Continuity and Differentiability — JEE MCQs with answers

Answer: A — The chain rule gives $-\sin(xy)$ times the derivative of the inner function $xy$, and the product rule gives $\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$. Hence the answer is $-\sin(xy)\left(y + x\dfrac{dy}{dx}\right)$ — option A. Option B keeps both terms of the product rule but divides by the chain factor instead of multiplying by it, writing $\dfrac{x}{\dfrac{dy}{dx}}$ where $x\dfrac{dy}{dx}$ belongs. Option C differentiates $xy$ as though it were $x + y$, giving $1 + \dfrac{dy}{dx}$. Option D makes the same product-rule error as C and in addition loses the minus sign that comes from differentiating the cosine.

Answer: D — $\log y = \cos x\,\log\sin x$ gives $\frac{1}{y}\frac{dy}{dx} = \cos x\cdot\frac{\cos x}{\sin x} - \sin x\,\log\sin x = \cos x\cot x - \sin x\log\sin x$; multiplying back by $y$ gives the printed answer. The answer is option D. Option A is taking $\frac{d}{dx}\cos x$ as $+\sin x$. Option B is stopping at $\frac{1}{y}\frac{dy}{dx}$ and never multiplying back by $y$. Option C is both slips at once: the sign of $\frac{d}{dx}\cos x$ wrong AND the result never multiplied back by $y$.

Answer: B — Option B is correct: $x\sqrt{x}=x^{3/2}$, so the derivative is $\dfrac{1}{\sqrt{1-x^{3}}}\cdot\dfrac{3}{2}x^{1/2}=\dfrac{3\sqrt{x}}{2\sqrt{1-x^{3}}}$, which is the printed $\dfrac{3}{2}\sqrt{\dfrac{x}{1-x^{3}}}$. Option A drops the inner derivative $\dfrac{3}{2}x^{1/2}$ entirely. Option C uses $1-x^{2}$ under the root instead of $1-u^{2}$ with $u=x^{3/2}$. Option D makes both slips at once and so returns the derivative of $\sin^{-1}x$ itself.

Answer: D — Writing $\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x}$ and using the Class XI standard limit gives $\lim_{x \to 0^{-}} \frac{\sin 3x}{x} = 3$, so option D is the key. The right hand limit and the value are unchanged at $\lim_{x \to 0^{+}}(x+1) = 1 = f(0)$, so the change turns $x = 0$ into a point of discontinuity. Option A, $1$, is the left hand limit of the original function and is what comes of ignoring the $3$ inside the sine. Option B, $\frac{1}{3}$, divides by $3$ where one should multiply, the reciprocal slip. Option C, $0$, puts $x = 0$ straight into the numerator, $\sin 0 = 0$, and reads the quotient as $0$.

Answer: B — p.127 prints $\log_{b}:\mathbf{R}^{+}\to\mathbf{R}$, which is B - "a function from positive real numbers to all real numbers". A is the exponential's direction, domain $\mathbf{R}$ and positive values. C would deny the negative values $\log x$ takes for $0<x<1$. D would give the logarithm a domain containing $0$ and negative numbers, which log feature (1) rules out.

Answer: A — A. The sentence reads 'Further, we introduce a new class of functions called exponential and logarithmic functions.' B and D rest on the same wrong premise: both name function types the paragraph does mention, and treat a mention as the announcement of the new class. Inverse trigonometric functions are named only as functions whose differentiation will be learnt, and polynomial and trigonometric functions only as Class XI material. C names two functions that $\S$5.2 uses heavily as worked Examples — Examples 3, 7, 15 and 20 — but that the Introduction never mentions at all.

Answer: D — $\frac{dx}{d\theta} = -3a\cos^{2}\theta\sin\theta$, which is zero exactly when $\cos\theta = 0$ or $\sin\theta = 0$. Of the four values offered, only $\theta = \frac{\pi}{2}$ makes $\cos\theta = 0$, so **option D** is the answer. At $\theta = \frac{\pi}{6}$ (**option A**) the value is $-\frac{9a}{8}$, at $\theta = \frac{\pi}{4}$ (**option B**) it is $-\frac{3\sqrt{2}a}{4}$ and at $\theta = \frac{\pi}{3}$ (**option C**) it is $-\frac{3\sqrt{3}a}{8}$; none of these is zero, so the rule gives $-\tan\theta$ without trouble. Notice that $\frac{dy}{d\theta}$ also vanishes at $\theta = \frac{\pi}{2}$; the rule still gives nothing, because its condition is on the denominator.

Answer: D — Using the table of $\S 5.3.3$, differentiating with respect to $x$ gives $\dfrac{1}{\sqrt{1 - x^{2}}} + \dfrac{1}{\sqrt{1 - y^{2}}}\dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = -\dfrac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$ — option D. On this curve $0 < x < 1$ and $0 < y < 1$, so both radicals are real and positive and the derivative is negative, as it must be. Option A has the sign of a sum instead of the sign obtained after moving a term across. Option B interchanges the two radicals, the slip of dividing the wrong way round when solving for $\dfrac{dy}{dx}$. Option C differs in two places: it interchanges the radicals and drops the sign as well.

Answer: D — $\log y = \frac{1}{2}\left[\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5)\right]$; differentiating each logarithm gives $\frac{1}{x-k}$ with the sign it carries. The answer is option D. Option A is reading the quotient as a product, so all five logarithms are added. Option B is applying the factor $\frac{1}{2}$ a second time inside the bracket. Option C is both slips at once: the quotient read as a product AND the factor $\frac{1}{2}$ applied a second time.

Answer: D — Option D is correct. Section 5.6 derives only the first order rule $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$, valid whenever $\dfrac{dx}{dt}\neq0$, and the chapter prints no formula at all for the second order derivative of a parametric pair; a solver must differentiate $\dfrac{dy}{dx}$ with respect to $t$ and divide by $\dfrac{dx}{dt}$ a second time. Option A states a rule the chapter never gives, and it is false: for Q17 it yields $1.4395$ at $a=2$, $t=0.5$ instead of $1.4796$. Option B is incomplete as written and is not printed in Section 5.7, which treats only $y$ as a function of $x$. Option C is false as well: the Summary's five standard derivatives are those of $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$, $e^{x}$ and $\log x$, and it mentions neither parametric nor second order derivatives.