Limits and Derivatives — JEE MCQs with answers

Answer: B — With $u=x-\frac{\pi}{5}$, $\cos(u+h)-\cos u=-2\sin\left(u+\frac{h}{2}\right)\sin\frac{h}{2}$, and $\frac{\sin(h/2)}{h}\to\frac{1}{2}$, so $f'(x)=-\sin\left(x-\frac{\pi}{5}\right)$. Replacing $\frac{\sin(h/2)}{h}$ by $1$ gives $-2\sin\left(x-\frac{\pi}{5}\right)$. Dropping the factor $2$ in $-2\sin\left(u+\frac{h}{2}\right)\sin\frac{h}{2}$ gives $-\frac{1}{2}\sin\left(x-\frac{\pi}{5}\right)$. Splitting $\cos\left(x+h-\frac{\pi}{5}\right)$ as $\cos(x+h)+\cos\frac{\pi}{5}$, as if cosine added across a sum, reduces the quotient to $\frac{\cos(x+h)-\cos x}{h}$ and gives $-\sin x$.

Answer: A — Put $y=1+x$, so that $y\to 1$ as $x\to 0$, as NCERT does in Example 3(ii). The limit becomes $\lim_{y\to 1}\dfrac{y^{1/4}-1^{1/4}}{y-1}$, and by the Remark after Theorem 2 (rational $n$, positive $a$) this is $\frac{1}{4}\cdot 1^{-3/4}=\frac{1}{4}$, option A. Option C ($-\dfrac{3}{4}$) uses $(n-1)a^{n-1}$: $\left(-\frac{3}{4}\right)\cdot 1=-\frac{3}{4}$. Option D ($1$) reads the $\dfrac{0}{0}$ obtained by substituting $x=0$ as $1$. Option B ($0$) reads the $\dfrac{0}{0}$ obtained by substituting $x=0$ as $0$.

Answer: D — The step $\tfrac{1}{2}OA.CD < \tfrac{x}{2\pi}.\pi.(OA)^2 < \tfrac{1}{2}OA.AB$ uses the sector-area formula, which equals $\tfrac{x}{2\pi}$ of the full circle's area only when $x$ is the radian measure of the angle -- in degrees the fraction of the circle would be $x/360$, not $x/(2\pi)$. $\sin x,\cos x$ are defined for any angle unit; $OA$ is the unit circle's radius (length $1$), not $\pi$; and the radian requirement comes from this geometric step, not from the Sandwich Theorem.

Answer: A — A is correct: §12.2 states this definition directly, and then applies it to compute 9.8 m/s and 14.7 m/s for the first examples. B is wrong: §12.2 uses the DIFFERENCE of the distances (distance travelled BETWEEN t₁ and t₂), not their sum. C is wrong: §12.2 divides by (t₂ - t₁), the length of the time interval, not by t₂ alone. D is wrong: §12.2 divides by (t₂ - t₁); it does not multiply by it.

Answer: C — C is correct: This is the exact sentence printed in §12.1 describing Calculus. A is wrong: §12.1 says the points in the DOMAIN change, not the range; this reverses the printed wording. B is wrong: This reverses the direction §12.1 prints: it is the points in the domain that change, and the value of the function that changes with them. D is wrong: §12.1 speaks of a function's value changing as points in its domain change; it does not restrict the changing variable to time.

Answer: A — Following Example 12, $f'(x)=\lim_{h\to0}\frac{1}{h}\left[\frac{(x+3)-(x+h+3)}{(x+3)(x+h+3)}\right]=\lim_{h\to0}\frac{-1}{(x+3)(x+h+3)}=-\frac{1}{(x+3)^2}$. Combining the two fractions over the single denominator $x+h+3$ gives $\lim_{h\to0}\frac{-1}{x+h+3}=-\frac{1}{x+3}$. Copying Example 12's answer for $\frac{1}{x}$ without the shift by $3$ gives $-\frac{1}{x^2}$, the derivative of a different function. Simplifying correctly but then putting $h=1$ instead of letting $h\to0$ gives $-\frac{1}{(x+3)(x+4)}$.

Answer: B — NCERT's exact remark: "We also remark that we will not come across such limits in this course." This is an explicit scope statement fencing infinite limits out of the syllabus, even though this one instance is shown. Options A and C are wrong: NCERT says the opposite — such limits are set aside, and Theorem 1 is not applied to them. Option D is wrong: here the function is not even defined at 0.

Answer: C — Write $\frac{\sin 2x}{5x} = \frac{2}{5}\cdot\frac{\sin 2x}{2x}$. As $x\to0$, $2x\to0$, so $\frac{\sin 2x}{2x}\to1$ by part (i) of Theorem 5 of §12.4 (p.235), and the limit is $\frac{2}{5}$ (the method of EXERCISE 12.1 Q13, p.237). The other options: $\frac{5}{2}$ comes from swapping the two constants, i.e. working with $\frac{\sin 5x}{2x}$. $2$ comes from ignoring the constant $5$ in the denominator, i.e. using $\frac{\sin 2x}{x}$. $\frac{1}{2}$ comes from turning the fraction upside down and also ignoring the $5$.

Answer: B — B is correct: Fig 12.1 marks point A on the curve at t = 2, the same instant §12.2's whole discussion is built around. A is wrong: Fig 12.1 marks A at t = 2; t = 4 is the last instant in Table 12.1, not where A sits. C is wrong: 19.6 is the distance s(2) = 4.9 × 2² at A -- its vertical coordinate, not its time. D is wrong: Point A is marked at t = 2 in Fig 12.1, not at the origin t = 0.

Answer: D — D is correct: The epigraph is signed 'WHITEHEAD' in the printed text. A is wrong: A.L. Cauchy is named only in the chapter's closing Historical Note, not as the author of the epigraph. B is wrong: Newton is pictured in a portrait on the same page, but the epigraph itself is signed WHITEHEAD. C is wrong: Karl Weierstrass is named only in the chapter's closing Historical Note, not as the author of the epigraph.