JEE Mathematics Application of Derivatives MCQs Set 17

Answer: (d) |m| ≤ 1

Explanation:
1. y = ln(1 + x²), so m = 2x/(1 + x²).
2. Since 1 + x² ≥ 2|x|, |m| ≤ 1, with equality at x = ±1.

Teacher's Note:
1. Use the AM–GM inequality on 1 + x².

Answer: (c) −√3 ≤ a ≤ √3

Explanation:
1. dy/dx = 3x² − 2ax + 1 ≥ 0 for all x.
2. This needs discriminant 4a² − 12 ≤ 0, so |a| ≤ √3.

Teacher's Note:
1. A quadratic is non-negative everywhere when its discriminant is non-positive.

Answer: (b) 3

Explanation:
1. x² − x = 2 gives x = 2 (first quadrant) or x = −1.
2. dy/dx = 2x − 1 = 3 at x = 2.

Teacher's Note:
1. Pick the root with positive x.

Answer: (a) −1

Explanation:
1. y = x gives x² = 4 − x², x = √2 = y.
2. dy/dx = −x/√(4 − x²) = −x/y = −1.

Teacher's Note:
1. On a circle the slope is −x/y.

Answer: (d) −1

Explanation:
1. For x in [−π, 0], cos⁻¹(cos x) = −x.
2. So the slope is −1.

Teacher's Note:
1. cos⁻¹(cos x) = |x| on [−π, π].

Answer: (a) \( \frac12 \)

Explanation:
1. By the fundamental theorem of calculus, dy/dx = 1/(1 + x³).
2. At x = 1 this is 1/2.

Teacher's Note:
1. Differentiate an integral with respect to its upper limit.

Answer: (d) 0

Explanation:
1. dx/dt = eᵗ(sin t + cos t) and dy/dt = eᵗ(cos t − sin t).
2. At t = π/4, dy/dt = 0 and dx/dt ≠ 0, so the slope is 0.
3. The inclination is 0.

Teacher's Note:
1. Slope = (dy/dt)/(dx/dt).

Answer: (b) 1

Explanation:
1. The curve exists only for x, y ≥ 0.
2. dy/dx = −√(x/y) ≤ 0, so slope +1 is impossible; slope −1 requires x = y = a/2^(2/3), giving exactly one point.
3. So there is 1 such tangent.

Teacher's Note:
1. Both coordinates are non-negative, so the slope is never positive.

Answer: (a) (a⁴, a⁴)

Explanation:
1. dy/dx = −√(y/x). Slope −1 gives x = y.
2. Then 2√x = 2a², so x = a⁴: the point is (a⁴, a⁴).

Teacher's Note:
1. The slope −1 condition forces x = y.

Answer: (a) π/2

Explanation:
1. For x² = 5 − 4y the slope at (1, 1) is −x/2 = −1/2.
2. For y = x² the slope is 2.
3. m₁m₂ = −1, so the curves cut at right angles, π/2.

Teacher's Note:
1. Product of slopes −1 means perpendicular.

Answer: (d) none of these

Explanation:
1. At x = 1: y = e⁻¹ and dy/dx = −e⁻¹.
2. Tangent: y − 1/e = −(1/e)(x − 1), i.e. x + ey = 2, which is none of the given forms.

Teacher's Note:
1. For x > 0, |x| = x.

Answer: (a) \( \frac xa+\frac yb=1 \)

Explanation:
1. At x = 0, y = b.
2. dy/dx = −(b/a)e^(−x/a) = −b/a there.
3. Tangent: y − b = −(b/a)x, i.e. x/a + y/b = 1.

Teacher's Note:
1. The point of contact is (0, b).

Answer: (b) a

Explanation:
1. At (x₁, y₁) the tangent is x/√x₁ + y/√y₁ = √a.
2. The intercepts are √a·√x₁ and √a·√y₁, with sum √a(√x₁ + √y₁) = √a·√a = a.

Teacher's Note:
1. Write the tangent in intercept form.

Answer: (c) 4

Explanation:
1. The line is y = 3 − x.
2. c/(x + 1) = 3 − x gives c = (3 − x)(x + 1), i.e. x² − 2x + (c − 3) = 0.
3. Tangency needs equal roots: 4 − 4(c − 3) = 0, so c = 4.

Teacher's Note:
1. Tangency means the discriminant is zero.

Answer: (b) (a, b)

Explanation:
1. At (a, b): aⁿ/aⁿ + bⁿ/bⁿ = 2, so the point is on the curve, and a/a + b/b = 2 puts it on the line.
2. Differentiating xⁿ/aⁿ + yⁿ/bⁿ = 2 gives y′ = −(bⁿxⁿ⁻¹)/(aⁿyⁿ⁻¹), which at (a, b) equals −b/a.
3. That is also the slope of the line x/a + y/b = 2, so the line touches the curve at (a, b).

Teacher's Note:
1. Compute the slope of the curve at (a, b).

Answer: (a) 4

Explanation:
1. For √x + √y = √k the sum of intercepts of any tangent is k.
2. Here √k = 2, so k = 4.

Teacher's Note:
1. Write the tangent in intercept form, as for any curve √x + √y = constant.

Answer: (c) \( \frac12 \) unit²

Explanation:
1. The slope at (1, 0) is 1/x = 1, so the normal has slope −1: y = 1 − x.
2. It cuts the axes at (1, 0) and (0, 1), enclosing area ½·1·1 = 1/2.

Teacher's Note:
1. Normal slope is the negative reciprocal.

Answer: (b) 5v m/min

Explanation:
1. Distance apart D = √((3vt)² + (4vt)²) = 5vt.
2. dD/dt = 5v m/min.

Teacher's Note:
1. Use Pythagoras in terms of time.

Answer: (c) \( \frac{1}{\sqrt2} \)

Explanation:
1. dy/dx = x²/√(x² + 1).
2. At x = 1 this is 1/√2.

Teacher's Note:
1. Fundamental theorem of calculus.

Answer: (a) (−2, 2)

Explanation:
1. 3x² dx/dt = 12 dy/dt, so dy/dt = (x²/4) dx/dt.
2. The abscissa changes faster when |dy/dt| < |dx/dt|, i.e. x²/4 < 1, so −2 < x < 2.

Teacher's Note:
1. Compare the magnitudes of the two rates.

Answer: (b) 2x² − 3x + 1

Explanation:
1. Areas are y² and x². d(y²)/d(x²) = 2y(dy/dx)/(2x) = y(1 − 2x)/x.
2. With y = x(1 − x) this is (1 − x)(1 − 2x) = 2x² − 3x + 1.

Teacher's Note:
1. Divide the two derivatives with respect to x.

Answer: (b) 3k

Explanation:
1. V = x³, so dV/V = 3 dx/x.
2. The per cent error in V is 3k.

Teacher's Note:
1. Logarithmic differentiation: relative error multiplies by the exponent.

Answer: (c) \( \frac12-\frac{\alpha\sqrt3}{120} \)

Explanation:
1. 1′ = α/60 radian.
2. cos(60° + h) ≈ cos 60° − h sin 60° = 1/2 − (α/60)(√3/2) = 1/2 − α√3/120.

Teacher's Note:
1. Use the linear approximation f(a + h) ≈ f(a) + h f′(a).

Answer: (d) 2 and 4 only

Explanation:
1. f = ax² + bx + c with f(1) = 1 and f′(1) = 1, so 2a + b = 1 and c = a.
2. f(0) + f′(0) + f″(0) = a + (1 − 2a) + 2a = 1 + a, which is not always 1.
3. 2f(0) = 2a and 1 − f′(0) = 1 − b = 2a, so statement 4 holds; statement 2 is the tangency condition.

Teacher's Note:
1. Write the general parabola and impose the two tangency conditions.

Answer: (c) 4

Explanation:
1. Slope ±1: dy/dx = −4x/(9y) = ±1 gives y = ∓4x/9.
2. Substituting in 4x² + 9y² = 36: 4x² + 16x²/9 = 36, so x² = 81/13 and x = ±9/√13.
3. Both signs of x with the matching y give the points (9/√13, 4/√13), (9/√13, −4/√13), (−9/√13, 4/√13), (−9/√13, −4/√13): 4 points.

Teacher's Note:
1. Two slopes (+1 and −1) each give two points.