JEE Mathematics Application of Derivatives MCQs Set 25

Answer: (a) \( \frac{17}{4} \)

Explanation:
1. At x = 0: k = −1 and m = dy/dx = 4.
2. Tangent length \( =1\cdot\sqrt{1+\frac1{16}}=\frac{\sqrt{17}}4 \); normal length \( =\sqrt{1+16}=\sqrt{17} \).
3. Product = 17/4.

Teacher's Note:
1. Find k and m at x = 0.
2. Multiply the two lengths.

Answer: (a) \( \frac{10}{3} \)

Explanation:
1. f'(x) = \( ax^2+2(a+2)x+(3a-10) \). For a > 0 the minimum is at the larger root, which must be negative with two real roots. For a ≤ 0 the roots are either absent or both positive (or the function is not cubic), so no negative minimum occurs.
2. Roots of the same (negative) sign: product \( \frac{3a-10}{a}\gt0 \) gives a > 10/3 (sum \( -\frac{2(a+2)}{a}\lt0 \) holds automatically).
3. Real roots: \( 4(a+2)^2-4a(3a-10)\gt0 \) gives \( a^2-7a-2\lt0 \), so a < \( \frac{7+\sqrt{57}}{2} \). The set is \( \left(\frac{10}3,\frac{7+\sqrt{57}}2\right) \), so \( \alpha=\frac{10}{3} \).

Teacher's Note:
1. For a cubic with positive leading coefficient the minimum is at the larger root of f'.
2. Use sum and product of the roots of f' together with the discriminant.

Answer: (d) \( 4(\sqrt2-1) \)

Explanation:
1. By symmetry the centre is (0, k) with radius r. Distance to the line y = x is k/√2, so r = k/√2.
2. The circle touches the parabola at its vertex (0, 4) from below, so r = 4 − k.
3. k = √2 r, so r(1 + √2) = 4 and \( r=\frac4{1+\sqrt2}=4(\sqrt2-1) \).

Teacher's Note:
1. Use symmetry to put the centre on the y-axis.
2. Equate the two expressions for the radius.

Answer: (b) x = 2r

Explanation:
1. Constraint: 4x + 2πr = 2. Area A = x² + πr².
2. Using x = (1 − πr)/2: A = (1 − πr)²/4 + πr², and dA/dr = −π(1 − πr)/2 + 2πr = 0 gives 1 − πr = 4r, i.e. (1 − πr)/2 = 2r.
3. So x = 2r.

Teacher's Note:
1. Eliminate x using the length constraint.
2. x = (1 − pi r)/2 from the constraint.

Answer: (a) A and D

Explanation:
1. The derivative of \( \ln(\sqrt{1+x^2}-x) \) is \( -\frac1{\sqrt{1+x^2}} \).
2. f'(x) = \( 2-\frac1{1+x^2}-\frac1{\sqrt{1+x^2}} \) ≥ 2 − 1 − 1 = 0, with equality only at x = 0.
3. So f is increasing on the whole real line, hence A and D are true.

Teacher's Note:
1. Differentiate each term and bound the last two using 1 + x² ≥ 1.
2. Equality in the bound holds only at one point.

Answer: (a) Both true; Statement 2 is a correct explanation of Statement 1

Explanation:
1. f'(x) = 1/x + 2bx + a = 0 at x = −1 and x = 2 gives −1 − 2b + a = 0 and 1/2 + 4b + a = 0, so b = −1/4, a = 1/2 (Statement 2 true).
2. f''(x) = −1/x² + 2b = −1/x² − 1/2 < 0 for all x ≠ 0, so both points are local maxima (Statement 1 true).
3. The values of a and b are what determine the sign of f'', so Statement 2 explains Statement 1.

Teacher's Note:
1. Solve for a and b first.
2. Check the sign of f''.

Answer: (b) \( -\infty\lt f(x)\lt0 \)

Explanation:
1. Let g(x) = e⁻ˣ f(x). Then g'' = e⁻ˣ(f'' − 2f' + f) ≥ 1 > 0, so g is strictly convex.
2. g(0) = g(1) = 0, so a strictly convex g lies below the chord: g < 0 on (0, 1).
3. Hence f(x) = eˣ g(x) < 0 on (0, 1).

Teacher's Note:
1. Multiply f by e^(−x) to absorb the first-order terms.
2. A convex function lies below its chord.

Answer: (c) x = 1 is a point of maxima and x = −1 is a point of minima of f

Explanation:
1. f(x)/x³ → 2 as x → 0, so f(0) = f'(0) = f''(0) = 0 and f''' (0)/6 = 2. With f' having only the critical points ±1, f' = c x²(x² − 1), so \( f=c\left(\frac{x^5}5-\frac{x^3}3\right) \) and f/x³ → −c/3 = 2 gives c = −6.
2. f(x) = −6x⁵/5 + 2x³, an odd function. f'(x) = −6x²(x² − 1) > 0 for 0 < |x| < 1 and < 0 for |x| > 1: x = 1 is a maximum and x = −1 a minimum.
3. So the statement 'x = 1 is a point of minima and x = −1 is a point of maxima' is false; the others hold (f(1) = 4/5, f(−1) = −4/5, f(1) − 4f(−1) = 4).

Teacher's Note:
1. The limit gives the cubic term of f near 0.
2. Check the sign of f' near x = 1.

Answer: (d) A and C

Explanation:
1. Let g(x) = e⁻²ˣ f(x). Then g'(x) = e⁻²ˣ(f' − 2f) > 0, so g is increasing and g(x) > g(0) = 1 for x > 0, i.e. f(x) > e²ˣ (A true).
2. Then f'(x) > 2f(x) > 2e²ˣ > 0 so f is increasing (C true), and f'(x) > e²ˣ so B is false; D is false.

Teacher's Note:
1. Multiply by e^(−2x) to form an increasing function.
2. Use f' > 2f > 0.

Answer: (a) A and B

Explanation:
1. f is piecewise linear, with possible corners where |x + 2| = 2|x| (x = −2/3, 2), and at x = −2, 0.
2. Evaluating near the corners: f(−2) = 0 with f > 0 on both sides (local minimum); f(−2/3) = 8/3 is larger than values on both sides (local maximum); f is monotone through 2/3 and through 2.
3. Of the listed values, −2 and −2/3 give extrema.

Teacher's Note:
1. Piecewise linear functions have extrema only at corners.
2. Compare f just left and right of each candidate.

Answer: (d) B and C

Explanation:
1. g'(x) = f'(x/2) − f'(1 − x).
2. Since f'' < 0, f' is decreasing. So g'(x) > 0 when x/2 < 1 − x, i.e. x < 2/3, and g'(x) < 0 when x > 2/3.
3. Thus g increases on [0, 2/3] (C) and decreases on [2/3, 1] (B).

Teacher's Note:
1. Differentiate and compare f' at the two arguments.
2. f' is decreasing, so a smaller argument gives a larger f'.

Answer: (b) B, C and D

Explanation:
1. By Rolle's theorem f' has a root in each of (4, 5), (5, 6), (6, 7): B is true. f' has three real roots, so f'' has two real roots: A is false.
2. f is symmetric about x = 11/2 (substituting x → 11 − x permutes the factors), so f' is zero at 11/2 (C true) and f'' is symmetric about 11/2 (D true).

Teacher's Note:
1. Apply Rolle's theorem between consecutive zeros.
2. Check what x → 11 − x does to the four factors.

Answer: (a) 40

Explanation:
1. Fuel cost per hour \( =kv^2 \) with \( k=\frac{48}{256}=\frac3{16} \).
2. Cost per mile \( C(v)=\frac{3v}{16}+\frac{300}{v} \), so \( C'(v)=\frac3{16}-\frac{300}{v^2}=0 \) gives \( v^2=1600 \).
3. v = 40 mph (C'' > 0 so this is a minimum).

Teacher's Note:
1. Divide the hourly cost by the speed to get cost per mile.
2. Set the derivative to zero.

Answer: (b) \( \alpha \)

Explanation:
1. f(x) = f(4π − x) on [2π, 4π], so the graph is symmetric about x = 2π.
2. The largest interval of increase has length α = 2π.
3. Hence β = 2π = α.

Teacher's Note:
1. The reflection x → 4π − x fixes x = 2π.
2. Compare 2π with the length of the increasing interval.

Answer: (d) \( f'(1)\gt1 \)

Explanation:
1. f is strictly convex, so its derivative at the right end 1 exceeds the slope of the chord from 1/2 to 1.
2. By the mean value theorem the chord slope is (1 − 1/2)/(1 − 1/2) = 1 = f'(c) for some c in (1/2, 1), and f' is increasing, so f'(1) > f'(c) = 1.

Teacher's Note:
1. f'' > 0 means f' is increasing.
2. Compare with the chord slope.

Answer: (b) 36

Explanation:
1. Put \( y=\sqrt{1-(x-2)^2} \): the point (x, y) lies on the upper half of the circle with centre (2, 0) and radius 1, and the expression is the squared distance from (x, y) to (5, −4).
2. The distance from (5, −4) to the centre is 5, so the greatest distance to the circle is 6, reached at \( \left(\frac75,\frac45\right) \), which lies on the upper half with x in [1, 3].
3. The maximum value is 6² = 36.

Teacher's Note:
1. Interpret the expression as a squared distance from a fixed point.
2. Farthest point on a circle = centre distance + radius.

Answer: (b) \( \frac{3\sqrt2}8 \)

Explanation:
1. For the point (y², y), the distance to x − y + 1 = 0 is \( \frac{|y^2-y+1|}{\sqrt2}=\frac{y^2-y+1}{\sqrt2} \).
2. The numerator is least at y = 1/2, where it equals 3/4.
3. Distance \( =\frac3{4\sqrt2}=\frac{3\sqrt2}8 \).

Teacher's Note:
1. Write the distance from a general point of the parabola to the line.
2. The quadratic y² − y + 1 has its minimum at y = 1/2.

Answer: (d) decreasing on (0, 1) and increasing on (1, 2)

Explanation:
1. φ'(x) = f'(x) − f'(2 − x).
2. f'' > 0 means f' is increasing, so φ'(x) < 0 when x < 2 − x (x < 1) and φ'(x) > 0 when x > 1.
3. So φ decreases on (0, 1) and increases on (1, 2).

Teacher's Note:
1. Differentiate φ using the chain rule.
2. Compare f' at x and at 2 − x.

Answer: (b) A and C

Explanation:
1. Four squares of total area 100 give side x = 5. Let the sheet be 8k by 15k; V(x) = x(8k − 2x)(15k − 2x).
2. V'(x) = 12x² − 92kx + 120k², and V'(5) = 0 gives 120k² − 460k + 300 = 0, i.e. 6k² − 23k + 15 = 0, so k = 3 or k = 5/6.
3. k = 5/6 gives sides 20/3 and 12.5, too small to remove squares of side 5. So k = 3 and the sides are 24 and 45.

Teacher's Note:
1. The corner squares have side 5.
2. Set dV/dx = 0 at x = 5 to find k.

Answer: (a) \( \frac\beta2 \)

Explanation:
1. The function rises on [0, 2π] to f(2π) = (2π + 0)/2 = π and falls on [2π, 4π].
2. So the maximum value is π, and β = 2π gives π = β/2.

Teacher's Note:
1. The maximum is at the line of symmetry.
2. Evaluate f(2π).

Answer: (c) \( 2\sqrt2 \)

Explanation:
1. Put t = x − 1/x. Then x² + 1/x² = t² + 2, so h = (t² + 2)/t = t + 2/t.
2. h' = 1 − 2/t² = 0 gives t = ±√2; h'' = 4/t³ > 0 for t = √2 (a local minimum) and negative for t = −√2 (a local maximum).
3. Local minimum value = √2 + 2/√2 = 2√2.

Teacher's Note:
1. Write f in terms of g.
2. The sign of h'' = 4/t³ decides minima and maxima.