Differential Calculus Questions and Answers

Answer: C

The tangent is y - 1 = 2(x - 1).

The slope of \(y = x^{2}\) is \(dy/dx = 2x\), which is 2 at x = 1. The point-slope form through (1,1) is therefore y - 1 = 2(x - 1), that is, y = 2x - 1.

y = 2x - 2 has the right slope but passes through (1, 0), not (1, 1). x = 2y - 1 and x - 1 = 2(y - 1) have slope 1/2, which comes from swapping x and y, the tangent to the inverse curve.

Answer: C

The height is \(\frac{4}{3}R\).

Put the cone's apex on the sphere and its base circle on the sphere too. If the height is h, the base lies a distance h - R from the centre, so the base radius r satisfies \(r^{2} = R^{2} - (h - R)^{2} = 2Rh - h^{2}\).

\[ \begin{aligned} V &= \frac{\pi}{3}(2Rh^{2} - h^{3}) \\ \frac{dV}{dh} &= \frac{\pi}{3}(4Rh - 3h^{2}) = 0 \\ h &= \frac{4R}{3} \end{aligned} \]

The second derivative, \(4R - 6h\), is negative there, so this is a maximum. h = R puts the base through the centre (a cone of volume \(\pi R^{3}/3\), smaller than the optimum \(32\pi R^{3}/81\)), and 3R/2 and 5R/4 are other heights with smaller volume.

Answer: D

Rolle's theorem needs \(f(0) = f(\pi)\) and a point \(c\) in \((0, \pi)\) where \(f'(c) = 0\). Here \(f(0) = \cos 0 = 1\) and \(f(\pi) = \cos 4\pi = 1\), so the theorem applies.

\[ f'(x) = -4\sin 4x = 0 \]

\[ 4x = k\pi \Rightarrow x = \frac{k\pi}{4} \]

Inside the open interval this gives \(\pi/4\), \(\pi/2\) and \(3\pi/4\). Of the values offered, only \(\pi/4\) is one of them. The values \(0\) and \(\pi\) are the end points, not interior points, and \(\pi/3\) gives \(f'(\pi/3) = -4\sin(4\pi/3) \neq 0\). The answer is \(\pi/4\), option D.

Answer: B

Use the sum to product identity, or simply note that the first term is fixed because \(x + y = 6\).

\[ z = \cos 6 + \cos(x - y) \]

The only term that varies is \(\cos(x - y)\), whose largest value is \(1\) (reached when \(x - y = 0\), that is \(x = y = 3\)). So the maximum is \(\cos 6 + 1 = 1 + \cos 6\). The option \(1 - \cos 6\) has the wrong sign on the fixed term, \(1 + \sin 2\) and \(1 + \cos 3\) do not equal \(1 + \cos 6\), since \(\cos 3 \neq \cos 6\). The answer is option B.

Answer: B

The statement that is not true is that every continuous function is differentiable. The function \(f(x) = |x|\) is continuous at \(0\) but has a corner there, so its derivative does not exist at \(0\). The opposite statement, that every differentiable function is continuous, is a standard theorem and is true.

The function \(|x| + \sin|x-3|\) is a sum of continuous functions, so it is continuous, and it has corners at \(x = 0\) and \(x = 3\) where the absolute values change sign, so it is not differentiable there. Both of those statements are true, which leaves option B as the false one.

Answer: A

Take natural logarithms of \(x^y y^x = C\) to get \(y\ln x + x\ln y = \ln C\) and differentiate implicitly:

The negative sign is essential because the two sides of the constraint trade off against each other. The unsigned version in the second option and the options with x and y swapped (the third and fourth) come from errors in collecting the terms with y' or in inverting the fraction.

Answer: A

The nth derivative is \(\dfrac{(-1)^{n}n!}{(x + 3)^{n+1}}\).

Write \(y = (x + 3)^{-1}\) and differentiate repeatedly:

\[ \begin{aligned} y' &= -(x + 3)^{-2} \\ y'' &= 2(x + 3)^{-3} \\ y''' &= -6(x + 3)^{-4} \end{aligned} \]

Each step multiplies by the current power (giving \(1, 2, 6 = n!\)), flips the sign and raises the power of \((x+3)\) in the denominator by one, so \(y^{(n)} = (-1)^{n}n!/(x + 3)^{n+1}\).

The form with \((-1)^{n+1}\) has the wrong sign (it gives \(+1\) for the first derivative), and the forms with \((x + 3)^{n}\) or \((n + 1)!\) fail at \(n = 1\).

Answer: B

The roots lie on an elliptical curve.

The roots are the points where \(f(x, y) = 0\):

\[ \begin{aligned} \frac{x^{2}}{2} + \frac{y^{2}}{3} &= 5 \\ \frac{x^{2}}{10} + \frac{y^{2}}{15} &= 1 \end{aligned} \]

This is an ellipse in the xy-plane with semi-axes \(\sqrt{10}\) along x and \(\sqrt{15}\) along y.

An ellipse has infinitely many points, so the roots are not a finite set. A sphere would need three variables, and the function has only two. A hyperbola would need the two squared terms to have opposite signs, but both are positive here.

Answer: B

The minimum is at \(x = -\frac{1}{2}\).

Expand and differentiate:

\[ \begin{aligned} y &= (x + 3)(x - 2) = x^{2} + x - 6 \\ y' &= 2x + 1 = 0 \Rightarrow x = -\tfrac{1}{2} \\ y'' &= 2 \gt 0 \end{aligned} \]

The positive second derivative confirms a minimum, and \(-\tfrac{1}{2}\) lies inside \(-4 \lt x \lt 4\). A quicker route: the parabola opens upward and is symmetric about the midpoint of its roots \(-3\) and \(2\), which is \(-\tfrac{1}{2}\). The minimum value is \(y = -6.25\).

The value \(\tfrac{1}{2}\) has the sign wrong, and \(\pm\tfrac{3}{2}\) do not make the derivative zero.

Answer: C

\(A = 3\).

For continuity at x = 2, the left-hand limit must equal the value of the function there. The value and right-hand limit come from \(1 + x^{2}\), and the left-hand limit from \(A + x\):

\[ \begin{aligned} A + 2 &= 1 + 2^{2} = 5 \\ A &= 3 \end{aligned} \]

A = 2 would leave a jump of 1 at x = 2, and 2.5 and 3.5 leave jumps of 0.5. Only A = 3 makes the two pieces meet.