Integrals — JEE MCQs with answers

Answer: D — Formula 7.4 (5) is $\int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin^{-1}\frac{x}{a} + C$, and it carries no numerical factor in front. With $a = 4$ this gives option D, valid on $|x| < 4$. Option A puts $a^{2} = 16$ into the argument where $a = 4$ belongs. Option B attaches a factor $\frac{1}{a}$ that formula (5) does not have, borrowing it from formula (3). Option C makes both changes at once. Differentiating at $x = -2,\ \tfrac12,\ 1,\ 3$ returns $\frac{1}{\sqrt{16-x^{2}}}$ for option D alone.

Answer: C — The quotient is $1$ and the remainder is $(x^{2}+1)-(x^{2}-5x+6) = 5x-5$, so $\frac{x^{2}+1}{x^{2}-5x+6} = 1+\frac{5x-5}{x^{2}-5x+6}$, option C. The remaining fraction is proper and its denominator is $(x-2)(x-3)$. Option A subtracts the wrong way round, which reverses the sign of the remainder. Option B computes the constant of the remainder as $1+6$ in place of $1-6$. Option D makes both slips. A quick check at $x = 0$: the left side is $\frac{1}{6}$, and option C gives $1-\frac{5}{6} = \frac{1}{6}$, while A, B and D give $\frac{11}{6}$, $\frac{13}{6}$ and $-\frac{1}{6}$.

Answer: C — Option C is correct. Put $\sin^{-1}x = t$; then $\frac{dx}{\sqrt{1-x^{2}}} = dt$ and the integral is $\int t\,dt = \frac{t^{2}}{2} + \mathrm{C}$, and differentiating this with respect to $x$ returns the integrand. Slip 1 omits the $\frac{1}{2}$ from $\int t\,dt$. Slip 2 substitutes $\cos^{-1}x = t$ instead, whose differential is $-\frac{dx}{\sqrt{1-x^{2}}}$, and then ignores the minus sign. Option A is slip 1 alone, option B is slip 2 alone and option D is the two together. Although $\sin^{-1}x$ and $-\cos^{-1}x$ differ by the constant $\frac{\pi}{2}$, their squares do not: $\frac{1}{2}(\sin^{-1}x)^{2} - \frac{1}{2}(\cos^{-1}x)^{2} = \frac{\pi}{2}\sin^{-1}x - \frac{\pi^{2}}{8}$, which varies with $x$, so option B is a different function and not a second correct answer. Domain: $-1 < x < 1$.

Answer: B — B. The aside reads: here it is assumed that $f(x) > 0$ for $x \in [a, b]$. It then adds that the assertion made below is equally true for other functions as well, a claim the book makes without giving any reason. Option C opens the interval, which is not what is printed. Option A is a condition on the derivative; the curve in Fig 7.1 does rise, but the text does not assume it. Option D makes both changes.

Answer: D — Since $x^{2}+2x+5=(x+1)^{2}+4$, put $x+1=y$ to get $\int\sqrt{y^{2}+2^{2}}\,dy$. Result (ii) of Section 7.6.2 with $a=2$ gives $\frac{1}{2}y\sqrt{y^{2}+4}+\frac{4}{2}\log\left|y+\sqrt{y^{2}+4}\right| + C$, and returning to $x$ gives $\frac{1}{2}(x+1)\sqrt{x^{2}+2x+5}+2\log\left|x+1+\sqrt{x^{2}+2x+5}\right| + C$, valid for every real $x$: option D. Option B uses $a^{2}=4$ as the coefficient instead of $\frac{a^{2}}{2}=2$. Option C applies the result as if the variable were $x$ and never replaces $y$ by $x+1$ outside the radical. Option A makes both slips. At $x=0$ the integrand is $\sqrt{5}$, and only option D has that derivative.

Answer: A — A. The page prints $\frac{d}{dx}[\mathrm{F}(x) + \mathrm{C}] = f(x)$, $x \in \mathrm{I}$, and it is careful to keep the same $\mathrm{I}$ in the conclusion as in the hypothesis. Option B adds the constant to the right-hand side as well, which is false. Option C multiplies by the constant, giving a true identity about a different function, not the statement made. Option D is the trap worth naming: it copies the correct formula but widens the conclusion to every real $x$, although the hypothesis was only ever made on $\mathrm{I}$. Nothing is known about $\mathrm{F}$ outside $\mathrm{I}$, and the Remark on the next page depends on staying inside one interval.

Answer: A — Call the integral $I$. Here $a + b = 2 + 6 = 8$, so $P_3$ replaces $x$ by $8 - x$ and $I = \int_2^6 \frac{\sqrt{8-x}}{\sqrt{8-x}+\sqrt{x}}\,dx$. Adding the two forms gives $2I = \int_2^6 1\,dx = 6 - 2 = 4$, so $I = 2$: option A. Option B takes $I = 4$, forgetting that the sum is $2I$. Option C takes $\int_2^6 1\,dx$ as the upper limit $6$ instead of the length $4$ of the interval, as would be right only if the lower limit were $0$, and then halves. Option D makes both slips.

Answer: B — Here $dt=5x^{4}\,dx$, so $5x^{4}\sqrt{x^{5}+1}\,dx$ becomes $\sqrt{t}\,dt$. When $x=-1$, $t=0$ and when $x=1$, $t=2$, so $t$ varies from $0$ to $2$. The integral is therefore $\int_{0}^{2}\sqrt{t}\,dt$, which is option B; this is the transformed integral in the Alternatively part of Example 26. Option A keeps the limits $-1$ and $1$, which belong to $x$ and not to $t$. Option C takes $dt=x^{4}\,dx$ and so leaves a factor 5 behind. Option D makes both of these mistakes.

Answer: B — No definite integral in this chapter is evaluated as the limit of a sum. Theorem 2 in Section 7.8 gives the value as the difference of the values of F at the end points, and Section 7.9 evaluates definite integrals by substitution, again through an anti derivative. The first of the two ways is mentioned only twice in the whole chapter, here and in Remark (ii) after Theorem 2, which recommends the theorem precisely because it spares the reader that calculation; it is announced and never carried out. That is option B. Options A, C and D each credit a section with a development it does not contain: Section 7.8 states the two fundamental theorems and introduces the area function, Section 7.9 is the substitution method for definite integrals, and the Summary only repeats what the body has already said.

Answer: D — The Introduction describes the Fundamental Theorem of Calculus as a connection between indefinite integral and definite integral which makes the definite integral a practical tool for science and engineering, which is option D. Option A misdescribes it twice over: the Introduction calls it a connection, not an existence criterion, and the chapter states its two fundamental theorems without proving them, saying that their proofs are beyond its scope. Option B is the guidance on the choice of the first function, which belongs to Section 7.6 and is given there as a Remark. Option C is partial fraction decomposition, the method of Section 7.5. Neither of those two has anything to do with the sentence in the Introduction.