JEE Mathematics Application of Derivatives MCQs Set 27

Answer: (a) x = 2r

Explanation:
1. The wire gives \( 4x + 2\pir = 2 \), so \( x = (1 - \pir)/2 \) and dx/dr = −π/2.
2. Total area \( A = x^{2} + \pir^{2} \), so dA/dr = \( 2x(-\pi/2) + 2\pir = -\pix + 2\pir \).
3. Setting dA/dr = 0 gives \( x = 2r \). Also d²A/dr² = \( \pi^{2}/2 + 2\pi \) > 0, so this is a minimum.

Teacher's Note:
1. Use the length condition to eliminate one variable, then set dA/dr = 0 and check the sign of the second derivative.

Answer: (a) \( (2^{-0.4},\,1) \)

Explanation:
1. \( f'(x) \) has the sign of \( (1 + x^{0.6})x^{0.4} - (1 + x) = x^{0.4} + x - 1 - x = x^{0.4} - 1 \le 0 \) on [0, 1], so \( f \) is decreasing.
2. \( K = f(0) = 1 \) and \( k = f(1) \) = 2^0.6/2 = 2^−0.4.

Teacher's Note:
1. Check the sign of the derivative to see that \( f \) is monotonic.

Answer: (a) α = 2, β = −1/2

Explanation:
1. \( f'(x) = \alpha/x + 2\betax + 1 = 0 \) at \( x = -1 \): \( -\alpha - 2\beta + 1 = 0 \), and at \( x = 2 \): \( \alpha/2 + 4\beta + 1 = 0 \).
2. From the first, \( \alpha = 1 - 2\beta \). Then \( (1 - 2\beta)/2 + 4\beta + 1 = 0 \) gives \( 3\beta = -3/2 \), so \( \beta = -1/2 \) and \( \alpha = 2 \).

Teacher's Note:
1. At an extreme point of a differentiable function, \( f' = 0 \).

Answer: (c) local maximum at π and local minimum at 2π

Explanation:
1. \( f'(x) = \sqrt{x} \sin x \), which is positive on (0, π), negative on (π, 2π) and positive on \( (2\pi, 5\pi/2) \).
2. So \( f' \) changes sign from + to − at π (local maximum) and from − to + at 2π (local minimum).

Teacher's Note:
1. The sign changes of \( f' \) decide maxima and minima.

Answer: (c) −1

Explanation:
1. From the first piece, \( f(-1) = k + 2 \). For \( x \le -1, f(x) = k - 2x \ge k + 2 \), so the left side never goes below \( f(-1) \).
2. For \( x \) > \( -1, f(x) = 2x + 3 \) takes values just above 1 near \( x = -1 \). For \( f(-1) \) to be the smallest nearby value we need \( k + 2 \le 1 \), i.e. \( k \le -1 \).
3. Among the options only \( k = -1 \) satisfies \( k \le -1 \).

Teacher's Note:
1. A local minimum at \( x = -1 \) needs \( f(-1) \) to be at most the values on both sides.

Answer: (a) P(−1) is not the minimum but P(1) is the maximum of P

Explanation:
1. \( P'(0) = 0 \) gives \( c = 0 \), so P'(x) = x(4x² + 3ax \( + 2b \)) and the quadratic has no real root; it is positive (leading coefficient 4).
2. So \( P' \) < 0 for \( x \) < 0 and \( P' \) > 0 for \( x \) > 0: \( P \) decreases then increases, with its minimum at \( x = 0 \).
3. On [−1, 1] the maximum is at an end point; since \( P(-1) \) < \( P(1) \) it is \( P(1) \). \( P(-1) \) is not the minimum (that is \( P(0) \)).

Teacher's Note:
1. One critical point and a positive leading coefficient give a single minimum.

Answer: (a) 122

Explanation:
1. \( x^{2} - 11x + 30 \le 0 \) gives \( S = [5, 6] \).
2. \( f'(x) = 9(x - 1)(x - 3) \) > 0 for \( x \) > 3, so \( f \) is increasing on [5, 6].
3. Maximum = \( f(6) = 648 - 648 + 162 - 40 = 122 \).

Teacher's Note:
1. Find the interval first, then use monotonicity on it.

Answer: (c) 2

Explanation:
1. Let \( g(x) = x^{2} - x \sin x - \cos x \); \( g \) is even and \( g(0) = -1 \).
2. \( g'(x) = 2x - \sin x - x \cos x = x(2 - \cos x) - \sin x \) > 0 for \( x \) > 0 (since \( x \) > \( \sin x \) and \( 2 - \cos x \ge 1 \)), so \( g \) increases from −1 on (0, ∞) and crosses 0 exactly once; by symmetry once on (−∞, 0).
3. Total: 2 solutions.

Teacher's Note:
1. Use evenness and show that \( g \) is monotonic for \( x \) > 0.

Answer: (d) a = b = c

Explanation:
1. Each of \( f, g, h \) has a non-negative derivative on [0, 1] (e.g. \( h' \) = 2xe^x²(1 \( + x^{2} \)) − 2xe^−x² ≥ 0), so each is increasing.
2. The maximum of each is at \( x = 1 \): \( f(1) = g(1) = h(1) = e + 1/e \).
3. Hence \( a = b = c \).

Teacher's Note:
1. On \( [0, 1], x \) and \( x^{2} \) do not exceed 1, so the e^x² term dominates; each function is increasing.

Answer: (a) a local maximum

Explanation:
1. Near \( 0 (x \ne 0), f(x) = |x| \) < \( 1 = f(0) \).
2. So \( f(0) \) is larger than all nearby values: a local maximum.

Teacher's Note:
1. Compare \( f(0) \) with the nearby values.

Answer: (b) 1/4

Explanation:
1. Let \( y = x^{25}(1 - x)^{75} \); \( y' = x^{24}(1 - x)^{74}[25(1 - x) - 75x] = x^{24}(1 - x)^{74}(25 - 100x) \).
2. \( y' \) changes from + to − at \( x = 1/4 \), where the maximum occurs.

Teacher's Note:
1. Factor out the common powers before solving \( y' = 0 \).

Answer: (b) −2/3

Explanation:
1. Use \( a + b - |a - b| = 2 \min(a, b) \) with \( a = |x + 2| \) and \( b = 2|x| \). So \( f(x) = 2 \min(|x + 2|, 2|x|) \).
2. \( |x + 2| = 2|x| \) at \( x = -2/3 \) and \( x = 2 \). Hence \( f = 2|x + 2| \) for \( x \le -2/3, f = 4|x| \) for \( -2/3 \le x \le 2 \), and \( f = 2(x + 2) \) for \( x \ge 2 \).
3. \( f \) increases (slope 2) up to \( x = -2/3 \), where \( f = 8/3 \), and then decreases (slope −4) towards \( x = 0 \). So \( x = -2/3 \) is a local maximum. (The other turning points \( x = -2 \) and \( x = 0 \) are local minima with \( f = 0 \), and at \( x = 2 \) there is no extremum.)
4. Of the given options, only \( x = -2/3 \) is a point of local extremum.

Teacher's Note:
1. Rewrite the expression as twice the smaller of two simple functions and compare slopes on each piece.

Answer: (b) 1

Explanation:
1. \( f(x) = x^{7} + 14x^{5} + 16x^{3} + 30x - 560 \) has \( f'(x) = 7x^{6} + 70x^{4} + 48x^{2} + 30 \) > 0, so \( f \) is strictly increasing.
2. An increasing continuous function of odd degree crosses zero exactly once: one real solution.

Teacher's Note:
1. A strictly increasing polynomial of odd degree has exactly one real root.

Answer: (d) 2

Explanation:
1. By AM–GM, \( x + y \) ≥ 2√(xy) = 2, with equality at \( x = y = 1 \).

Teacher's Note:
1. AM–GM gives the minimum directly.

Answer: (a) 7

Explanation:
1. \( x^{2} - 9x + 20 \le 0 \) gives \( A = [4, 5] \).
2. \( f'(x) = 6(x - 2)(x - 3) \) > 0 for \( x \) > 3, so \( f \) is increasing on [4, 5].
3. Maximum = \( f(5) = 250 - 375 + 180 - 48 = 7 \).

Teacher's Note:
1. Use monotonicity on the given interval.

Answer: (a) \( \left[-4,\frac{3-\sqrt{21}}2\right]\cup(1,\infty) \)

Explanation:
1. With \( c = \sqrt{a + 4}/(1 - a) - 1, f'(x) \) = 5cx⁴ − 3 ≤ 0 for all \( x \) forces \( c \le 0 \) (otherwise \( f' \) > 0 for large \( x \)).
2. If a > 1 then \( 1 - a \) < 0 and \( c \) < −1, which is fine. If \( -4 \le a \) < 1: \( \sqrt{a + 4} \le 1 - a \) gives \( a + 4 \le (1 - a)^{2} \), i.e. \( a^{2} - 3a - 3 \ge 0 \), so \( a \le (3 - \sqrt{21})/2 \).
3. The set is \( [-4, (3 - \sqrt{21})/2] \) ∪ (1, ∞).

Teacher's Note:
1. The sign of the coefficient of \( x^{4} \) decides whether \( f' \) can be positive.

Answer: (b) 2

Explanation:
1. For \( x \) < \( 2 (x \ne 0) \): \( f = (2 - x)/x^{2}, f' = (x - 4)/x^{3} \), with no zero in \( x \) < 2.
2. For \( x \) > 2: \( f = (x - 2)/x^{2}, f' = (4 - x)/x^{3} \), which vanishes at \( x = 4 \).
3. \( x = 2 \) is a critical point where \( f \) is not differentiable, and \( x = 0 \) is not in the domain. Critical points: \( x = 2 \) and \( x = 4 \).

Teacher's Note:
1. Critical points include points where \( f' \) does not exist (inside the domain).

Answer: (d) decreasing on [0, ∞)

Explanation:
1. \( f'(x) \) has the sign of \( \ln(e + x)/(\pi + x) - \ln(\pi + x)/(e + x) \), i.e. of \( (e + x)\ln(e + x) - (\pi + x)\ln(\pi + x) \).
2. The function \( t \ln t \) is increasing for \( t \) > \( 1/e \), and \( e + x \) < \( \pi + x \), so this expression is negative.
3. Hence \( f' \) < 0 and \( f \) is decreasing on [0, ∞.)

Teacher's Note:
1. Compare \( (e + x)\ln(e + x) \) with \( (\pi + x)\ln(\pi + x) \) using the monotonicity of \( t \ln t \).

Answer: (b) a ≠ c

Explanation:
1. The slopes are \( -a - c \) for \( x \) < \( 0, -a + c \) for 0 < \( x \) < \( b/a \), and \( a + c \) for \( x \) > \( b/a \).
2. If \( c \) < a the function decreases until \( b/a \): minimum only at \( b/a \). If \( c \) > a it increases from 0: minimum only at 0. If \( c = a \) the function is constant on \( [0, b/a] \) and the minimum is attained on a whole interval.
3. So the minimum is at exactly one point iff \( a \ne c \).

Teacher's Note:
1. A piecewise linear function has a unique minimum only if no piece is flat at the lowest level.

Answer: (a) increasing in (−∞, ∞)

Explanation:
1. Here \( \log\sqrt{1 + x^{2} - x} \) means \( \frac{1}{2} \ln(x^{2} - x + 1) \), so \( f'(x) = 2 - 1/(1 + x^{2}) + (2x - 1)/(2(x^{2} - x + 1)) \).
2. Writing this with a common denominator gives \( f'(x) = 2 - 1/(1 + x^{2}) + (x - 1/2)/(x^{2} - x + 1) \), which is positive for all real \( x \) (it equals 1.78, 1, 0.5, 1.2, 2 at \( x = -5, -1, 0, 0.5, 1 \)).
3. So \( f \) is increasing on \( R \).

Teacher's Note:
1. Differentiate and show \( f' \) > 0 everywhere; note that \( \cot^{-1}x \) has derivative \( -1/(1 + x^{2}) \).

Answer: (d) a² − 3b > 0, a > 0, b > 0

Explanation:
1. \( f'(x) = 3x^{2} \) + 2ax \( + b \) has two distinct real roots iff \( a^{2} - 3b \) > 0; the local minimum is at the larger root.
2. The larger root is negative iff both roots are negative: sum \( -2a/3 \) < 0 gives a > 0, and product \( b/3 \) > 0 gives \( b \) > 0.

Teacher's Note:
1. A local minimum of a cubic with positive leading coefficient is at the larger root of \( f' \).

Answer: (a) −1/√3

Explanation:
1. \( f'(\alpha) = \tan \pi/6 = 1/\sqrt{3}, f'(\beta) = \sqrt{3} \) and \( f'(\gamma) = 1 \).
2. First integral = \( [f'^{2}/2] \) from β to \( \gamma = (1 - 3)/2 = -1 \). Second integral = \( f'(\gamma) - f'(\alpha) = 1 - 1/\sqrt{3} \).
3. Sum = \( -1 + 1 - 1/\sqrt{3} = -1/\sqrt{3} \).

Teacher's Note:
1. ∫f'f'' \( dx = f'^{2}/2 \) and ∫f'' \( dx = f' \).

Answer: (b) \( \dfrac{a^2-b^2}{2} \)

Explanation:
1. Put \( t = \tan x \) > 0: the expression is ab(a² \( - b^{2})t/(a^{2}t^{2} + b^{2} \)).
2. By AM–GM, \( a^{2}t^{2} + b^{2} \) ≥ 2abt, so the expression is at most ab(a² − b²)/(2ab) = \( (a^{2} - b^{2})/2 \), attained at \( t = b/a \).

Teacher's Note:
1. Dividing by \( \cos^{2}x \) turns the expression into a function of \( \tan x \).

Answer: (d) 1/4

Explanation:
1. By AM–GM, 1 + x^2m ≥ 2x^m, so x^m/(1 + x^2m) ≤ 1/2, and similarly for \( y \).
2. The product is at most 1/4, attained at \( x = y = 1 \).

Teacher's Note:
1. Bound each factor separately with AM–GM.

Answer: (d) 2

Explanation:
1. f'(x) = 6(x² − 3ax \( + 2a^{2}) = 6(x - a)(x - 2a \)), so the maximum is at \( p = a \) and the minimum at q = 2a (f'' < 0 at a, > 0 at \( 2a \)).
2. \( p^{2} = q \) gives \( a^{2} = 2a \), so \( a = 2 \) (as a > 0).

Teacher's Note:
1. Factor \( f' \) to read off the two critical points.

Answer: (c) \( 2\sqrt3\pi \)

Explanation:
1. With height \( h \) and radius \( r \): \( r^{2} = 9 - h^{2} \), so \( V = (\pi/3)(9 - h^{2})h \).
2. dV/dh = \( (\pi/3)(9 - 3h^{2}) = 0 \) gives \( h = \sqrt{3}, a \) maximum.
3. \( V = (\pi/3)(6)(\sqrt{3}) = 2\sqrt{3}\pi \).

Teacher's Note:
1. Express the volume in terms of the height alone.

Answer: (a) 7

Explanation:
1. \( f' \) changes sign at \( x = 1 \), so \( f'(1) = 0 \): \( 3 - 6(a - 2) + 3a = 0 \), giving \( a = 5 \). Then \( f(x) = x^{3} - 9x^{2} + 15x + 7 \) and \( f(1) = 14 \).
2. \( f(x) - 14 = x^{3} - 9x^{2} + 15x - 7 = (x - 1)^{2}(x - 7) \).
3. So the required root is \( x = 7 \).

Teacher's Note:
1. A point where \( f' \) changes sign from + to − has \( f' = 0 \).

Answer: (c) \( (1+2^{n/2})^2 \)

Explanation:
1. For a, \( b \) > \( 0, (1 + 1/a)(1 + 1/b) = 1 + 1/a + 1/b \) + 1/(ab) ≥ 1 + 2/√(ab) + 1/(ab) (= 1 + 1/√(ab))².
2. Take \( a \) = sinⁿα, \( b \) = cosⁿα. Then √(ab) = (\( \sin \alpha \cos \) α)^n/2 = (\( \frac{1}{2} \sin \) 2α)^n/2 ≤ 2^−n/2, so 1/√(ab) ≥ 2^n/2.
3. Hence the expression is at least (1 + 2^n/2)², and equality holds at \( \alpha = \pi/4 \).

Teacher's Note:
1. Use AM–GM on \( 1/a + 1/b \) and the bound \( \sin \alpha \cos \alpha \le 1/2 \).

Answer: (a) 27ab² ≥ 4c³

Explanation:
1. The minimum of ax² \( + b/x \) for \( x \) > 0 occurs where 2ax = \( b/x^{2} \), i.e. \( x^{3} = b/(2a) \).
2. There the value is ax² \( + b/x \) = 3ax² = 3a(b/2a)^2/3.
3. The inequality holds for all \( x \) iff this minimum is ≥ \( c \): \( 27a^{3}(b/2a)^{2} \ge c^{3} \), i.e. 27ab² ≥ \( 4c^{3} \).

Teacher's Note:
1. The condition says the minimum of the left side is at least \( c \).

Answer: (d) 1/2

Explanation:
1. \( f(1) = 0 = f(0) \), so we need continuity at 0 and differentiability on (0, 1).
2. x^α \( \ln x \to 0 \) as \( x \to 0^{+} \) only if α > 0, so negative values and \( \alpha = 0 (\ln x \to -\infty) \) fail.
3. For \( \alpha = 1/2, f \) is continuous on [0, 1] and differentiable on (0, 1).

Teacher's Note:
1. Continuity at 0 requires α > 0.

Answer: (a) (−1/2, 0)

Explanation:
1. For −1/2 < \( x \) < 0: \( |x| = -x \) and \( {x} = x + 1 \), so \( f' = -2x - 1 \), which is negative for \( x \) > −1/2: decreasing.
2. For \( 0 \le x \) < 1: \( f' = x - x = 0 \) (constant), and for \( 1 \le x \) < 2: \( f' = x - (x - 1) = 1 \) > 0 (increasing).
3. So \( f \) is decreasing exactly in \( (-1/2, 0) \) among the options.

Teacher's Note:
1. Write \( |x| \) and \( {x} \) separately on each unit interval.