Rate Measure, Increasing and Decreasing Functions — JEE Main Maths PYQ MCQs with Solutions

Solution: Volume of ice $V=\dfrac{4}{3}\pi[(r+x)^3-r^3]$ with $r=10$. So $\dfrac{dV}{dt}=4\pi(r+x)^2\dfrac{dx}{dt}=-50$. At $x=5$, $r+x=15$, giving $-50=4\pi(225)\dfrac{dx}{dt}\Rightarrow \dfrac{dx}{dt}=-\dfrac{1}{18\pi}$. So the thickness decreases at $\dfrac{1}{18\pi}$ cm/min.

Solution: $h(x)=e^{x^2-x}-(x^2-x)$, so $h'(x)=(2x-1)\left(e^{x^2-x}-1\right)$, which has the same sign as $(2x-1)(x^2-x)$ since $e^t-1$ has the sign of $t$. Solving $(2x-1)(x^2-x)\ge0$ gives $x\in\left[0,\tfrac12\right]\cup[1,\infty)$.

Solution: With semi-vertical angle $\theta$, $\tan\theta=\dfrac12$, so the radius $r=h\tan\theta=\dfrac{h}{2}$. Volume $V=\dfrac13\pi r^2h=\dfrac{\pi h^3}{12}$, so $\dfrac{dV}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}$. At $h=10$: $5=\dfrac{100\pi}{4}\dfrac{dh}{dt}\Rightarrow \dfrac{dh}{dt}=\dfrac{1}{5\pi}$.

Solution: $\varphi'(x)=f'(x)-f'(2-x)$. Since $f''>0$, $f'$ is strictly increasing. For $x<1$, $x<2-x$ so $f'(x)<f'(2-x)$, giving $\varphi'(x)<0$; for $x>1$, $x>2-x$ so $\varphi'(x)>0$. Hence $\varphi$ decreases on $(0,1)$ and increases on $(1,2)$.

Solution: Since $f$ changes from increasing to decreasing at $x=1$, $f'(1)=0$: $3-6(a-2)+3a=0\Rightarrow a=5$. Then $f(x)=x^3-9x^2+15x+7$, so $f(x)-14=x^3-9x^2+15x-7=(x-1)^2(x-7)$. Thus the required root is $x=7$.

Solution: For any nonzero constant $c$, the function $t\mapsto \dfrac{t}{\sqrt{c^2+t^2}}$ has derivative $\dfrac{c^2}{(c^2+t^2)^{3/2}}>0$, so it is strictly increasing in $t$. Each term of $f$ is such a function of a linear (increasing) expression in $x$, so $f'(x)=\dfrac{a^2}{(a^2+x^2)^{3/2}}+\dfrac{b^2}{(b^2+(x-d)^2)^{3/2}}>0$ for all $x$; hence $f$ is increasing.

Solution: $g(-u)=2\tan^{-1}(e^{-u})-\dfrac{\pi}{2}=2\left(\dfrac{\pi}{2}-\tan^{-1}(e^u)\right)-\dfrac{\pi}{2}=-\left(2\tan^{-1}(e^u)-\dfrac{\pi}{2}\right)=-g(u)$, so $g$ is odd. Also $g'(u)=\dfrac{2e^u}{1+e^{2u}}>0$ for all $u$, so $g$ is strictly increasing on $(-\infty,\infty)$.

Solution: $f'(x)=3x^2+2bx+c$ has discriminant $4b^2-12c$. Since $0<b^2<c$, we get $4b^2-12c<4c-12c=-8c<0$, so $f'(x)>0$ for all $x$ (as the leading coefficient is positive and $f'$ has no real roots). Hence $f$ is strictly increasing everywhere.

Solution: $3\sin x-4\sin^3x=\sin3x$, which increases when $3x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, i.e. $x\in\left(-\dfrac{\pi}{6},\dfrac{\pi}{6}\right)$, an interval of length $\dfrac{\pi}{3}$.

Solution: $f'(x)=e^{x-x^2}\left[1+x(1-2x)\right]=e^{x-x^2}(1-x)(1+2x)$. This is $\ge0$ exactly when $-\tfrac12\le x\le1$, so $f$ increases on $\left[-\tfrac12,1\right]$.

Solution: Let $h(x)=x-\log_e(1+x)$. Then $h(0)=0$ and $h'(x)=1-\dfrac{1}{1+x}=\dfrac{x}{1+x}>0$ for $x\in(0,1)$, so $h$ is increasing and $h(x)>0$, giving $\log_e(1+x)<x$.

Solution: By the Fundamental Theorem of Calculus, $f'(x)=e^x(x-1)(x-2)$. Since $e^x>0$, $f'(x)<0$ exactly when $(x-1)(x-2)<0$, i.e. $1<x<2$; so $f$ decreases on $(1,2)$.

Solution: $f(x)=1-\tfrac12\sin^22x=\tfrac34+\tfrac14\cos4x$, so $f'(x)=-\sin4x$, which is positive when $\sin4x<0$. For $x\in\left(\tfrac{\pi}{4},\tfrac{3\pi}{8}\right)$, $4x\in(\pi,\tfrac{3\pi}{2})$, where $\sin4x<0$ throughout, so $f$ increases on this whole interval (unlike the other listed intervals, where $\sin4x$ changes sign).

Solution: $f'(x)=\dfrac{\sin x-x\cos x}{\sin^2x}$; with $h(x)=\sin x-x\cos x$, $h(0)=0$ and $h'(x)=x\sin x>0$, so $h(x)>0$ and $f$ is increasing. For $g(x)=x\cot x$, $g'(x)=\dfrac{\tfrac12\sin2x-x}{\sin^2x}$; with $k(x)=\tfrac12\sin2x-x$, $k(0)=0$ and $k'(x)=\cos2x-1\le0$, so $k(x)<0$ for $x>0$, making $g'(x)<0$, i.e. $g$ is decreasing. So only $f$ is increasing.

Solution: With $u=e+x>e$ and $k=\pi-e>0$, $f=\dfrac{\ln(u+k)}{\ln u}$. Its derivative has the sign of $\psi(u)-\psi(u+k)$, where $\psi(u)=u\ln u$. Since $\psi'(u)=\ln u+1>0$ for $u>e$, $\psi$ is increasing there, so $\psi(u)<\psi(u+k)$, making the derivative negative. Hence $f$ is decreasing on $(0,\infty)$.

Solution: Since $f$ is increasing and $g$ is decreasing, $h=f\circ g$ is decreasing on $[0,\infty)$, so $h(0)$ is its maximum value, i.e. $h(x)\le h(0)=0$ for all $x\ge0$. But $h(x)=f(g(x))\ge0$ since $f$'s range is $[0,\infty)$. So $h(x)=0$ for every $x\ge0$, meaning $h(x)-h(1)=0$ always -- neither always negative, always positive, nor strictly increasing. The answer is 'None of these'.

Solution: Let $g(x)=f(x)e^{-2x}$. Then $g'(x)=e^{-2x}(f'(x)-2f(x))>0$, so $g$ is increasing; with $g(0)=1$, $g(x)>1$ for $x>0$, giving $f(x)>e^{2x}$. Then $f'(x)>2f(x)>2e^{2x}>e^{2x}$. Also $f'(x)>2f(x)>0$ (since $f(x)>e^{2x}>0$), so $f$ is increasing on $(0,\infty)$.

Solution: Swapping limits directly gives $f(1/x)=-f(x)$, so (c) holds. By Leibniz's rule, $f'(x)=\dfrac{2}{x}\cosh\!\left(x-\dfrac1x\right)>0$ for all $x>0$, so $f$ is strictly increasing on all of $(0,\infty)$ -- so (a) holds but (b) fails. For $\phi(x)=f(2^x)$, $\phi(-x)=f(2^{-x})=f(1/2^x)=-f(2^x)=-\phi(x)$, so $\phi$ is odd, confirming (d).

Solution: $h'(x)=f'(x)\left[1-2f(x)+3f(x)^2\right]$. The quadratic $3t^2-2t+1$ has discriminant $4-12=-8<0$ with positive leading coefficient, so it is always positive. Hence $h'(x)$ has the same sign as $f'(x)$: $h$ increases exactly when $f$ increases, and decreases exactly when $f$ decreases.