Q.Integrate the following function: $\int \sec x (\sec x + \tan x) dx$

Showing the 12 most recent of 18 on this concept.

1. 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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2. 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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3. 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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4. 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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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The coefficient of $x^{12}$ in the expansion of $(x^2 + 2x + 2)^8$ is (A) 1120 (B) 2240 (C) 2576 (D) 4152

   ›Reveal solutionSolution

   The coefficient of $x^{12}$ in $(x^2+2x+2)^8$ is found by expanding the trinomial using the multinomial theorem, selecting terms where the total exponent of $x$ is 12, and summing the contributions. The result is 2576, which corresponds to option (C).

   We need the coefficient of $x^{12}$ in $(x^2 + 2x + 2)^8$. This is a trinomial raised to a power, so the multinomial theorem is the natural tool. The key idea: each term in the expansion comes from picking, for each of the 8 factors, one of the three terms ($x^2$, $2x$, or $2$), multiplying them together, and summing over all choices. The exponent of $x$ in a term is the sum of contributions: each time we pick $x^2$ we get 2, each time we pick $2x$ we get 1, and picking $2$ gives 0. So we need to count how many ways we can get a total exponent of 12.

   Let’s denote:

   • $a$ = number of times we pick $x^2$ (contributes $2a$ to the exponent)
   • $b$ = number of times we pick $2x$ (contributes $b$ to the exponent)
   • $c$ = number of times we pick $2$ (contributes $0$ to the exponent)

   We have $a + b + c = 8$ (since we choose from 8 factors), and the exponent condition is $2a + b = 12$.

   1. Find possible integer triples $(a,b,c)$

      From $2a + b = 12$ and $a+b+c=8$, we can eliminate $b$:

      $b = 12 - 2a$.

      Then $c = 8 - a - b = 8 - a - (12 - 2a) = a - 4$.

      Since $b \ge 0$ and $c \ge 0$, we need $12 - 2a \ge 0 \Rightarrow a \le 6$, and $a - 4 \ge 0 \Rightarrow a \ge 4$.

      So $a$ can be 4, 5, or 6.

      • If $a=4$: $b = 12 - 8 = 4$, $c = 4 - 4 = 0$.
      • If $a=5$: $b = 12 - 10 = 2$, $c = 5 - 4 = 1$.
      • If $a=6$: $b = 12 - 12 = 0$, $c = 6 - 4 = 2$.

   2. Compute the coefficient for each triple

      The multinomial coefficient for a given $(a,b,c)$ is $\frac{8!}{a!\,b!\,c!}$, and each factor contributes its coefficient: $x^2$ has coefficient 1, $2x$ has coefficient 2, and $2$ has coefficient 2. So the term’s full coefficient is:

   $\frac{8!}{a!\,b!\,c!} \cdot (1)^a \cdot (2)^b \cdot (2)^c$

   Simplify: $(2)^b \cdot (2)^c = 2^{b+c}$.

   • For $(4,4,0)$: …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The coefficient of $x^{12}$ in the expansion of $(x^2 + 2x + 2)^8$ is (A) 2576 (B) 2240 (C) 1120 (D) 4152

   ›Reveal solutionSolution

   The coefficient of $x^{12}$ in $(x^2+2x+2)^8$ is found by treating the expansion as a multinomial and counting contributions where the exponents of $x$ sum to 12; the result is 2576, which corresponds to option (A).

   We need the coefficient of $x^{12}$ in $(x^2 + 2x + 2)^8$.

   The expression is a sum of three terms: $x^2$, $2x$, and $2$. When raised to the 8th power, each term in the expansion is of the form

   $\frac{8!}{a!\,b!\,c!}\,(x^2)^a\,(2x)^b\,(2)^c$

   where $a+b+c = 8$. The exponent of $x$ in such a term is $2a + b$. We want $2a + b = 12$.

   Why multinomial?

   Because we have three distinct “parts” inside the parentheses, the binomial theorem doesn’t directly apply; the multinomial theorem handles sums of more than two terms cleanly.

   ---

   1. Set up the conditions Let $a$ = number of factors choosing $x^2$, $b$ = number choosing $2x$, $c$ = number choosing $2$. Then:

   $a + b + c = 8, \quad 2a + b = 12.$

   2. Solve for possible integer triples From $2a + b = 12$ we have $b = 12 - 2a$. Substitute into $a + b + c = 8$:

   $a + (12 - 2a) + c = 8 \quad\Rightarrow\quad 12 - a + c = 8 \quad\Rightarrow\quad c = a - 4.$

   Since $b \ge 0$ and $c \ge 0$, we need:

   $12 - 2a \ge 0 \;\Rightarrow\; a \le 6, \qquad a - 4 \ge 0 \;\Rightarrow\; a \ge 4.$

   So $a$ can be 4, 5, or 6.

   3. List the triples

      • $a=4$: $b = 12 - 8 = 4$, $c = 4 - 4 = 0$ → $(4,4,0)$
      • $a=5$: $b = 12 - 10 = 2$, $c = 5 - 4 = 1$ → $(5,2,1)$
      • $a=6$: $b = 12 - 12 = 0$, $c = 6 - 4 = 2$ → $(6,0,2)$

   4. Compute the coefficient for each triple

      The general term’s coefficient (ignoring the $x$-power) is:

   $\frac{8!}{a!\,b!\,c!} \cdot (1)^a \cdot (2)^b \cdot (2)^c = \frac{8!}{a!\,b!\,c!} \cdot 2^{b+c}.$

   • For $(4,4,0)$: $$$$

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7. 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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8. 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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9. 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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10. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.The number of integral terms in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is (A) 32 (B) 33 (C) 34 (D) 35

    ›Reveal solutionSolution

    To find the number of integral terms, we determine the conditions on the exponent $r$ in the general term such that the radical parts become integers. This requires $r$ to be a multiple of the least common multiple of the denominators of the fractional exponents. For $(\sqrt{3} + \sqrt[8]{5})^{256}$, the terms are integral when $r$ is a multiple of 8, leading to 33 integral terms.

    The problem asks us to find the number of integral terms in the binomial expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$. An "integral term" is a term whose numerical value is an integer.

    The core idea here is to examine the general term of the binomial expansion. For a term to be an integer, any radical expressions within it must resolve to an integer. The binomial coefficient $\binom{n}{r}$ is always an integer, so we only need to focus on the powers of the base numbers.

    Let's write the given expression in a form that makes the exponents clear:

    $\sqrt{3} = 3^{1/2}$

    $\sqrt[8]{5} = 5^{1/8}$

    So the expansion is $(3^{1/2} + 5^{1/8})^{256}$.

    1. Write the general term of the expansion. The general term, $T_{r+1}$, in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. In our case, $n=256$, $a=3^{1/2}$, and $b=5^{1/8}$. Substituting these values, the general term is:

    $T_{r+1} = \binom{256}{r} (3^{1/2})^{256-r} (5^{1/8})^r$

    2. Simplify the exponents. Using the power rule $(x^a)^b = x^{ab}$, we simplify the exponents:

    $T_{r+1} = \binom{256}{r} 3^{\frac{256-r}{2}} 5^{\frac{r}{8}}$

    Here, $r$ is an integer ranging from $0$ to $256$ (i.e., $0 \le r \le 256$).
    

    3. Determine the conditions for an integral term.

    For $T_{r+1}$ to be an integral term, the powers of 3 and 5 must be non-negative integers. The binomial coefficient $\binom{256}{r}$ is always an integer.

    Therefore, we need:

    * $\frac{256-r}{2}$ to be an integer.

    * $\frac{r}{8}$ to be an integer.

    4. Analyze the conditions on $r$.
       • From the second condition, $\frac{r}{8}$ must be an integer. This means $r$ must be a multiple of 8. So, $r \in \{0, 8, 16, 24, \dots\}$. …

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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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