12th Standard Mathematics — Applications of Differential Calculus: Online Practice Test

For a sphere \( V=\frac{4}{3}\pi r^{3} \), so differentiating with respect to time, \[ \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}. \] Substituting the given rate \( \frac{dV}{dt}=3\pi \) and \( r=\frac{1}{2} \): \[ 3\pi = 4\pi\left(\tfrac{1}{2}\right)^{2}\frac{dr}{dt}=\pi\frac{dr}{dt}. \] Cancelling \( \pi \) gives \( \frac{dr}{dt}=3 \) cm/s, which is option (1).

Let \( y \) be the height of the balloon. With the observer 40 m from the launch point, \( \tan\theta=\frac{y}{40} \). Differentiating, \[ \sec^{2}\theta\,\frac{d\theta}{dt}=\frac{1}{40}\frac{dy}{dt}. \] When \( y=30 \) the line of sight has length \( \sqrt{40^{2}+30^{2}}=50 \), so \( \cos\theta=\frac{40}{50}=\frac{4}{5} \) and \( \sec^{2}\theta=\frac{25}{16} \). Using \( \frac{dy}{dt}=10 \): \[ \frac{25}{16}\frac{d\theta}{dt}=\frac{10}{40}=\frac{1}{4}\;\Rightarrow\;\frac{d\theta}{dt}=\frac{1}{4}\cdot\frac{16}{25}=\frac{4}{25}\text{ rad/s}. \] Option (2).

The particle is at rest when its velocity is zero. From \( s(t)=3t^{2}-2t-8 \), the velocity is \( v(t)=s'(t)=6t-2 \). Setting \( 6t-2=0 \) gives \( t=\frac{1}{3} \), option (2).

At the highest point the upward velocity is momentarily zero. From \( x=80t-16t^{2} \), \( \frac{dx}{dt}=80-32t \). Setting this equal to zero gives \( t=\frac{80}{32}=2.5 \) seconds, option (2).

\"Changes 8 times as fast\" means \( \frac{dy}{dt}=8\frac{dx}{dt} \). Differentiating \( 6y=x^{3}+2 \) with respect to time gives \( 6\frac{dy}{dt}=3x^{2}\frac{dx}{dt} \). Substituting, \[ 6(8)\frac{dx}{dt}=3x^{2}\frac{dx}{dt}\Rightarrow 48=3x^{2}\Rightarrow x^{2}=16\Rightarrow x=\pm4. \] For \( x=4 \): \( 6y=64+2=66\Rightarrow y=11 \), giving the point \( (4,11) \). (The choice \( x=-4 \) gives a point not among the options.) Option (1).

Write \( f(x)=(8-2x)^{1/2} \). Then \[ f'(x)=\frac{1}{2}(8-2x)^{-1/2}(-2)=\frac{-1}{\sqrt{8-2x}}. \] Setting the slope equal to \( -0.25=-\frac{1}{4} \): \[ \frac{-1}{\sqrt{8-2x}}=-\frac{1}{4}\Rightarrow\sqrt{8-2x}=4\Rightarrow 8-2x=16\Rightarrow x=-4. \] Option (2).

Differentiate: \( f'(x)=-8\sin 4x \). At \( x=\frac{\pi}{12} \) we have \( 4x=\frac{\pi}{3} \) and \( \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2} \), so the tangent slope is \( -8\cdot\frac{\sqrt{3}}{2}=-4\sqrt{3} \). The normal is perpendicular to the tangent, so its slope is \[ -\frac{1}{-4\sqrt{3}}=\frac{1}{4\sqrt{3}}=\frac{\sqrt{3}}{12}. \] Option (3).

A tangent is vertical where \( \frac{dy}{dx} \) is undefined. Differentiating \( y^{2}-xy+9=0 \) implicitly: \[ 2y\,y'-\left(y+x\,y'\right)=0\Rightarrow y'(2y-x)=y\Rightarrow y'=\frac{y}{2y-x}. \] This is undefined when the denominator vanishes, i.e. \( x=2y \). Substituting into the curve: \[ y^{2}-(2y)y+9=0\Rightarrow -y^{2}+9=0\Rightarrow y^{2}=9\Rightarrow y=\pm 3. \] Option (4).

Find each curve's tangent direction at the origin. For \( y^{2}=x \): \( 2y\,y'=1\Rightarrow y'=\frac{1}{2y} \), undefined at \( y=0 \) — the tangent is the vertical \( y \)-axis. For \( x^{2}=y \): \( y'=2x \), which is \( 0 \) at \( x=0 \) — the tangent is the horizontal \( x \)-axis. A vertical and a horizontal line meet at right angles, so the angle is \( \frac{\pi}{2} \). Option (3).

Combine over a common denominator: \[ \cot x-\frac{1}{x}=\frac{x\cos x-\sin x}{x\sin x}, \] which is the indeterminate form \( \frac{0}{0} \) as \( x\to 0 \). Apply L'Hopital's rule: the numerator derivative is \( \cos x-x\sin x-\cos x=-x\sin x \) and the denominator derivative is \( \sin x+x\cos x \); this is still \( \frac{0}{0} \). Applying the rule once more, \[ \lim_{x\to0}\frac{-\sin x-x\cos x}{2\cos x-x\sin x}=\frac{0}{2}=0. \] Option (1).

Simplify with identities: \[ \sin^{4}x+\cos^{4}x=(\sin^{2}x+\cos^{2}x)^{2}-2\sin^{2}x\cos^{2}x=1-\tfrac{1}{2}\sin^{2}2x=\tfrac{3}{4}+\tfrac{1}{4}\cos 4x. \] Then \( f'(x)=-\sin 4x \). The function increases where \( f'(x)>0 \), i.e. where \( \sin 4x<0 \). On \( \left[\frac{\pi}{4},\frac{\pi}{2}\right] \), the angle \( 4x \) runs through \( [\pi,2\pi] \), where the sine is negative — so \( f \) is increasing there. Option (3).

The function \( f(x)=x^{3}-3x^{2} \) is a polynomial (continuous and differentiable everywhere), and \( f(0)=0=f(3) \), so Rolle's theorem applies on \( [0,3] \). It guarantees a \( c\in(0,3) \) with \( f'(c)=0 \). Here \( f'(x)=3x^{2}-6x=3x(x-2) \), giving \( x=0 \) or \( x=2 \). Only \( x=2 \) lies in the open interval, so \( c=2 \). Option (4).

The Mean Value Theorem guarantees a \( c\in(1,9) \) with \( f'(c)=\frac{f(9)-f(1)}{9-1} \). For \( f(x)=\frac{1}{x} \), \( f'(x)=-\frac{1}{x^{2}} \), and the average rate of change is \[ \frac{\tfrac{1}{9}-1}{8}=-\frac{1}{9}. \] Setting \( -\frac{1}{c^{2}}=-\frac{1}{9} \) gives \( c^{2}=9 \), so \( c=3 \) (the value lying in \( [1,9] \)). Option (3).

An absolute value is never negative, so \( |3-x|\ge 0 \), with its smallest value \( 0 \) attained at \( x=3 \). Adding the constant \( 9 \), the least value of \( |3-x|+9 \) is \( 0+9=9 \). Option (4).

The tangent's slope is \[ m(x)=\frac{dy}{dx}=e^{x}\sin x+e^{x}\cos x=e^{x}(\sin x+\cos x). \] To maximise the slope, differentiate it again: \[ m'(x)=e^{x}(\sin x+\cos x)+e^{x}(\cos x-\sin x)=2e^{x}\cos x. \] Setting \( m'(x)=0 \) gives \( \cos x=0 \), i.e. \( x=\frac{\pi}{2} \) or \( x=\frac{3\pi}{2} \) on \( [0,2\pi] \). Comparing the slope, \( m\!\left(\frac{\pi}{2}\right)=e^{\pi/2}>0 \) while \( m\!\left(\frac{3\pi}{2}\right)=-e^{3\pi/2}<0 \); the slope is greatest at \( x=\frac{\pi}{2} \). Option (2).

Differentiate \( f(x)=x^{2}e^{-2x} \) using the product rule: \[ f'(x)=2x\,e^{-2x}-2x^{2}e^{-2x}=2x(1-x)e^{-2x}. \] For \( x>0 \) this is zero only at \( x=1 \); \( f' \) is positive on \( (0,1) \) and negative for \( x>1 \), so \( x=1 \) gives a maximum. The maximum value is \( f(1)=1\cdot e^{-2}=\frac{1}{e^{2}} \). Option (3).

Minimise the squared distance \( D=(x-6)^{2}+y^{2} \). On the curve \( y^{2}=x^{2}-4 \), so \[ D=(x-6)^{2}+x^{2}-4=2x^{2}-12x+32. \] Then \( \frac{dD}{dx}=4x-12=0\Rightarrow x=3 \), and \( y^{2}=3^{2}-4=5\Rightarrow y=\pm\sqrt{5} \). The nearest points are \( (3,\pm\sqrt{5}) \); the one in the list is \( (3,\sqrt{5}) \). Option (3).

Let the numbers be \( x,y>0 \) with \( x^{2}+y^{2}=200 \); we maximise \( P=xy \). It is easier to maximise \( P^{2}=x^{2}y^{2}=x^{2}(200-x^{2}) \). Putting \( u=x^{2} \), \( P^{2}=200u-u^{2} \), whose derivative \( 200-2u=0 \) gives \( u=100 \). Hence \( x^{2}=100 \) and \( y^{2}=200-100=100 \), so \( x=y=10 \) and the maximum product is \( P=10\cdot 10=100 \). Option (1).

Concavity is governed by the second derivative: \( y'=4ax^{3}+2bx \) and \( y''=12ax^{2}+2b \). A point of inflection requires \( y''=0 \), i.e. \( x^{2}=-\frac{b}{6a} \). Since \( ab>0 \), the ratio \( \frac{b}{a}>0 \), so \( -\frac{b}{6a}<0 \) and there is no real solution — the curve has no points of inflection. (It does have a horizontal tangent at \( x=0 \), and whether it is concave up or down depends on the common sign of \( a \) and \( b \), so only option (4) is always true.) Option (4).

Differentiate twice: \( y'=3(x-1)^{2} \) and \( y''=6(x-1) \). Then \( y''=0 \) at \( x=1 \), and \( y'' \) changes sign there (negative for \( x<1 \), positive for \( x>1 \)), confirming a point of inflection. At \( x=1 \), \( y=(1-1)^{3}=0 \), so the point of inflection is \( (1,0) \). Option (3).