JEE Mathematics Maxima and Minima MCQs Set 09

Answer: (a) f(x) is bounded, takes both its bounds, and its range contains exactly one integral point

Explanation:
1. Put \( t=\tan x\ge 0 \). Dividing numerator and denominator by \( t^n \) gives \( f=\dfrac{1}{(t^n+t^{-n})+(t^{n-1}+t^{-(n-1)})+\dots+(t+t^{-1})+1} \).
2. Each bracket is at least 2 (AM-GM), so the denominator is at least \( 2n+1 \) and \( f\le\frac{1}{2n+1} \), with equality at t = 1.
3. Also f(0) = 0.
4. Range is \( [0,\frac{1}{2n+1}] \), which contains only the integer 0.

Teacher's Note:
1. Dividing by \( t^n \) turns the denominator into terms of the form \( t^k+t^{-k} \).
2. Both bounds are attained, at x = 0 and at \( x=\pi/4 \).

Answer: (a) F F T F

Explanation:
1. S-1 is false: without continuity at c, the value f(c) could be lower than the nearby values, so c need not be a local maximum.
2. S-2 is false: for example \( f(x)=-\sqrt{-x} \) (x < 0), \( f(x)=-x^2 \) (x >= 0) has a local maximum at 0 and the concavity changes there.
3. S-3 is true: \( f(x)=x^2|x|=\pm x^3 \) and \( f''(x)=6|x| \), which exists and equals 0 at x = 0.
4. S-4 is false: \( f(x)=x^3 \) is bijective and differentiable at 0, but \( f^{-1}(x)=x^{1/3} \) is not differentiable at 0.

Teacher's Note:
1. Test each statement with a simple counterexample where possible.
2. A vertical tangent breaks differentiability of the inverse.

Answer: (c) \( (-\infty,4) \)

Explanation:
1. \( f'(x)=\dfrac{(x-4)(3-2x)-(t+3x-x^2)}{(x-4)^2}=\dfrac{-x^2+8x-(12+t)}{(x-4)^2} \).
2. For a maximum and a minimum, \( x^2-8x+(12+t)=0 \) must have two distinct real roots: \( 64-4(12+t)>0 \).
3. \( 16-12-t>0 \Rightarrow t<4 \).

Teacher's Note:
1. Local extrema need f' to change sign at two distinct points.
2. That needs a positive discriminant for the numerator quadratic.

Answer: (a) Exactly one statement is correct

Explanation:
1. I is false: at a maximum \( f''(c)\le 0 \), for example \( f=-x^4 \) at 0 has \( f''(0)=0 \).
2. II is false: the antiderivative of \( f(x)=1 \) is \( x+C \), not periodic.
3. III is true: the integral of a periodic function over any full period is the same.
4. IV is false: if f' were increasing up to c with \( f'(c)=0 \), then f' < 0 on the left of c, which contradicts a maximum.
5. Only III is correct.

Teacher's Note:
1. Test each statement with a simple counterexample.
2. Statement III is a standard property of periodic functions.

Answer: (d) \( (-3\sqrt3,-2\sqrt3)\cup(2\sqrt3,3\sqrt3) \)

Explanation:
1. \( f'(x)=a^2-3x^2=0 \Rightarrow x=\pm\frac{|a|}{\sqrt3} \). \( f''(x)=-6x>0 \) only for the negative root, so the minimum is at \( x=-\dfrac{|a|}{\sqrt3} \).
2. Since \( x^2+x+2>0 \) for all x, the inequality means \( (x+2)(x+3)<0 \), i.e. \( -3<x<-2 \).
3. So \( 2<\dfrac{|a|}{\sqrt3}<3 \), i.e. \( 2\sqrt3<|a|<3\sqrt3 \).
4. Therefore \( a\in(-3\sqrt3,-2\sqrt3)\cup(2\sqrt3,3\sqrt3) \).

Teacher's Note:
1. The numerator is always positive, so only the denominator's sign matters.
2. Remember a can be negative, since only \( a^2 \) appears in f.

Answer: (a) \( \dfrac{6+\pi}{6} \)

Explanation:
1. Let the width be 2r and the height h. Then \( P=\pi r+2h+4r \), so \( h=\frac{P-(\pi+4)r}{2} \).
2. Light \( L=3(2rh)+\frac{\pi r^2}{2}=3Pr-\left(\frac{5\pi}{2}+12\right)r^2 \).
3. \( L'(r)=0 \Rightarrow r=\dfrac{3P}{5\pi+24} \), a maximum since \( L''<0 \).
4. Then \( P=\frac{(5\pi+24)r}{3} \) and \( h=\frac{(\pi+6)r}{3} \).
5. Height divided by width \( =\dfrac{(\pi+6)r/3}{2r}=\dfrac{\pi+6}{6} \).

Teacher's Note:
1. Write P and the light transmitted in terms of r alone.
2. A quadratic in r is maximised at its vertex.

Answer: (a) \( x=\frac14\left(b+\sqrt{b^2-1}\right) \)

Explanation:
1. \( f'(x)=\frac{1}{8x}-b+2x=0 \Rightarrow 16x^2-8bx+1=0 \), so \( x=\frac{b\pm\sqrt{b^2-1}}{4} \).
2. \( f'(x)=\frac{16x^2-8bx+1}{8x} \) is positive outside the roots and negative between them.
3. So f' changes from - to + at the larger root: local minimum there; the smaller root is a local maximum.

Teacher's Note:
1. Set f' = 0 to get a quadratic in x.
2. The sign pattern of f' decides which root is a minimum.

Answer: (a) \( \frac{3\sqrt3}{4}r^2 \)

Explanation:
1. Let the chord QR be at distance x from the centre on the side opposite to P. Its half-length is \( \sqrt{r^2-x^2} \), height from P is \( r+x \).
2. Area \( =(r+x)\sqrt{r^2-x^2} \).
3. Maximise: derivative gives \( 2x^2+rx-r^2=0 \), so \( x=\frac r2 \).
4. Area \( =\frac{3r}{2}\cdot\frac{\sqrt3 r}{2}=\frac{3\sqrt3}{4}r^2 \).

Teacher's Note:
1. Use the distance of the chord from the centre as the variable.
2. The maximum triangle is equilateral.

Answer: (a) \( \dfrac{4}{3\sqrt3} \) sq. units

Explanation:
1. Q = (1, 0), radius of the second circle = r, so \( S=(1-r,0) \). R lies on both circles: if \( R=(\cos\phi,\sin\phi) \), then \( r^2=2-2\cos\phi \).
2. Area \( =\frac12\cdot QS\cdot\sin\phi=\frac12 r\cdot r\sqrt{1-\frac{r^2}{4}} \), so \( A^2=\frac{r^4}{4}\left(1-\frac{r^2}{4}\right) \).
3. Let \( u=r^2 \): \( \frac{d}{du}\left(u^2-\frac{u^3}{4}\right)=2u-\frac{3u^2}{4}=0 \Rightarrow u=\frac83 \).
4. \( A^2=\frac14\cdot\frac{64}{9}\cdot\frac13=\frac{16}{27} \), so \( A=\frac{4}{3\sqrt3} \).

Teacher's Note:
1. Express the area in terms of r using the intersection point.
2. Maximising \( A^2 \) avoids the square root.

Answer: (a) \( \dfrac{ud}{\sqrt{v^2-u^2}} \)

Explanation:
1. Landing at distance x from A: \( T=\frac{\sqrt{d^2+x^2}}{u}+\frac{L-x}{v} \).
2. \( T'=\frac{x}{u\sqrt{d^2+x^2}}-\frac1v=0 \Rightarrow vx=u\sqrt{d^2+x^2} \).
3. Squaring: \( x^2(v^2-u^2)=u^2d^2 \), so \( x=\frac{ud}{\sqrt{v^2-u^2}} \).
4. T'' > 0, so this is a minimum.

Teacher's Note:
1. Total time is swimming time plus walking time.
2. Check that the landing point lies within the stretch from A to the house.

Answer: (a) \( \pm(\sqrt2,\sqrt2) \)

Explanation:
1. Use polar coordinates: \( r^2(5-4\sin2\theta)=4 \), so \( r^2=\frac{4}{5-4\sin2\theta} \).
2. r is maximum when \( \sin2\theta=1 \): \( r^2=4 \), \( r=2 \), at \( \theta=\frac\pi4 \) (and \( \frac{5\pi}{4} \)).
3. The points are \( (2\cos45^\circ,2\sin45^\circ)=(\sqrt2,\sqrt2) \) and its negative.
4. (The minimum distance \( \frac23 \) occurs at \( \pm(\frac{\sqrt2}{3},-\frac{\sqrt2}{3}) \).)

Teacher's Note:
1. In polar form the distance is r itself.
2. The maximum of r corresponds to the maximum of \( \sin 2\theta \).

Answer: (a) \( 3\sqrt[3]{4} \) cm

Explanation:
1. Let the base side be a and the height h: \( \frac{a^2h}{3}=36\sqrt2 \). The slant height is \( s=\sqrt{h^2+\frac{a^2}{4}} \) and the surface area \( S=a^2+2as \).
2. Writing \( h=\frac{108\sqrt2}{a^2} \) and setting \( dS/da=0 \) gives \( h=\sqrt2\,a \).
3. Then \( \frac{\sqrt2a^3}{3}=36\sqrt2 \Rightarrow a^3=108 \Rightarrow a=3\sqrt[3]{4} \) cm.

Teacher's Note:
1. Eliminate h using the volume and minimise the surface area.
2. The optimum shape has \( h=\sqrt2\,a \), independent of the volume.

Answer: (a) \( \frac{4}{\sqrt{21}} \) miles

Explanation:
1. Landing x miles from the nearest point: \( T=\frac{\sqrt{4+x^2}}{2}+\frac{6-x}{5} \).
2. \( T'=\frac{x}{2\sqrt{4+x^2}}-\frac15=0 \Rightarrow 5x=2\sqrt{4+x^2} \).
3. Squaring: \( 25x^2=16+4x^2 \Rightarrow x^2=\frac{16}{21} \), \( x=\frac{4}{\sqrt{21}} \).
4. \( T''>0 \) so this is a minimum.

Teacher's Note:
1. Time is rowing distance divided by 2 plus walking distance divided by 5.
2. Check the answer lies between 0 and 6.

Answer: (b) \( \frac{\pi}{6} \)

Explanation:
1. The tangent is \( \frac{x\cos\theta}{3\sqrt3}+y\sin\theta=1 \), with intercepts \( 3\sqrt3\sec\theta \) and \( \csc\theta \).
2. \( S=3\sqrt3\sec\theta+\csc\theta \), \( S'=3\sqrt3\sec\theta\tan\theta-\csc\theta\cot\theta=0 \Rightarrow \tan^3\theta=\frac{1}{3\sqrt3} \).
3. \( \tan\theta=\frac{1}{\sqrt3} \), so \( \theta=\frac{\pi}{6} \) (S' changes from - to +).

Teacher's Note:
1. Write the tangent in intercept form to read off the intercepts.
2. Set the derivative of their sum to zero.

Answer: (d) \( \frac52 \)

Explanation:
1. \( f'(x)=12x^2-2x-2=2(3x+1)(2x-1) \). On [0, 1], f decreases on \( [0,\frac12] \) and increases afterwards; \( f(0)=1,\ f(\frac12)=\frac14,\ f(1)=2 \).
2. For \( x\le\frac12 \), the minimum over [0, x] is f(x), so \( g(\frac14)=f(\frac14)=\frac1{16}-\frac1{16}-\frac12+1=\frac12 \).
3. For \( \frac12\le x\le1 \), the minimum stays \( f(\frac12)=\frac14 \), so \( g(\frac34)=\frac14 \).
4. \( g(\frac54)=3-\frac54=\frac74 \).
5. Sum \( =\frac12+\frac14+\frac74=\frac52 \).

Teacher's Note:
1. A running minimum follows the function while it falls and then stays at its lowest value.
2. Find the turning point of f first.

Answer: (b) 36

Explanation:
1. Put \( x-2=\cos\phi \). Then \( \sqrt{-3+4x-x^2}=\sqrt{1-(x-2)^2}=\sin\phi \) with \( \phi\in[0,\pi] \).
2. The expression becomes \( (\sin\phi+4)^2+(\cos\phi-3)^2=26+8\sin\phi-6\cos\phi \).
3. Its maximum is \( 26+\sqrt{8^2+6^2}=36 \), attained when \( \sin\phi=\frac45 \), \( \cos\phi=-\frac35 \), which lies in the allowed range.

Teacher's Note:
1. The substitution \( x-2=\cos\phi \) removes the square root.
2. The maximum of \( a\sin\phi+b\cos\phi \) is \( \sqrt{a^2+b^2} \).

Answer: (a) \( \frac32+\frac{1}{\sqrt2}-2\sqrt3 \)

Explanation:
1. \( f'(x)=3\sin x+4\cos x \). At \( \frac{5\pi}{4} \) and \( \frac{4\pi}{3} \) both sine and cosine are negative, so \( f'<0 \) on the whole interval: f is decreasing, and its least value is at \( x=\frac{4\pi}{3} \).
2. \( f(x)=\left[-3\cos t+4\sin t\right]_{5\pi/4}^{x}=-3\cos x+4\sin x+\frac{1}{\sqrt2} \).
3. At \( x=\frac{4\pi}{3} \): \( -3(-\frac12)+4(-\frac{\sqrt3}{2})+\frac{1}{\sqrt2}=\frac32-2\sqrt3+\frac{1}{\sqrt2} \).

Teacher's Note:
1. Check the sign of the integrand on the interval to see where the minimum is.
2. Evaluate the antiderivative at both limits.

Answer: (b) \( [2,4] \)

Explanation:
1. Let the vertices divide the sides of the unit square into parts \( x_i \) and \( 1-x_i \) (i = 1..4). Each side of the quadrilateral satisfies \( \text{side}^2=x_i^2+(1-x_{i+1})^2 \), so the sum is \( \sum\left[x_i^2+(1-x_i)^2\right] \).
2. For each i, \( x^2+(1-x)^2 \) has minimum \( \frac12 \) (at \( x=\frac12 \)) and maximum 1 on [0, 1].
3. So the sum lies between \( 4\cdot\frac12=2 \) and \( 4\cdot1=4 \): the interval [2, 4].

Teacher's Note:
1. Express every side by Pythagoras using the segments on the square's sides.
2. The sum splits into four terms of the form \( x^2+(1-x)^2 \).

Answer: (c) 2

Explanation:
1. On \( (-3,-1] \), \( (2+x)^3 \) is increasing, with \( f(-1)=1 \).
2. On \( (-1,0) \), \( x^{2/3} \) is decreasing from 1 to 0; on \( (0,2) \), it is increasing.
3. So f has a local maximum at x = -1 and a local minimum at x = 0 (a cusp).
4. The total number is 2.

Teacher's Note:
1. Check the join at x = -1 and the point x = 0 where \( x^{2/3} \) has a cusp.
2. Look at the increasing/decreasing pattern.

Answer: (b) x = 2 and x = 4

Explanation:
1. For \( x<2 \): \( f=\frac{2-x}{x^2} \), \( f'=\frac{x-4}{x^3} \), which has no zero in \( x<2,\ x\ne0 \).
2. For \( x>2 \): \( f=\frac{x-2}{x^2} \), \( f'=\frac{4-x}{x^3}=0 \) at \( x=4 \).
3. At x = 2 the function has a corner (\( f'(2^-)=-\frac14 \), \( f'(2^+)=\frac14 \)), so f' does not exist there, making it a critical point. x = 0 is not in the domain.
4. Critical points: x = 2 and x = 4.

Teacher's Note:
1. Critical points include places where the derivative does not exist but the function is defined.
2. x = 0 is outside the domain.

Answer: (b) (1), (2) and (4) only

Explanation:
1. For \( x\le0 \), \( f=\frac{1}{1-x}+\frac{1}{2-x} \) is increasing; for \( 0\le x\le1 \), \( f=\frac{1}{1+x}+\frac{1}{2-x} \), with \( f(0)=f(1)=\frac32 \) and \( f(\frac12)=\frac43 \); for \( x\ge1 \), f is decreasing.
2. So f has local (and global) maxima \( \frac32 \) at x = 0 and x = 1, and a local minimum \( \frac43 \) at \( x=\frac12 \).
3. As \( |x|\to\infty \), \( f\to0 \) but never equals 0, so there is no absolute minimum.
4. Correct: (1), (2) and (4).

Teacher's Note:
1. Treat the three regions \( x\le0 \), \( 0\le x\le1 \), \( x\ge1 \) separately.
2. Think about the behaviour as \( x\to\pm\infty \).