JEE Mathematics Maxima and Minima MCQs Set 10

Answer: (a) (1), (2) and (3) only

Explanation:
1. The domain is \( 0<x<2 \). \( f(2-x)=\log(2x-x^2)+\sin\left(\pi-\frac{\pi x}{2}\right)=f(x) \): the graph is symmetric about x = 1.
2. \( \log(2x-x^2)\le\log1=0 \) and \( \sin\frac{\pi x}{2}\le1 \), both with equality at x = 1, so the maximum is 1.
3. As \( x\to0^+ \) or \( 2^- \), \( f\to-\infty \): no absolute minimum.
4. f is defined only on (0, 2), so it is not periodic.

Teacher's Note:
1. Check the symmetry by replacing x with 2 - x.
2. The logarithm term is unbounded below near the end points.

Answer: (d) All four

Explanation:
1. \( f=x^{4/3}-x^{1/3} \), \( f'=\frac13x^{-2/3}(4x-1) \), \( f''=\frac29x^{-5/3}(2x+1) \).
2. f' fails to exist at 0 (vertical tangent), so (3) holds.
3. f' changes sign only at \( x=\frac14 \) (at 0, \( f'\to-\infty \) on both sides): one extremum, a minimum, so (2) holds.
4. f'' changes sign at \( x=-\frac12 \) and at x = 0: two points of inflection, so (1) holds.
5. The minimum value is \( f(\frac14)=-\frac34\cdot4^{-1/3}=-3\times2^{-8/3} \) and f is unbounded above, so (4) holds. All four are true.

Teacher's Note:
1. Compute f' and f'' in the form \( x^{-2/3}(\dots) \) and \( x^{-5/3}(\dots) \).
2. The cube root causes a vertical tangent at 0.

Answer: (b) (2), (3) and (4) only

Explanation:
1. Near x = 0 the values are close to 1, while \( f(0)=2 \), so x = 0 is a local maximum: (2) is true and (1) is false.
2. At x = 1, the left values tend to 0 and the right values are at least \( \sqrt2 \), so f(1) = \( \sqrt2 \) is neither a local maximum nor a minimum: (3) is true.
3. \( \sqrt{x+1}\to\infty \), so there is no global maximum: (4) is true.

Teacher's Note:
1. Compare f(0) with the values on both sides.
2. A function unbounded above has no global maximum.

Answer: (d) is equal to -1

Explanation:
1. \( f(x)=1-\dfrac{2}{x^2+1} \).
2. \( x^2+1\ge1 \), so \( \dfrac{2}{x^2+1}\le2 \) and \( f(x)\ge-1 \).
3. Equality holds at x = 0, so the minimum value -1 is attained.

Teacher's Note:
1. Rewrite the fraction as 1 minus a positive term.
2. The extreme value occurs at x = 0.

Answer: (a) a local maximum at \( x=1+\ln2 \) and a local minimum at \( x=e \)

Explanation:
1. \( g'(x)=f(x) \). On \( (1,2] \), \( f=2-e^{x-1} \) is positive for \( x<1+\ln2 \) and negative afterwards: g' changes from + to - at \( 1+\ln2 \), a local maximum.
2. On \( (2,3] \), \( f=x-e \) is negative for \( x<e \) and positive for \( x>e \): g' changes from - to + at e, a local minimum.
3. At x = 1 f stays positive (it jumps from e to 1) and at x = 2, f is continuous and negative, so there is no extremum there.

Teacher's Note:
1. The extrema of g are where its derivative f changes sign.
2. A jump in f does not produce an extremum unless the sign changes.

Answer: (b) 9

Explanation:
1. Let the vertex on the arc be \( (R\cos\phi,R\sin\phi) \) with \( R=\frac12 \); the leg along the diameter runs from the foot of the perpendicular to the end of the diameter, of length \( R(1+\cos\phi) \), and the other leg is \( R\sin\phi \).
2. Area \( =\frac12R^2(1+\cos\phi)\sin\phi \). Its derivative gives \( \cos\phi+\cos^2\phi-\sin^2\phi=0 \), so \( \cos\phi=\frac12 \), \( \phi=\frac\pi3 \).
3. \( M=\frac12\cdot\frac14\cdot\frac32\cdot\frac{\sqrt3}{2}=\frac{3\sqrt3}{32} \).
4. \( 32\sqrt3M=32\sqrt3\cdot\frac{3\sqrt3}{32}=9 \).

Teacher's Note:
1. Use the angle at the centre as the variable.
2. Solve \( 2\cos^2\phi+\cos\phi-1=0 \) for the maximising angle.

Answer: (b) 1

Explanation:
1. \( g=e^f \) is increasing in f, so g has a local maximum exactly where f does.
2. The sign of f' changes only at the roots of odd multiplicity, x = 2009 and x = 2011.
3. Reading from the right: f' > 0 for \( x>2011 \) (crossing 2012 changes nothing), f' < 0 on (2009, 2011) (crossing 2011 flips the sign; crossing 2010 does not), and f' > 0 for \( x<2009 \).
4. So x = 2009 is a local maximum (+ to -) and x = 2011 is a local minimum. The number of local maxima is 1.

Teacher's Note:
1. A monotonic increasing transformation preserves local extrema.
2. Even powers do not change the sign of f'.

Answer: (b) 2

Explanation:
1. Let \( g(x)=x^4-4x^3+12x^2+x-1 \). \( g''(x)=12x^2-24x+24 \) has discriminant \( 576-1152<0 \), so \( g''>0 \) and g is convex: it can have at most 2 real roots.
2. \( g(0)=-1<0 \) while \( g(x)\to\infty \) as \( x\to\pm\infty \), so there is a root on each side of 0.
3. Exactly 2 distinct real roots.

Teacher's Note:
1. A strictly convex function crosses the x-axis at most twice.
2. A negative value at 0 guarantees two crossings.

Answer: (a) \( (-2,-1)\cup[1,\infty) \)

Explanation:
1. f(1) = -1 and on [1, 3], \( 2x-3\ge-1 \).
2. On [0, 1), \( f(x)=-x^3+k \) with \( k=\frac{b^3-b^2+b-1}{b^2+3b+2} \). We need \( -x^3+k\ge-1 \) for all \( x\in[0,1) \), i.e. \( k\ge x^3-1 \), whose supremum is 0, so \( k\ge0 \).
3. \( k=\dfrac{(b-1)(b^2+1)}{(b+1)(b+2)}\ge0 \), so \( \dfrac{b-1}{(b+1)(b+2)}\ge0 \).
4. Sign chart: positive for \( -2<b<-1 \) and for \( b\ge1 \) (b = 1 gives 0).

Teacher's Note:
1. Factor the numerator: \( b^3-b^2+b-1=(b-1)(b^2+1) \).
2. Remember b cannot be -1 or -2.

Answer: (a) a local maximum

Explanation:
1. For \( 0<|x|\le2 \) we have \( f(x)=|x| \), which is close to 0 near x = 0.
2. \( f(0)=1 \), and for small \( |h|\ne0 \), \( f(h)=|h|<1=f(0) \).
3. So f(0) is larger than all nearby values: a local maximum.

Teacher's Note:
1. Compare the value at the point with the values in a small neighbourhood.
2. The function is not continuous at 0, but extrema still make sense.

Answer: (a) 0

Explanation:
1. \( f'(x)=2x\log x+x=x(2\log x+1)>0 \) for \( x\ge1 \).
2. f is increasing on [1, e], so its minimum is \( f(1)=0 \).

Teacher's Note:
1. On [1, e] we have \( \log x\ge0 \).
2. An increasing function has its minimum at the left end.

Answer: (c) 1

Explanation:
1. Let \( t=\sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right) \), so \( t\in[1,\sqrt2] \) and \( \sin2x=t^2-1 \).
2. \( f=\dfrac{t^2-1}{t/\sqrt2}=\sqrt2\left(t-\frac1t\right) \), which is increasing in t.
3. The greatest value is at \( t=\sqrt2 \): \( \sqrt2\left(\sqrt2-\frac{1}{\sqrt2}\right)=1 \).

Teacher's Note:
1. Use \( (\sin x+\cos x)^2=1+\sin2x \).
2. A function of the form \( t-\frac1t \) is increasing for t > 0.

Answer: (b) \( a=2,\ b=-\frac12 \)

Explanation:
1. \( y'=\dfrac ax+2bx+1 \) vanishes at x = -1 and x = 2.
2. x = -1: \( -a-2b+1=0 \). x = 2: \( \frac a2+4b+1=0 \).
3. From the first, \( a=1-2b \); then \( \frac12-b+4b+1=0\Rightarrow b=-\frac12 \), \( a=2 \).

Teacher's Note:
1. Extrema occur where the first derivative is zero.
2. Solve the two linear equations.

Answer: (b) \( \frac14 \)

Explanation:
1. Let \( f=x^{25}(1-x)^{75} \). \( f'(x)=x^{24}(1-x)^{74}\left[25(1-x)-75x\right]=x^{24}(1-x)^{74}(25-100x) \).
2. f' vanishes at \( x=\frac14 \), positive before and negative after: a maximum. At 0 and 1, f = 0.

Teacher's Note:
1. Take out the common factors before setting the derivative to zero.
2. The end points give f = 0.

Answer: (d) (0, 1]

Explanation:
1. The leading coefficient \( 1+b^2>0 \), so the minimum is at \( x=-\frac{b}{1+b^2} \).
2. \( m(b)=1-\dfrac{b^2}{1+b^2}=\dfrac{1}{1+b^2} \).
3. As \( b^2 \) runs over \( [0,\infty) \), \( m(b) \) takes all values in (0, 1], with 1 at b = 0.

Teacher's Note:
1. The minimum of \( px^2+qx+r \) is \( r-\frac{q^2}{4p} \).
2. m(b) never reaches 0.

Answer: (d) infinite

Explanation:
1. \( f'(x)=\sin\frac\pi x-\frac\pi x\cos\frac\pi x \). Setting this to 0 gives \( \tan\frac\pi x=\frac\pi x \).
2. With \( u=\frac\pi x\in(\pi,\infty) \), the equation \( \tan u=u \) has exactly one root in each interval \( \left(n\pi,n\pi+\frac\pi2\right) \).
3. Infinitely many such roots exist, so f' vanishes at infinitely many points of (0, 1).

Teacher's Note:
1. Substitute \( u=\pi/x \) to turn the interval (0, 1) into \( (\pi,\infty) \).
2. The graphs of tan u and u intersect infinitely often.

Answer: (a) \( \left(0,\frac1e\right) \)

Explanation:
1. For \( 0<x<1 \): \( -\ln x=px \) has exactly one root when \( p>0 \) (the left side decreases from \( \infty \) to 0 and the right side increases from 0).
2. For \( x>1 \): \( \ln x=px \), i.e. \( p=\frac{\ln x}{x} \). The function \( \frac{\ln x}{x} \) rises from 0 to a maximum \( \frac1e \) at x = e and then falls to 0, so there are two roots exactly when \( 0<p<\frac1e \).
3. Total three roots for \( p\in\left(0,\frac1e\right) \).

Teacher's Note:
1. Split the modulus into the cases x < 1 and x > 1.
2. Study the graph of \( \ln x/x \).

Answer: (c) 15

Explanation:
1. With n extra trees, the output is \( P(n)=(50+n)(800-10n) \).
2. \( P'(n)=800-10n-10(50+n)=300-20n=0\Rightarrow n=15 \), and \( P''=-20<0 \).

Teacher's Note:
1. Write the total output as a product of the number of trees and the output per tree.
2. A downward parabola has its maximum at the vertex.

Answer: (c) (3) and (4) only

Explanation:
1. On [1, 2): \( f=x^2+1 \) rises from 2 to 5 (excluding 5). At x = 2, \( f=\frac52 \): a drop. So f is neither increasing nor decreasing; (1) and (2) are false.
2. On [2, 3): \( f=\frac{x^2+1}{2} \) (from 2.5 up to 5); on [3, 3.9]: \( f=\frac{x^2+1}{3} \), reaching \( \frac{3.9^2+1}{3}=\frac{16.21}{3} \) at x = 3.9. This is larger than all earlier values (below 5), so (3) is true.
3. The least value is \( f(1)=2 \): (4) is true.

Teacher's Note:
1. Treat each interval where [x] is constant separately.
2. Check what happens at the jump points x = 2 and x = 3.

Answer: (a) (1) and (3) only

Explanation:
1. Near x = 1, \( (x-2)^n\approx(-1)^n \) and \( (x-1)^4\ge0 \), so \( f\approx(-1)^n(x-1)^4 \): a local maximum when n is odd, a local minimum when n is even. So (1) is true and (2) is false.
2. Near x = 2, \( (x-1)^4\approx1>0 \) and \( (x-2)^n \) has an even power when n is even: f \( \ge0 \) with f(2) = 0: local minimum, so (3) is true.
3. When n is odd, \( (x-2)^n \) changes sign at 2: no extremum, so (4) is false.

Teacher's Note:
1. Look at the sign of the other factor near the point.
2. A factor with an odd power changes sign and gives no extremum.

Answer: (a) (1), (2) and (3) only

Explanation:
1. \( f'(x)=4x^3-4=0\Rightarrow x=1 \), which is the only critical point, a minimum with \( f(1)=1-4-1=-4 \). So (4) is false.
2. f(0) = -1 < 0, \( f(x)\to\infty \) as \( x\to\pm\infty \), and f decreases for x < 1 and increases for x > 1.
3. Hence there is exactly one root on each side of 0: one negative and one positive: (1), (2) and (3) are true.

Teacher's Note:
1. Find the minimum value first; it is below zero.
2. A function that decreases then increases crosses zero exactly twice.

Answer: (a) g(x) is increasing in [0, 2]

Explanation:
1. \( g'(x)=2x\left[f'(x^2-2)-f'(6-x^2)\right] \). Since \( f''<0 \), f' is decreasing, so \( f'(u)-f'(v)>0 \) when \( u<v \).
2. Here \( u=x^2-2<v=6-x^2 \) iff \( x^2<4 \). For \( 0<x<2 \): both factors positive, so \( g'>0 \). For \( x>2 \): \( g'<0 \).
3. For \( -2<x<0 \): \( g'<0 \); for \( x<-2 \): \( g'>0 \). So g is increasing on [0, 2], decreasing beyond 2, and has local maxima at x = -2 and x = 2 (a minimum at 0), so the other three options are false.

Teacher's Note:
1. Use the fact that f' is decreasing to compare \( f'(u) \) and \( f'(v) \).
2. Analyse the sign in four regions of x.