Application of Derivatives — JEE MCQs with answers

Answer: D — Option D is correct. Marginal revenue is $\dfrac{dR}{dx}=100-0.5x=85$ at $x=30$; marginal cost is $\dfrac{dC}{dx}=x+40=70$ at $x=30$. The marginal profit is $85-70=15$, so profit is still rising at that output. Option A subtracts the wrong way round, computing marginal cost minus marginal revenue. Option B differentiates $-0.25x^{2}$ to $-0.25x$, giving a marginal revenue of $92.5$. Option C makes both slips: the same power-rule error, subtracted the wrong way round.

Answer: C — $f'(x) = 2x - 1 < 0$ exactly when $x < \frac{1}{2}$, so inside $(-1, 1)$ the decreasing part is $\left(-1, \frac{1}{2}\right)$ — option (C). Option (D) is the increasing part, obtained by reading the sign of $f'$ backwards. Option (A) comes from solving $2x + 1 = 0$ and placing the turning point at $-\frac{1}{2}$. Option (B) applies both slips at once. Check: $f'(0) = -1 < 0$ and $f'\left(\frac{3}{4}\right) = \frac{1}{2} > 0$.

Answer: C — Because the definition is stated only for an interior point, an end point of the domain can never be called a point of local maxima or local minima by it; C is the key. The other three options all put a condition on the derivative, which Definition 4 never mentions at all. A is not excluded — Example 19 (p.166) finds a local minimum of $3 + |x|$ at $x = 0$, where $f$ is not differentiable. B belongs to Theorem 3(iii), not to Definition 4, which compares values of $f$ and never differentiates. D describes the ordinary case rather than an exclusion.

Answer: A — A. The Introduction says the student has learnt to find the derivative of composite functions, inverse trigonometric functions, implicit functions, exponential functions and logarithmic functions. B replaces 'implicit' by 'parametric'; C replaces 'inverse trigonometric' by 'hyperbolic'; D carries both substitutions at once. Neither 'parametric' nor 'hyperbolic' occurs in the paragraph.

Answer: D — $f'(x) = 3x^2 + 2ax + 3$. A quadratic with a positive leading coefficient is $\ge 0$ for every real $x$ exactly when its discriminant is $\le 0$: $(2a)^2 - 4(3)(3) \le 0$, i.e. $4a^2 \le 36$, i.e. $-3 \le a \le 3$ — option (D). Sympy confirms the boundary: at $a = 3$ and at $a = -3$ the minimum of $f'$ over $\mathbf{R}$ is exactly $0$, while at $a = 4$ and at $a = -4$ it is $-\frac{7}{3} < 0$. Option (A) reverses the discriminant inequality, which is the condition for $f'$ to take negative values. Option (B) keeps only the upper branch. Option (C) comes from computing the discriminant's $4ac$ term as $4 \times 3 = 12$ instead of $4 \times 3 \times 3 = 36$: that turns $4a^2 \le 36$ into $4a^2 \le 12$, i.e. $a^2 \le 3$ and $-\sqrt{3} \le a \le \sqrt{3}$.

Answer: C — $f$ is increasing, so on $[0, 1]$ its largest value is $f(1) = 3$: option C. Option A interchanges the value and the point. Option B gives $f(0) = 2$, which is the absolute minimum value on $[0, 1]$, not the maximum. Option D keeps the correct value but attaches it to the point where the minimum occurs.

Answer: D — D. Example 5 states 'by marginal cost we mean the instantaneous rate of change of total cost at any level of output', and the solution repeats that marginal cost is the rate of change of total cost with respect to the output. B keeps the right quantity but differentiates with respect to time, which the section never does for cost. A keeps the right independent variable but replaces cost by revenue, which is the section's definition of marginal revenue instead. C carries both substitutions.

Answer: D — D. The sentence runs '(iii) to find turning points on the graph of a function which in turn will help us to locate points at which largest or smallest value (locally) of a function occurs'. A names the roots of the function, which the paragraph never mentions. B names the points of non-differentiability, a notion that enters only much later, in $\S$6.4. C states a different announced use altogether: the paragraph lists locating increasing and decreasing intervals as a separate item, not as the purpose of turning points.

Answer: B — $f'(x) = \cos x - \sin x = 0$ gives $x = \dfrac{\pi}{4}$ in $[0,\pi]$. The three candidates are $f(0) = 1$, $f\left(\dfrac{\pi}{4}\right) = \sqrt{2}$ and $f(\pi) = -1$, so the least value is $-1$, at the right end point, and the greatest is $\sqrt{2}$, at the interior critical point. So option B is correct. Option A: the least value of $\sin x + \cos x$ over the whole of $\mathbf{R}$, which needs $x$ to reach $\dfrac{5\pi}{4}$, outside $[0,\pi]$. Option C: $f\left(\dfrac{3\pi}{4}\right)$, the point of $[0,\pi]$ where $f$ vanishes, mistaken for its least value. Option D: the absolute MAXIMUM on $[0,\pi]$, given in place of the minimum.

Answer: A — Option A is correct. Differentiating $x^{2}+y^{2}=25$ gives $\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}$, and with $x=4$ m, $y=3$ m and $\dfrac{dx}{dt}=2$ cm/s this is $-\dfrac{8}{3}$ cm/s. The printed key drops the minus sign and states the magnitude, because the question itself already says the height is decreasing. A stem must therefore say which of the two it wants. Option B contradicts the question, which states that the height is decreasing. Option C names the given datum instead of the answer; the foot moves at $2$ cm/s. Option D describes $\dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac43$, a different number with a different unit.