JEE Mathematics Maxima and Minima MCQs Set 14

Answer: (d) \(n\pi,\ n=1,2,\dots\)

Explanation:
1. \(f'(x)=\frac{\sin x}{x}\).
2. \(f'(x)=0\) gives \(\sin x=0\), so \(x=n\pi\).
3. The sign of \(f'\) changes at each \(n\pi\), so these are the points of extrema.

Teacher's Note:
1. A derivative that is a ratio vanishes where its numerator vanishes. Check that the sign really changes before calling it an extremum.

Answer: (a) \(x=\frac{\pi}{4},\ \frac{5\pi}{4}\)

Explanation:
1. The maximum value of \(\sin 2x\) is 1, which occurs when \(2x=\frac{\pi}{2}+2k\pi\).
2. So \(x=\frac{\pi}{4}+k\pi\); in \([0,2\pi]\) this gives \(x=\frac{\pi}{4}\) and \(\frac{5\pi}{4}\).

Teacher's Note:
1. The period of \(\sin 2x\) is \(\pi\), so the maximum repeats every \(\pi\). List all values inside the given interval.

Answer: (d) 120

Explanation:
1. Let \(f(x)=x^4-62x^2+ax+9\). Then \(f'(x)=4x^3-124x+a\).
2. The maximum is at the interior point \(x=1\), so \(f'(1)=0\): \(4-124+a=0\).
3. Hence \(a=120\).

Teacher's Note:
1. An interior maximum or minimum on a closed interval must be a critical point. Put the given point into \(f'(x)=0\).

Answer: (a) \(2\pi\text{ and }0\)

Explanation:
1. \(f'(x)=1+2\cos 2x=0\) gives \(\cos 2x=-\frac12\), so \(x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\).
2. Compare the values at the end points and critical points: \(f(0)=0\), \(f(2\pi)=2\pi\), and the interior values lie between 0 and \(2\pi\) (for example \(f(\frac{\pi}{3})\approx1.91\), \(f(\frac{5\pi}{3})\approx4.37\)).
3. So the maximum is \(2\pi\) and the minimum is 0.

Teacher's Note:
1. On a closed interval always compare the end-point values with the values at critical points. Here the extreme values occur at the end points.

Answer: (c) \(1\)

Explanation:
1. Let \(g(x)=x^2-x+1\). Then \(g'(x)=2x-1=0\) at \(x=\frac12\), where \(g=\frac34\) is the minimum.
2. At the end points \(g(0)=g(1)=1\), so the maximum of \(g\) is 1.
3. The cube root is increasing, so the maximum of the given function is \(1^{1/3}=1\).

Teacher's Note:
1. A monotonic increasing function of g has its maximum where g has its maximum. Check end points of the interval.

Answer: (d) neither a local maximum nor a local minimum

Explanation:
1. \(f'(x)=12x^2-36x+27=3(2x-3)^2\).
2. \(f'(x)\ge0\) for all \(x\) and does not change sign at \(x=\frac32\).
3. So the function is increasing everywhere and has no local extremum.

Teacher's Note:
1. If \(f'\) has a repeated root, the sign of \(f'\) does not change there. Such a point is a point of inflexion, not an extremum.

Answer: (a) maxima at \(x=0,-5\) and minima at \(x=-3\)

Explanation:
1. \(f'(x)=-3x^3-24x^2-45x=-3x(x+3)(x+5)\), so the critical points are \(0,-3,-5\).
2. The sign of \(f'\): positive for \(x<-5\), negative on \((-5,-3)\), positive on \((-3,0)\), negative for \(x>0\).
3. So \(f'\) changes from + to - at \(-5\) and \(0\) (maxima) and from - to + at \(-3\) (minimum).

Teacher's Note:
1. Factorise \(f'\) fully and test the sign in each interval. A + to - change gives a maximum and a - to + change gives a minimum.

Answer: (c) 2

Explanation:
1. \(f'(x)=a\cos x+\cos 3x\).
2. At \(x=\frac{\pi}{3}\): \(\frac a2+\cos\pi=0\), so \(\frac a2=1\).
3. Hence \(a=2\).

Teacher's Note:
1. An extremum at a differentiable point needs \(f'=0\) there. Substitute the point and solve for the parameter.

Answer: (d) \(3-\frac{1}{e^2}\)

Explanation:
1. On \([-2,0]\), \(|x|=-x\), so \(h(x)=f(|x|)-g(x)=1-x-e^x\).
2. \(h'(x)=-1-e^x<0\), so \(h\) is decreasing and its maximum is at \(x=-2\).
3. \(h(-2)=3-e^{-2}=3-\frac{1}{e^2}\).

Teacher's Note:
1. Replace
2. x
3. by the correct sign on the given interval first. Then use the sign of the derivative to locate the end-point maximum.

Answer: (b) 1/4

Explanation:
1. \(f'(x)=x^{24}(1-x)^{74}[25(1-x)-75x]=x^{24}(1-x)^{74}(25-100x)\).
2. \(f'=0\) in \((0,1)\) at \(x=\frac14\), with \(f'\) changing from + to -.
3. So the maximum is at \(x=\frac14\).

Teacher's Note:
1. For \(x^p(1-x)^q\) the maximum is at \(x=\frac{p}{p+q}\). Here \(\frac{25}{100}=\frac14\).

Answer: (c) \(\frac1e\)

Explanation:
1. \(\ln f=-x\ln x\), so \(\frac{f'}{f}=-(\ln x+1)\).
2. \(f'=0\) gives \(x=\frac1e\); \(f'>0\) before it and \(f'<0\) after it.
3. So the maximum is at \(x=\frac1e\).

Teacher's Note:
1. Use logarithmic differentiation for variable powers. The sign change of \(f'\) confirms a maximum.

Answer: (d) \(\sqrt{\frac ba}\)

Explanation:
1. \(f'(x)=a-\frac{b}{x^2}=0\) gives \(x=\sqrt{\frac ba}\).
2. \(f''(x)=\frac{2b}{x^3}>0\), so this is a minimum.
3. The least value is \(2\sqrt{ab}\).

Teacher's Note:
1. AM-GM gives the same result quickly: \(ax+\frac bx\ge2\sqrt{ab}\) with equality when \(ax=\frac bx\).

Answer: (c) \(\frac16\)

Explanation:
1. \(f'(x)=\frac{(4+x+x^2)-x(1+2x)}{(4+x+x^2)^2}=\frac{4-x^2}{(4+x+x^2)^2}\).
2. On \([-1,1]\), \(4-x^2>0\), so \(f\) is increasing.
3. The maximum is at \(x=1\): \(f(1)=\frac16\).

Teacher's Note:
1. If \(f'\) keeps one sign on the interval, the extremes occur at the end points.

Answer: (a) \(e^2\)

Explanation:
1. \(f'(x)=2x\ln x+x=x(2\ln x+1)>0\) on \([1,e]\), so \(f\) is increasing.
2. The greatest value is \(f(e)=e^2\cdot1=e^2\).

Teacher's Note:
1. Check the sign of \(f'\) on the whole interval before looking for critical points.

Answer: (b) 10 and 25

Explanation:
1. Let \(y=35-x\) and \(f=x^2(35-x)^5\).
2. \(f'=x(35-x)^4[2(35-x)-5x]=x(35-x)^4(70-7x)\), which is zero at \(x=10\).
3. Then \(y=25\), and \(f'\) changes from + to -, so it is a maximum.

Teacher's Note:
1. For \(x^py^q\) with fixed sum \(S\), the maximum is at \(x=\frac{pS}{p+q}\). Here \(x=\frac{2\cdot35}{7}=10\).

Answer: (c) 8 and 8

Explanation:
1. Let the numbers be \(x\) and \(16-x\), and \(S=x^3+(16-x)^3\).
2. \(S'=3x^2-3(16-x)^2=0\) gives \(x=16-x\), so \(x=8\).
3. \(S''=6x+6(16-x)=96>0\), so the sum of cubes is minimum at 8 and 8.

Teacher's Note:
1. For a fixed sum, symmetric expressions like the sum of cubes are minimum when the numbers are equal.

Answer: (a) 3 cm

Explanation:
1. \(V=x(18-2x)^2\).
2. \(V'=(18-2x)^2-4x(18-2x)=(18-2x)(18-6x)=0\) gives \(x=3\) (\(x=9\) gives zero volume).
3. \(V''<0\) at \(x=3\), so the volume is maximum.

Teacher's Note:
1. Reject the root that makes the box vanish. Confirm with the second derivative.

Answer: (b) the diameter of the base

Explanation:
1. Surface area \(S=2\pi r^2+2\pi rh\), so \(h=\frac{S-2\pi r^2}{2\pi r}\).
2. \(V=\pi r^2h=\frac{r(S-2\pi r^2)}{2}\), and \(V'=0\) gives \(S=6\pi r^2\).
3. Then \(2\pi rh=4\pi r^2\), so \(h=2r\), the diameter.

Teacher's Note:
1. Eliminate h using the fixed surface area, then differentiate with respect to r.

Answer: (c) \(\sqrt2\) times the radius of the base

Explanation:
1. Curved surface \(S=\pi rl\) with \(V=\frac13\pi r^2h\) fixed. Minimise \(S^2=\pi^2r^2(r^2+h^2)\).
2. Put \(h=\frac{3V}{\pi r^2}\): \(S^2=\pi^2r^4+\frac{9V^2}{r^2}\). Then \(\frac{d(S^2)}{dr}=4\pi^2r^3-\frac{18V^2}{r^3}=0\), so \(2\pi^2r^6=9V^2\).
3. Since \(9V^2=\pi^2r^4h^2\), this gives \(2r^2=h^2\), so \(h=\sqrt2\,r\).

Teacher's Note:
1. Minimising \(S^2\) avoids the square root. The answer \(h=\sqrt2\,r\) is worth remembering.

Answer: (a) \((2\sqrt2,4)\)

Explanation:
1. For a point \((x,y)\) with \(x^2=2y\), \(d^2=x^2+(y-5)^2=2y+(y-5)^2\).
2. \(\frac{d(d^2)}{dy}=2+2(y-5)=0\) gives \(y=4\), so \(x^2=8\), \(x=\pm2\sqrt2\).
3. The point \((2\sqrt2,4)\) is among the options, and the second derivative is positive, so it is the minimum.

Teacher's Note:
1. Minimise the square of the distance and write it in one variable. Both \(x=\pm2\sqrt2\) give the same distance.

Answer: (b) double the radius of the circle

Explanation:
1. Let the radius be \(r\) and the side \(a\): \(2\pi r+4a=k\), \(A=\pi r^2+a^2\).
2. Differentiating the constraint: \(2\pi\,dr+4\,da=0\), so \(\frac{da}{dr}=-\frac\pi2\).
3. \(\frac{dA}{dr}=2\pi r+2a\frac{da}{dr}=2\pi r-\pi a=0\), so \(a=2r\).

Teacher's Note:
1. Use the constraint relation between the small changes of the two variables. The side of the square comes out as the diameter of the circle.

Answer: (b) length \(=\frac{20}{\pi+4}\), breadth \(=\frac{10}{\pi+4}\)

Explanation:
1. Let the semicircle have radius \(r\), so the width is \(2r\), and let the height of the rectangle be \(h\). The perimeter gives \(2r+2h+\pi r=10\).
2. Area \(A=2rh+\frac{\pi r^2}{2}=r(10-2r-\pi r)+\frac{\pi r^2}{2}\), and \(A'=0\) gives \(10=(4+\pi)r\), so \(r=\frac{10}{\pi+4}\).
3. Then \(h=\frac{10}{\pi+4}\), and the width \(2r=\frac{20}{\pi+4}\).
4. So the length (width of the window) is \(\frac{20}{\pi+4}\) and the breadth (height) is \(\frac{10}{\pi+4}\).

Teacher's Note:
1. Maximum light means maximum area. Write the area in terms of r using the perimeter condition.

Answer: (d) \(\frac{4r}{3}\)

Explanation:
1. Let the height be \(h\) and base radius \(x\). From the sphere, \(x^2=h(2r-h)\).
2. \(V=\frac13\pi x^2h=\frac\pi3(2rh^2-h^3)\), so \(V'=0\) gives \(4rh=3h^2\).
3. Hence \(h=\frac{4r}{3}\).

Teacher's Note:
1. Use the relation \(x^2=h(2r-h)\) from the chord property of a circle. The maximum volume is \(\frac{32}{81}\) of the sphere's volume.

Answer: (a) \(\frac{2R}{\sqrt3}\text{ and }\frac{4\pi R^3}{3\sqrt3}\)

Explanation:
1. Let the height be \(h\) and base radius \(r\): \(r^2=R^2-\frac{h^2}{4}\).
2. \(V=\pi r^2h=\pi\left(R^2h-\frac{h^3}{4}\right)\), and \(V'=0\) gives \(R^2=\frac{3h^2}{4}\), so \(h=\frac{2R}{\sqrt3}\).
3. Then \(V=\pi\cdot\frac{2R^2}{3}\cdot\frac{2R}{\sqrt3}=\frac{4\pi R^3}{3\sqrt3}\).

Teacher's Note:
1. Relate r and h through the right triangle formed with the sphere's radius. Then differentiate the volume.

Answer: (b) 12

Explanation:
1. Slope \(=y'=-3x^2+6x+9\).
2. Let \(s=y'\). Then \(s'=-6x+6=0\) at \(x=1\), and \(s''=-6<0\).
3. The maximum slope is \(s(1)=-3+6+9=12\).

Teacher's Note:
1. To maximise the slope, differentiate the slope again. This is a maximisation of the first derivative.

Answer: (d) \(\frac\pi2\)

Explanation:
1. Area \(=\frac12\,PQ\cdot PR\sin\theta\) with fixed sides.
2. \(\sin\theta\) is maximum (equal to 1) when \(\theta=\frac\pi2\).

Teacher's Note:
1. With two sides fixed the area depends only on \(\sin\theta\). No calculus is needed.

Answer: (b) 24

Explanation:
1. By AM-GM, \(2a+3b\ge2\sqrt{6ab}\), so \(ab\ge2\sqrt{6}\sqrt{ab}\).
2. Hence \(\sqrt{ab}\ge2\sqrt6\) and \(ab\ge24\).
3. Equality holds at \(2a=3b\); with \(ab=2a+3b\) this gives \(a=6,b=4\).

Teacher's Note:
1. Replace the right side by its AM-GM bound and solve the resulting inequality in \(\sqrt{ab}\). Verify equality is attainable.

Answer: (b) \(a+b\)

Explanation:
1. Put \(x=a\sec\theta\), \(y=b\csc\theta\). Then \(r^2=a^2\sec^2\theta+b^2\csc^2\theta=a^2+b^2+a^2\tan^2\theta+b^2\cot^2\theta\).
2. By AM-GM \(a^2\tan^2\theta+b^2\cot^2\theta\ge2ab\), so \(r^2\ge a^2+b^2+2ab=(a+b)^2\).
3. The minimum length is \(a+b\).

Teacher's Note:
1. A parametric form turns the problem into an AM-GM minimisation. Equality needs \(\tan^2\theta=\frac ba\).

Answer: (d) \(\frac p4\)

Explanation:
1. Perimeter \(p=2r+r\theta\), so \(r\theta=p-2r\).
2. Area \(A=\frac12r^2\theta=\frac12r(p-2r)\).
3. \(A'=\frac12(p-4r)=0\) gives \(r=\frac p4\), and \(A''<0\).

Teacher's Note:
1. Write the arc length in terms of r using the perimeter. The maximum area is \(\frac{p^2}{16}\).

Answer: (b) \(2c\sqrt{ab}\)

Explanation:
1. By AM-GM, \(ax+by\ge2\sqrt{abxy}=2\sqrt{abc^2}\).
2. So the minimum is \(2c\sqrt{ab}\), attained when \(ax=by\).

Teacher's Note:
1. The product xy is fixed, so AM-GM gives the minimum of the sum directly.

Answer: (c) \(\frac{c^3}{\sqrt{2ab}}\)

Explanation:
1. By AM-GM, \(c^6=a^2x^4+b^2y^4\ge2abx^2y^2\).
2. So \((xy)^2\le\frac{c^6}{2ab}\), giving \(xy\le\frac{c^3}{\sqrt{2ab}}\).
3. Equality holds when \(ax^2=by^2\).

Teacher's Note:
1. Bound the sum of two terms by twice their geometric mean to link it with the product xy.