionSolution

   The integral simplifies by noticing that the numerator is exactly the derivative of the argument inside the sec² term, leading to a simple substitution. The result is $\frac{1}{3} \tan \sqrt{9x^2 - 12x + 1} + c$, so the correct option is (D).

   The key insight here is that the integrand looks messy, but the structure hints at a chain-rule derivative in reverse. When you see $\sec^2(\text{something})$ multiplied by something that looks like the derivative of that "something," you should immediately think of the derivative of $\tan(\text{something})$. The trick is to check whether the numerator is exactly the derivative of the expression inside the square root.

   Let’s work through it step by step.

   1. Simplify the expression under the square root. Notice that $9x^2 - 12x + 1$ can be rewritten by completing the square:

   $9x^2 - 12x + 1 = 9\left(x^2 - \frac{4}{3}x\right) + 1 = 9\left[\left(x - \frac{2}{3}\right)^2 - \frac{4}{9}\right] + 1 = 9\left(x - \frac{2}{3}\right)^2 - 4 + 1 = (3x-2)^2 - 3.$

   So $\sqrt{9x^2 - 12x + 1} = \sqrt{(3x-2)^2 - 3}$. This confirms the two square-root expressions in the problem are identical. Let’s denote

   $u = \sqrt{9x^2 - 12x + 1}.$

   2. Find the derivative of $u$ with respect to $x$.

   $u = (9x^2 - 12x + 1)^{1/2} \quad \Rightarrow \quad \frac{du}{dx} = \frac{1}{2}(9x^2 - 12x + 1)^{-1/2} \cdot (18x - 12) = \frac{9x - 6}{\sqrt{9x^2 - 12x + 1}}.$

   Factor a 3:

   $\frac{du}{dx} = \frac{3(3x - 2)}{\sqrt{9x^2 - 12x + 1}}.$

   3. Compare with the numerator of the integrand. The integrand is

   $\frac{3x \sec^2 u - 2 \sec^2 u}{\sqrt{9x^2 - 12x + 1}} = \frac{(3x - 2) \sec^2 u}{\sqrt{9x^2 - 12x + 1}}.$

   We have $(3x - 2)$ in the numerator, but $\frac{du}{dx}$ contains $3(3x - 2)$. So the numerator is exactly $\frac{1}{3} \cdot \frac{du}{dx} \cdot \sqrt{9x^2 - 12x + 1}$? Wait, let’s check carefully:

   $\frac{du}{dx} = \frac{3(3x-2)}{\sqrt{9x^2 - 12x + 1}} \quad \Rightarrow \quad (3x-2) = \frac{1}{3} \cdot \frac{du}{dx} \cdot \sqrt{9x^2 - 12x + 1}.$

   Substituting this into the integrand:

   $$$$

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2. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.$\int \sqrt{x^2 + x + 1} \, dx \times \int \frac{1}{\sqrt{x^2 + x + 1}} \, dx =$ (A) $x + C$ (B) $\left( \frac{2x+1}{4} \sqrt{x^2 + x + 1} + \frac{3}{8} \sinh^{-1} \frac{2x+1}{\sqrt{3}} \right) \sinh^{-1} \left( \frac{2x+1}{\sqrt{3}} \right) + C$ (C) $\frac{2x+1}{2} \sinh^{-1} \left( \sqrt{x^2 + x + 1} \right) + \left( \frac{3}{8} \sinh^{-1} \frac{2x+1}{\sqrt{3}} \right)^2 + C$ (D) $\frac{2x+1}{2} \left( \sinh^{-1} \frac{2x+1}{\sqrt{3}} \right)^2 + C$

   ›Reveal solutionSolution

   The product of the two integrals is not a simple closed form; the given options are distractors. The correct approach is to recognise that the product of an indefinite integral and another indefinite integral is not a single constant or simple function — the problem is ill-posed as written. None of the options are correct.

   The question presents a product of two indefinite integrals: $\int \sqrt{x^2 + x + 1} \, dx$ multiplied by $\int \frac{1}{\sqrt{x^2 + x + 1}} \, dx$. This is a trap. Indefinite integrals are families of functions differing by a constant. Their product is not a well-defined function — it depends on the arbitrary constants of integration. The options try to make you think you can compute each integral separately and multiply, but that ignores the constants.

   Let’s see why this is a conceptual dead end.

   1. Each integral is a family, not a single function.

      $\int f(x)\,dx = F(x) + C_1$ and $\int g(x)\,dx = G(x) + C_2$. Their product is $(F(x)+C_1)(G(x)+C_2)$, which expands to $F(x)G(x) + C_1 G(x) + C_2 F(x) + C_1 C_2$. This is not a single expression — it depends on two arbitrary constants. No option includes two independent constants, so none can represent the general product.

   2. Even if we set both constants to zero (the “particular” antiderivatives), the product is not among the options.

      Let’s compute the two integrals with zero constants. Complete the square:

      $x^2 + x + 1 = \left(x + \frac12\right)^2 + \frac34$.

      Let $u = x + \frac12$, so $du = dx$. Then:

      • $\int \sqrt{u^2 + \frac34}\,du$
      • $\int \frac{1}{\sqrt{u^2 + \frac34}}\,du$

   3. Standard forms.

      Recall:

      $\int \sqrt{u^2 + a^2}\,du = \frac{u}{2}\sqrt{u^2 + a^2} + \frac{a^2}{2}\sinh^{-1}\frac{u}{a} + C$

      $\int \frac{1}{\sqrt{u^2 + a^2}}\,du = \sinh^{-1}\frac{u}{a} + C$

      Here $a^2 = \frac34$, so $a = \frac{\sqrt3}{2}$.

   4. Compute each integral (constants zero).

      • $I_1 = \int \sqrt{u^2 + \frac34}\,du = \frac{u}{2}\sqrt{u^2 + \frac34} + \frac{3}{8}\sinh^{-1}\frac{2u}{\sqrt3}$

        Back-substitute $u = x + \frac12$:

        $I_1 = \frac{x+\frac12}{2}\sqrt{x^2+x+1} + \frac{3}{8}\sinh^{-1}\frac{2x+1}{\sqrt3}$

        $= \frac{2x+1}{4}\sqrt{x^2+x+1} + \frac{3}{8}\sinh^{-1}\frac{2x+1}{\sqrt3}$

      • $I_2 = \int \frac{1}{\sqrt{u^2 + \frac34}}\,du = \sinh^{-1}\frac{2u}{\sqrt3} = \sinh^{-1}\frac{2x+1}{\sqrt3}$

   5. Their product (with zero constants) is: …

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3. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.$\lim_{n \to \infty} \frac{(2n(2n-1)\ldots(n+2)(n+1))^{1/n}}{n} =$ (A) $\int_{0}^{1} \log x \, dx$ (B) $\int_{0}^{1} x \log x \, dx$ (C) $\int_{0}^{1} (x+1) \log(x+1) \, dx$ (D) $\int_{0}^{1} \log(1+x) \, dx$

   ›Reveal solutionSolution

   Taking the natural log turns the expression into a Riemann sum for $\int_0^1 \log(1+x)\,dx$. Correct option: (D).

   Setting up. The numerator is the product of the $n$ integers from $n+1$ to $2n$:

   $(2n)(2n-1)\cdots(n+1)=\prod_{k=1}^{n}(n+k).$

   Divide inside the $n$-th root by $n$ (there are $n$ factors, so $n$ powers of $n$):

   $\frac{\big(\prod_{k=1}^{n}(n+k)\big)^{1/n}}{n}=\left(\prod_{k=1}^{n}\frac{n+k}{n}\right)^{1/n}=\left(\prod_{k=1}^{n}\Big(1+\frac{k}{n}\Big)\right)^{1/n}.$

   Take logarithms. Writing $L$ for the limit,

   $\log L=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\log\!\Big(1+\frac{k}{n}\Big).$

   This is exactly the Riemann sum (with $x_k=k/n,\ \Delta x=1/n$) for the definite integral …

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4. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.Consider the following Assertion (A): $\int \sqrt{x-3} \left( \sin^{-1}(\log x) + \cos^{-1}(\log x) \right) dx = \frac{\pi}{3} (x-3)^{\frac{3}{2}} + c$ Reason (R): $\sin^{-1}(f(x)) + \cos^{-1}(f(x)) = \frac{\pi}{2}, |f(x)| < 1$ (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   The Reason (the inverse-trig identity) is true, but the Assertion's integral is defined on an empty domain, so it is false.

   The Reason states the standard identity sin⁻¹(f(x))+cos⁻¹(f(x))=π/2 for |f(x)|≤1 — true. For the Assertion, the factor sin⁻¹(log x)+cos⁻¹(log x) requires |log x|≤1, i.e. x∈[1/e, e]. But √(x−3) requires x≥3. The intersection of [1/e, e] and [3, ∞) is empty, so the integrand exists …

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5. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.Consider the following Assertion (A): $\int \sqrt{x-3} \left( \sin^{-1}(\log x) + \cos^{-1}(\log x) \right) dx = \frac{\pi}{3} (x-3)^{\frac{3}{2}} + c$ Reason (R): $\sin^{-1}(f(x)) + \cos^{-1}(f(x)) = \frac{\pi}{2},\ |f(x)| < 1$ (A) is false, but (R) is true (B) (A) is true, but (R) is false (C) Both (A) and (R) are true, (R) is not the correct explanation of (A) (D) Both (A) and (R) are true, (R) is the correct explanation of (A)

   ›Reveal solutionSolution

   The key idea is that $\sin^{-1}(\log x) + \cos^{-1}(\log x) = \frac{\pi}{2}$ only when $|\log x| < 1$, which is not true for all $x$ in the domain of the integrand. The integration in (A) incorrectly assumes this identity holds universally, so (A) is false, while (R) is a true statement about the identity's condition.

   The problem tests two things: the domain restrictions on inverse trigonometric identities, and whether you blindly apply a formula without checking where it works. The identity $\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}$ is valid only for $t \in [-1, 1]$. Here $t = \log x$, so the identity holds only when $|\log x| \leq 1$, i.e., $x \in [1/e, e]$. But the integrand $\sqrt{x-3}$ requires $x \geq 3$ for real values. Since $3 > e$, there is no $x$ where both conditions overlap — the identity never applies in the domain of integration. So (A) is built on a false premise.

   1. Check the domain of the integrand.

      The square root $\sqrt{x-3}$ is real only when $x \geq 3$. So the integration in (A) is meant over $x \geq 3$.

   2. Check where the identity in (R) holds.

      $\sin^{-1}(f(x)) + \cos^{-1}(f(x)) = \frac{\pi}{2}$ is true only if $|f(x)| \leq 1$. Here $f(x) = \log x$, so we need $|\log x| \leq 1$, which means $1/e \leq x \leq e$.

   3. Compare the two domains.

      The integration domain is $x \geq 3$, but the identity works only for $x \in [1/e, e]$. Since $3 > e$, there is no overlap. For every $x \geq 3$, $\log x > 1$, so $\sin^{-1}(\log x)$ is not even defined (its argument exceeds 1). The integrand itself is undefined for $x \geq 3$ because $\sin^{-1}(\log x)$ has no real value there.

   4. Consequence for Assertion (A). …

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\int \frac{\cos\left(\frac{7x}{2}\right)}{\cos\left(\frac{x}{2}\right)} dx = a \sin x + b \sin 2x + c \sin^3 x - dx + k$, then $a + b + c + d =$ (A) $1$ (B) $-\frac{2}{3}$ (C) $\frac{4}{3}$ (D) $2$

   ›Reveal solutionSolution

   The key is to rewrite the integrand using sum‑to‑product identities, integrate term by term, match coefficients to the given form, and sum them. The result is $a+b+c+d = \frac{4}{3}$, so option (C) is correct.

   ---

   We start with the integral

   $\int \frac{\cos\left(\frac{7x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \, dx.$

   The presence of half‑angles suggests using a product‑to‑sum identity to simplify the ratio. The trick: express the numerator as a sum of cosines whose arguments are multiples of $x$, so that division by $\cos(x/2)$ becomes straightforward.

   1. Use the identity

   Recall:

   $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}.$

   We want to write $\cos(7x/2)$ as something times $\cos(x/2)$. Notice that if we set $A+B = 7x$ and $A-B = x$, then $A = 4x$, $B = 3x$. Then

   $\cos 4x + \cos 3x = 2\cos\frac{7x}{2}\cos\frac{x}{2}.$

   Hence

   $\cos\frac{7x}{2} = \frac{\cos 4x + \cos 3x}{2\cos(x/2)}.$

   But we have $\cos(7x/2)$ divided by $\cos(x/2)$, so

   $\frac{\cos(7x/2)}{\cos(x/2)} = \frac{\cos 4x + \cos 3x}{2\cos^2(x/2)}.$

   That still has a $\cos^2(x/2)$ in the denominator — not ideal. Let’s try a different pairing.

   2. A better pairing

   We want the denominator $\cos(x/2)$ to cancel directly. Write

   $\cos\frac{7x}{2} = \cos\left(4x - \frac{x}{2}\right) = \cos 4x \cos\frac{x}{2} + \sin 4x \sin\frac{x}{2}.$

   Then

   $\frac{\cos(7x/2)}{\cos(x/2)} = \cos 4x + \sin 4x \tan\frac{x}{2}.$

   That introduces a tangent — not simpler.

   3. Use sum‑to‑product in reverse

   Instead, consider writing $\cos(7x/2)$ as a sum of cosines of multiples of $x$ times $\cos(x/2)$. We can use the identity:

   $\cos n\theta = 2\cos((n-1)\theta)\cos\theta - \cos((n-2)\theta).$

   Let $\theta = x/2$, then $\cos(7x/2) = \cos 7\theta$. Applying the recurrence:

   $\cos 7\theta = 2\cos 6\theta \cos\theta - \cos 5\theta.$

   But $\cos 6\theta = \cos 3x$, $\cos 5\theta = \cos(5x/2)$ — still half‑angles. This path gets messy.

   4. The clean approach: use the identity

   $\cos A = \cos\left(\frac{A+B}{2} + \frac{A-B}{2}\right)$

   and sum with a cleverly chosen term. Actually, the classic trick is:

   $\frac{\cos(7x/2)}{\cos(x/2)} = \frac{\cos(4x + 3x/2)}{\cos(x/2)} \quad \text{no}.$

   Better: Write

   $\cos\frac{7x}{2} = \cos\left(3x + \frac{x}{2}\right) = \cos 3x \cos\frac{x}{2} - \sin 3x \sin\frac{x}{2}.$

   Then

   $\frac{\cos(7x/2)}{\cos(x/2)} = \cos 3x - \sin 3x \tan\frac{x}{2}.$

   Still not a polynomial in sines and cosines of $x$.

   5. The winning identity

   Use:

   $\cos\frac{7x}{2} = \cos\left(4x - \frac{x}{2}\right) = \cos 4x \cos\frac{x}{2} + \sin 4x \sin\frac{x}{2}.$

   Divide by $\cos(x/2)$:

   $\frac{\cos(7x/2)}{\cos(x/2)} = \cos 4x + \sin 4x \tan\frac{x}{2}.$

   Now use $\tan(x/2) = \frac{\sin x}{1+\cos x}$ or better: express $\sin 4x$ in terms of $\sin x$ and $\cos x$ and use the half‑angle tangent identity. But there’s a more direct route.

   6. Use the identity

   $\frac{\cos(7x/2)}{\cos(x/2)} = \frac{2\cos(7x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{\cos 4x + \cos 3x}{1+\cos x}$

   since $2\cos^2(x/2) = 1+\cos x$. So

   $I = \int \frac{\cos 4x + \cos 3x}{1+\cos x} \, dx.$

   Now use $\cos 4x = 2\cos^2 2x - 1$ and $\cos 3x = 4\cos^3 x - 3\cos x$, but denominator $1+\cos x$ suggests using the substitution $t = \tan(x/2)$ or expressing everything in terms of $\cos x$.

   7. Express numerator in terms of $\cos x$

   We have:

   $\cos 4x = 8\cos^4 x - 8\cos^2 x + 1,$

   $\cos 3x = 4\cos^3 x - 3\cos x.$

   So numerator = $8\cos^4 x + 4\cos^3 x - 8\cos^2 x - 3\cos x + 1$.

   Denominator = $1+\cos x$. Perform polynomial division in $\cos x$:

   Divide $8u^4 + 4u^3 - 8u^2 - 3u + 1$ by $u+1$ (where $u=\cos x$).

   Synthetic division with root $-1$:

   • Bring down 8.
   • $8 \times (-1) = -8$, add to 4 → $-4$. …

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7. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Term independent of $x$ in the expansion of $\left(5x + \frac{6}{x}\right)^4 \left(\frac{1}{1-2x}\right)$ is (A) $(201)6^3$ (B) $(27)6^4$ (C) $(157)5^4$ (D) $(27)5^6$

   ›Reveal solutionSolution

   Collect terms of $(5x+6/x)^4$ with non-positive power, weight each by $2^n$ from $\tfrac{1}{1-2x}=\sum 2^nx^n$: total $=5400+17280+20736=43416=(201)6^3$.

   General term of $\left(5x+\tfrac{6}{x}\right)^4$: $\binom{4}{r}5^{4-r}6^{r}x^{4-2r}$.

   $r$	power	coefficient
   $1$	$2$	$3000$
   $2$	$0$	$5400$
   $3$	$-2$	$4320$
   $4$	$-4$	$1296$

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8. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.The term independent of $x$ in the expansion of $\left(x - \dfrac{2}{\sqrt{x}}\right)^{21}$ is (A) $^{21}C_{15}(-2)^{15}$ (B) $^{21}C_{14}2^{14}$ (C) $-^{21}C_{7}(2)^{7}$ (D) $-^{21}C_{7}2^{14}$

   ›Reveal solutionSolution

   The general term has power $x^{21-\frac{3k}{2}}$; independence needs $k=14$, giving $^{21}C_{14}\,2^{14}$ — option (B).

   General term of $\left(x-\dfrac{2}{\sqrt{x}}\right)^{21}$:

   $T_{k+1}={}^{21}C_k\,x^{21-k}\left(-\frac{2}{\sqrt{x}}\right)^{k}={}^{21}C_k\,(-2)^{k}\,x^{\,21-k-\frac{k}{2}}={}^{21}C_k\,(-2)^{k}\,x^{\,21-\frac{3k}{2}}.$

   Term independent of $x$: set the exponent to zero, …

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9. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.The coefficient of $x^3$ in the expansion of $\left(1 - \frac{3}{4}x\right)^{\frac{1}{2}}$ is (A) $\frac{27}{1024}$ (B) $-\frac{27}{1024}$ (C) $\frac{81}{1024}$ (D) $-\frac{81}{1024}$

   ›Reveal solutionSolution

   Using $(1+u)^{1/2}$ with $u=-\tfrac{3}{4}x$, the $x^3$ coefficient is $\binom{1/2}{3}\left(-\tfrac34\right)^3=-\dfrac{27}{1024}$ — option (B).

   Binomial (general index) expansion. For $\left(1-\dfrac{3}{4}x\right)^{1/2}$ the coefficient of $x^3$ is

   $\binom{1/2}{3}\left(-\frac{3}{4}\right)^{3}.$

   Evaluate the binomial coefficient. …

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10. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.Coefficient of $x^3$ in the expansion of $\dfrac{(1-2x^2)^{1/3}}{(2+x)^{1/2}}$ is (A) $\dfrac{17\sqrt{2}}{384}$ (B) $\dfrac{17\sqrt{2}}{768}$ (C) $\dfrac{49\sqrt{2}}{768}$ (D) $\dfrac{49\sqrt{2}}{384}$

    ›Reveal solutionSolution

    Use the binomial expansion for rational exponents on each factor separately, then multiply the series and collect the $x^3$ term. The coefficient is $\frac{49\sqrt{2}}{768}$, which is option (C).

    The problem asks for the coefficient of $x^3$ in a product of two expressions, each raised to a fractional power. The direct approach — expanding each factor as a series using the general binomial theorem — is the natural path. For $(1 + u)^n$ where $n$ is any rational number, the expansion is an infinite series, and we only need terms up to $x^3$ from the product.

    Let’s write the expression as:

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

    We handle each factor separately.

    1. Expand $(1-2x^2)^{1/3}$

    The general binomial series for $|u|<1$ is:

    $(1+u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \cdots$

    Here $n = \frac13$ and $u = -2x^2$. So:

    $(1-2x^2)^{1/3} = 1 + \frac13(-2x^2) + \frac{\frac13\left(\frac13-1\right)}{2}(-2x^2)^2 + \frac{\frac13\left(\frac13-1\right)\left(\frac13-2\right)}{6}(-2x^2)^3 + \cdots$

    Compute term by term:

    • Constant term: $1$.
    • $x^2$ term: $\frac13(-2x^2) = -\frac{2}{3}x^2$.
    • $x^4$ term: $\frac{\frac13\left(-\frac23\right)}{2} \cdot 4x^4 = \frac{-\frac{2}{9}}{2} \cdot 4x^4 = -\frac{1}{9} \cdot 4x^4 = -\frac{4}{9}x^4$.
    • $x^6$ term: $\frac{\frac13\left(-\frac23\right)\left(-\frac53\right)}{6} \cdot (-8x^6) = \frac{\frac{10}{27}}{6} \cdot (-8x^6) = \frac{10}{162} \cdot (-8x^6) = \frac{5}{81} \cdot (-8x^6) = -\frac{40}{81}x^6$.

    We only need up to $x^3$ from the product, so terms of degree $x^4$ and higher from this factor will combine with negative powers from the other factor to produce $x^3$ — so we must keep them. In fact, we need all terms up to $x^6$ from this factor because the other factor will contribute $x^{-1}$, $x^{-2}$, etc. Let’s see.

    2. Expand $(2+x)^{-1/2}$

    Rewrite as:

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

    Now expand $\left(1 + \frac{x}{2}\right)^{-1/2}$ with $n = -\frac12$ and $u = \frac{x}{2}$:

    $\left(1 + \frac{x}{2}\right)^{-1/2} = 1 + \left(-\frac12\right)\frac{x}{2} + \frac{(-\frac12)(-\frac32)}{2}\left(\frac{x}{2}\right)^2 + \frac{(-\frac12)(-\frac32)(-\frac52)}{6}\left(\frac{x}{2}\right)^3 + \cdots$

    Compute:

    • Constant: $1$.
    • $x$ term: $-\frac12 \cdot \frac{x}{2} = -\frac{x}{4}$.
    • $x^2$ term: $\frac{\frac34}{2} \cdot \frac{x^2}{4} = \frac{3}{8} \cdot \frac{x^2}{4} = \frac{3x^2}{32}$.
    • $x^3$ term: $\frac{(-\frac12)(-\frac32)(-\frac52)}{6} \cdot \frac{x^3}{8} = \frac{-\frac{15}{8}}{6} \cdot \frac{x^3}{8} = -\frac{15}{48} \cdot \frac{x^3}{8} = -\frac{5}{16} \cdot \frac{x^3}{8} = -\frac{5x^3}{128}$.

    So:

    $(2+x)^{-1/2} = \frac{1}{\sqrt{2}} \left(1 - \frac{x}{4} + \frac{3x^2}{32} - \frac{5x^3}{128} + \cdots \right).$

    3. Multiply the two series

    We need the coefficient of $x^3$ in:

    $\left(1 - \frac{2}{3}x^2 - \frac{4}{9}x^4 - \frac{40}{81}x^6 + \cdots \right) \cdot \frac{1}{\sqrt{2}} \left(1 - \frac{x}{4} + \frac{3x^2}{32} - \frac{5x^3}{128} + \cdots \right)$

    Multiply term by term, collecting only $x^3$ contributions. Let’s list all pairs whose exponents sum to 3: …

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11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.$\sinh\left(\log\left(3+\sqrt{8}\right)\right)=$ (A) $3^{\frac{3}{2}}$ (B) $2^{\frac{3}{2}}$ (C) $8^{\frac{2}{3}}$ (D) $3^{\frac{1}{2}}$

    ›Reveal solutionSolution

    The key idea is to rewrite $\sinh(\log x)$ as $\frac{x - 1/x}{2}$, then simplify with $x = 3 + \sqrt{8}$ using its reciprocal property. The final value is $\sqrt{2}$.

    The problem asks for $\sinh(\log(3+\sqrt{8}))$. The direct approach — plugging into the definition of $\sinh$ — is the cleanest path. Recall that for any real number $t$,

    $\sinh t = \frac{e^t - e^{-t}}{2}.$

    Here $t = \log(3+\sqrt{8})$, so $e^t = 3+\sqrt{8}$ and $e^{-t} = \frac{1}{3+\sqrt{8}}$. The expression becomes

    $\frac{(3+\sqrt{8}) - \frac{1}{3+\sqrt{8}}}{2}.$

    The trick is to simplify $\frac{1}{3+\sqrt{8}}$. Multiply numerator and denominator by the conjugate $3-\sqrt{8}$:

    $\frac{1}{3+\sqrt{8}} \cdot \frac{3-\sqrt{8}}{3-\sqrt{8}} = \frac{3-\sqrt{8}}{9-8} = 3-\sqrt{8}.$

    So the reciprocal is simply $3-\sqrt{8}$ — a neat symmetry.

    Now the numerator becomes $(3+\sqrt{8}) - (3-\sqrt{8}) = 2\sqrt{8}$. Therefore

    $\sinh(\log(3+\sqrt{8})) = \frac{2\sqrt{8}}{2} = \sqrt{8}.$

    But $\sqrt{8} = 2\sqrt{2} = 2^{3/2}$. That matches one of the options directly. …

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12. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
    Q.If the coefficient of $x^4$ in the expansion of $\dfrac{x}{(x-1)^2(x-2)}$ is $\dfrac{m}{n}$ and $|m|, |n|$ are coprimes, then $\sqrt{|m+n|}=$ (A) $9$ (B) $\sqrt{33}$ (C) $7$ (D) $6\sqrt{2}$

    ›Reveal solutionSolution

    Partial fractions give the $x^4$ coefficient $=-\tfrac{49}{16}$, so $|m|=49,\ |n|=16$, $|m+n|=33$ and $\sqrt{|m+n|}=\sqrt{33}$.

    Resolve into partial fractions:

    $\frac{x}{(x-1)^2(x-2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}.$

    From $x=A(x-1)(x-2)+B(x-2)+C(x-1)^2$: at $x=1,\ B=-1$; at $x=2,\ C=2$; matching $x^2$, $A+C=0\Rightarrow A=-2$.

    So $f(x)=-\dfrac{2}{x-1}-\dfrac{1}{(x-1)^2}+\dfrac{2}{x-2}$. Expand each about $x=0$:

    $-\frac{2}{x-1}=2\sum_{n\ge0}x^n\ \Rightarrow\ [x^4]=2,$

    $-\frac{1}{(x-1)^2}=-\sum_{n\ge0}(n+1)x^n\ \Rightarrow\ [x^4]=-5,$ …

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