JEE Mathematics Application of Derivatives MCQs Set 26

Answer: (b) A, B and D only

Explanation:
1. f'(0)² = 85 − f(0)² ≥ 81, so f'(0) ≠ 0 and f is strictly monotonic near 0: (A) true.
2. By the mean value theorem on [−4, 0], |f'(x₀)| = |f(0) − f(−4)|/4 ≤ 4/4 = 1: (B) true.
3. (C) is not forced: f(x) = 2 sin(...) type functions with no limit satisfy the conditions: false.
4. (D) Let h = f² + f'². There are points of (−4, 0) and (0, 4) where h ≤ 1 + 4 = 5, while h(0) = 85, so h has an interior maximum at some α with h(α) ≥ 85. There h' = 2f'(f + f'') = 0 and f'(α)² ≥ 81, so f + f'' = 0 and f' ≠ 0: true.

Teacher's Note:
1. Use f' at 0 and the mean value theorem.
2. Consider h = f² + f'² and its maximum on (−4, 4).

Answer: (a) A and C

Explanation:
1. f(x) = (x − 1)(x − 2)/((x + 3)(x − 1)); the denominator vanishes at x = 1 and x = −3, so 1 is not in the domain: (A) true.
2. For x ≠ 1, −3: f(x) = (x − 2)/(x + 3), f'(x) = 5/(x + 3)² > 0, so f has no critical point and no maximum or minimum: (C) true, (B) false.
3. The domain is not an interval (it excludes −3 and 1), and f jumps from +∞ to −∞ around x = −3, so f is not increasing on its whole domain: (D) is not correct.

Teacher's Note:
1. Cancel the common factor, remembering the excluded point.
2. Check the derivative of the simplified form.

Answer: (a) 2.83

Explanation:
1. Expanding, 2 sin x + tan x = 3x + (9/60)x⁵ + ... > 3x for small x > 0, so 3x/(2 sin x + tan x) lies just below 1 and its integer part is 0. The base is 2.
2. tan³x − sin³x = (x + x³/3 + ...)³ − (x − x³/6 + ...)³ = x³ + x⁵ − x³ + x⁵/2 + ... = (3/2)x⁵ + ..., so the exponent tends to 3/2.
3. The limit is 2^(3/2) = 2√2 ≈ 2.83.

Teacher's Note:
1. Decide which of 3x and 2 sin x + tan x is larger for small x > 0 (the hint).
2. Expand tan x and sin x to the x⁵ order.

Answer: (c) \( \left(-\frac{13}4,3\right) \)

Explanation:
1. For a given x < 0 the inequality means x² + x − 3 < a < −x² + x + 3.
2. The lower bound x² + x − 3 has least value −13/4 (at x = −1/2); the upper bound −x² + x + 3 increases to 3 as x → 0⁻ (it never exceeds 3 for x < 0).
3. The union of these intervals over x < 0 is (−13/4, 3).

Teacher's Note:
1. Rewrite |x − a| < 3 − x² as a double inequality in a.
2. Find the smallest lower bound and the largest upper bound for x < 0.

Answer: (b) 9.43

Explanation:
1. The orthocentre is at B only if the angle at B is a right angle, so BC is perpendicular to BA (which is vertical): C = (c, b) where B = (0, b) and b = 1 + 7c²/36, with b = 1 + 2t.
2. Area = (1/2) b c, with c = 6√((b − 1)/7). At t = 7/2, b = 8 and c = 6.
3. db/dt = 2 and dc/dt = (dc/db)(db/dt) = (3/√(7(b − 1)))·2 = (3/7)·2 = 6/7.
4. dA/dt = (1/2)(c·db/dt + b·dc/dt) = (1/2)(12 + 48/7) = 66/7 ≈ 9.43.

Teacher's Note:
1. A right angle at B means BC is horizontal.
2. Write the area as (1/2) b c and differentiate with respect to t.

Answer: (a) 24

Explanation:
1. Let the common tangent be y = mx + c. Tangency with each parabola means the discriminant of x² + (4 − m)x + (8 − c) = 0 and of x² + (8 − m)x + (4 − c) = 0 is zero: (4 − m)² = 4(8 − c) and (8 − m)² = 4(4 − c).
2. Subtracting gives m = 8 and c = 4, so the common tangent is y = 8x + 4.
3. It touches the first parabola at x = (m − 4)/2 = 2, ordinate 4 + 8 + 8 = 20, and the second at x = (m − 8)/2 = 0, ordinate 4.
4. Sum of the ordinates = 20 + 4 = 24.

Teacher's Note:
1. Write the tangent as y = mx + c and set the discriminants to zero.
2. The contact abscissa is −(p − m)/2 for y = x² + px + q.

Answer: (c) 2

Explanation:
1. Parametrise the curve as x = 3t², y = 2t³ (then 4x³ = 108t⁶ = 27y²). Then dy/dx = t, and the tangent at t is y = tx − t³.
2. This tangent is normal at the point with parameter s if its slope t equals −1/(slope there) = −1/s, i.e. s = −1/t, and the point s lies on the line: 2s³ = 3ts² − t³.
3. With s = −1/t: −2/t³ = 3/t − t³, i.e. t⁶ − 3t² − 2 = 0, so (t² − 2)(t² + 1)² = 0 and t² = 2.
4. t = ±√2 gives two distinct lines.

Teacher's Note:
1. Use the parametrisation x = 3t², y = 2t³.
2. Impose both the slope condition and that the second point lies on the tangent line.

Answer: (b) B and D

Explanation:
1. f'(x) = [(1 + x tan x) − x(tan x + x sec²x)]/(1 + x tan x)² = (1 − x² sec²x)/(1 + x tan x)².
2. f' = 0 when x sec x = 1, i.e. x = cos x, which has exactly one root x₀ in (0, π/2).
3. For x < x₀, x² sec²x < 1 so f' > 0; for x > x₀, f' < 0. So f increases then decreases: exactly one maximum at x₀ with x₀ = cos x₀, and no minimum.
4. (B) and (D) are correct.

Teacher's Note:
1. Differentiate and simplify the numerator.
2. Study the sign of 1 − x² sec²x.

Answer: (c) \( -10 \)

Explanation:
1. Let s = 2ˣ + 2⁻ˣ ≥ 2. Then 4ˣ + 4⁻ˣ = s² − 2 and 8ˣ + 8⁻ˣ = s³ − 3s.
2. f = s³ − 3s − 4(s² − 2) = s³ − 4s² − 3s + 8 = g(s).
3. g'(s) = 3s² − 8s − 3 = (3s + 1)(s − 3), so for s ≥ 2 the minimum is at s = 3.
4. g(3) = 27 − 36 − 9 + 8 = −10.

Teacher's Note:
1. Introduce s = 2^x + 2^(-x) and note s ≥ 2.
2. Check that the critical point s = 3 is attainable.

Answer: (d) \( \left(f\left(\frac\pi{12}\right),f\left(\frac\pi4\right)\right)\text{ and }\left(f\left(\frac\pi4\right),f\left(\frac{5\pi}{12}\right)\right) \)

Explanation:
1. x/sin x is increasing on (0, π/2), so with a = f(π/12) < b = f(π/4) < c = f(5π/12).
2. Let h(x) = 2/(x − a) + 3/(x − b) + 4/(x − c). Just right of a, h → +∞ and just left of b, h → −∞, so h has a root in (a, b).
3. Just right of b, h → +∞ and just left of c, h → −∞, so h has a root in (b, c).
4. Hence the roots lie in (f(π/12), f(π/4)) and (f(π/4), f(5π/12)).

Teacher's Note:
1. Check that x/sin x is increasing, so the three points are in order.
2. Examine the signs of the sum near each pole.

Answer: (c) 29

Explanation:
1. f'(x) = 2cos 2x − 8(a + 1)cos x + 4a² + 8a − 14 = 4c² − 8(a + 1)c + 4(a² + 2a − 4), where c = cos x.
2. The quadratic in c has roots c = (a + 1) ± √5. We need it positive for every c in [−1, 1], so both roots lie outside [−1, 1] on the same side: a + 1 − √5 > 1 or a + 1 + √5 < −1.
3. This gives a > √5 or a < −2 − √5, so m = 2, n = 5.
4. m² + n² = 4 + 25 = 29.

Teacher's Note:
1. Put cos x = c and treat f' as a quadratic in c.
2. Positive on all of [−1, 1] means the roots are both outside that interval on the same side.

Answer: (d) 0.37

Explanation:
1. Consider the line y = px (p > 0) and the curve y = |ln x|. For x < 1 the curve −ln x meets the line exactly once.
2. For x > 1 the line meets y = ln x twice when 0 < p < 1/e, touches it once when p = 1/e (the line through the origin tangent to ln x touches at x = e), and misses it when p > 1/e.
3. So the number of solutions is 3 for 0 < p < 1/e, exactly 2 for p = 1/e, and 1 for p > 1/e.
4. p = 1/e ≈ 0.37.

Teacher's Note:
1. Treat the two branches x < 1 and x > 1 separately.
2. Find the slope of the tangent from the origin to y = ln x.

Answer: (b) \( [0,3] \)

Explanation:
1. Let g(x) = x⁴ + 4x³ − 8x²; then g'(x) = 4x(x + 4)(x − 1) with critical points −4, 0, 1.
2. g(−4) = −128 (minimum), g(0) = 0 (local maximum), g(1) = −3 (local minimum).
3. All four roots are real exactly when the line y = −k meets the curve four times (counting multiplicity): −3 ≤ −k ≤ 0.
4. So k ∈ [0, 3].

Teacher's Note:
1. Study the graph of x⁴ + 4x³ − 8x².
2. Four intersections with y = −k occur between the local maximum and the higher local minimum.

Answer: (c) S1 and S3

Explanation:
1. y' = −4/x³ + 4/x⁵ = 4(1 − x²)/x⁵ changes sign (for x > 0 at x = 1), so y is neither increasing nor decreasing: S1 true.
2. f(x) = eˣ is strictly increasing but eˣ = −1 has no solution: S2 false.
3. On the open interval (0, 1), f(x) = x has no local extremum: S3 true.
4. {x} = 0 at x = 6 is the minimum value, not the maximum: S4 false.

Teacher's Note:
1. Test each statement with a simple example.
2. Remember that a local extremum needs an interior point of the interval.

Answer: (b) k

Explanation:
1. From y² = 2x − 2x² with 0 ≤ x ≤ 1: d² = (x − k)² + 2x − 2x² = −x² + (2 − 2k)x + k².
2. The vertex is at x = 1 − k < 0 for k > 1, so d² is decreasing on [0, 1] and is maximum at x = 0.
3. d² = k², so the maximum distance is k.

Teacher's Note:
1. Write d² as a quadratic in x on the interval [0, 1].
2. Locate the vertex relative to the interval.

Answer: (b) 7.50

Explanation:
1. Let the ball have fallen s feet. By similar triangles, the light is at height 50 and the ball is at horizontal distance 30 and height 50 − s, so the shadow lies at distance x from the pole with x/50 = 30/s, i.e. x = 1500/s.
2. dx/dt = −(1500/s²)(ds/dt). At t = 1/2: s = 16(1/4) = 4 and ds/dt = 32t = 16.
3. dx/dt = −1500·16/16 = −1500 ft/s.
4. So 200|λ| = 1500 and |λ| = 7.5.

Teacher's Note:
1. Use similar triangles formed by the light, the ball and the ground.
2. Differentiate x = 1500/s with respect to t.

Answer: (d) \( \log_23\gt\log_35\gt\log_47 \)

Explanation:
1. Write each as log(n+1)/log n form: log₂3 = ln 3/ln 2, log₃5 = ln 5/ln 3, log₄7 = ln 7/ln 4.
2. Numerically ≈ 1.585, 1.465 and 1.404 respectively.
3. So log₂3 > log₃5 > log₄7.

Teacher's Note:
1. Compute the three values using natural logarithms.
2. The sequence of numbers decreases.

Answer: (b) \( \dfrac{2\pi}3 \)

Explanation:
1. The slant height is √3. If the base radius is r, the height is h = √(3 − r²).
2. V = (π/3) r² √(3 − r²). Maximise V² ∝ r⁴(3 − r²): derivative 12r³ − 6r⁵ = 0 gives r² = 2.
3. V = (π/3)(2)(1) = 2π/3.

Teacher's Note:
1. The slant height of the cone equals the radius of the sheet.
2. Maximise the square of the volume to avoid the root.

Answer: (b) \( \dfrac{9\sqrt3}{2\pi} \)

Explanation:
1. The function h(x) = sin x / x is concave on (0, π/2).
2. By Jensen's inequality, the sum ≤ 3 h((A + B + C)/3) = 3 h(π/3).
3. h(π/3) = (√3/2)/(π/3) = 3√3/(2π), so the maximum is 9√3/(2π), attained for the equilateral triangle.

Teacher's Note:
1. Check the concavity of sin x / x.
2. Apply Jensen's inequality with A + B + C = π.

Answer: (d) 4

Explanation:
1. The right side is at least 2 and equals 4 at |x| = log₂(2 + √3) ≈ 1.90; the left side is at most 4. So roots can only lie in |x| ≤ 1.90, where e^x lies in about [0.15, 6.7], and also need cos(e^x) ≥ 1/2.
2. Taking g(x) = 4cos(e^x) − 2^x − 2^(−x) and examining its sign changes on [−1.9, 1.9], g changes sign exactly four times.
3. The roots are approximately x = −1.88, 0.05, 1.78 and 1.87.
4. So the number of roots is 4.

Teacher's Note:
1. Bound both sides: the left side is at most 4 and the right side at least 2.
2. Study the sign of g(x) = 4cos(e^x) − 2^x − 2^(−x) between its critical regions.

Answer: (c) \( [4a,3a] \)

Explanation:
1. The domain is x(4a − x) ≥ 0, i.e. 4a ≤ x ≤ 0 (since a < 0).
2. Differentiating, \( f'(x)=\frac{2x(3a-x)}{\sqrt{4ax-x^2}} \).
3. For 4a < x < 3a we have x < 0 and 3a − x > 0, so f' < 0; for 3a < x < 0 both factors are negative and f' > 0.
4. So f decreases on [4a, 3a].

Teacher's Note:
1. First find the domain of the function.
2. Factor f' and study the sign on each side of x = 3a.

Answer: (d) \( \left(\frac12,\frac54\right) \)

Explanation:
1. Tangent at (t, t² + 1): y = 2tx + 1 − t².
2. It meets x = 0 at height 1 − t² and x = 1 at height 1 + 2t − t².
3. Area of the trapezium = (1/2)(1 − t² + 1 + 2t − t²)(1) = 1 + t − t².
4. This is maximum when t = 1/2, giving the point (1/2, 5/4).

Teacher's Note:
1. Find the heights of the tangent at x = 0 and x = 1.
2. The trapezium has parallel sides along x = 0 and x = 1 and width 1.

Answer: (a) \( \ln(1+x) \)

Explanation:
1. Let h(x) = (1 + x) ln(1 + x) − tan⁻¹x.
2. h(0) = 0 and h'(x) = ln(1 + x) + 1 − 1/(1 + x²).
3. For x > 0, ln(1 + x) > 0 and 1 − 1/(1 + x²) > 0, so h' > 0 and h(x) > 0.
4. Hence (1 + x) ln(1 + x) > tan⁻¹x, i.e. ln(1 + x) > tan⁻¹x/(1 + x).

Teacher's Note:
1. Cross-multiply by (1 + x) > 0 and define a difference function.
2. Show the difference is increasing from 0.

Answer: (a) \( |a|\gt2 \)

Explanation:
1. f(k) = 3, and the left branch is ≥ 3.
2. For x slightly greater than k, sin(h)/h < 1 (h = x − k > 0) tends to 1, so the right branch is slightly less than a² − 1.
3. We need a² − 2 + sin(h)/h ≥ 3 for small h > 0, i.e. a² ≥ 5 − sin(h)/h, which holds for all small h only if a² > 4 (at a² = 4 the right branch dips below 3).
4. Hence |a| > 2.

Teacher's Note:
1. Compare the right-hand values with f(k) = 3.
2. Remember sin(h)/h is slightly less than 1 for small h > 0.

Answer: (b) \( \left(0,\dfrac4\pi\right) \)

Explanation:
1. Put t = {x}/a, which takes every value in [0, 1/a).
2. sin t + cos t = √2 sin(t + π/4) reaches its maximum √2 at t = π/4.
3. So π/4 must lie in [0, 1/a), i.e. 1/a > π/4, i.e. a < 4/π.
4. Thus a ∈ (0, 4/π).

Teacher's Note:
1. Let t = {x}/a and find its range.
2. The maximum of sin t + cos t occurs at t = π/4.

Answer: (a) 2

Explanation:
1. f''(x) = 2ax + 2(a + 2) = 0 gives x = −(a + 2)/a (a ≠ 0), and f'' changes sign there.
2. This is negative when (a + 2)/a > 0, i.e. a < −2 or a > 0.
3. So α = −2, β = 0 and |α| + |β| = 2.

Teacher's Note:
1. The point of inflection is the root of f''.
2. Solve (a + 2)/a > 0.