Application of Derivatives Class 12 NCERT Solutions Chapter 6 Misc

Concept used. First Derivative Test for maximum. If f is continuous at a critical point c and f' changes sign from positive (left of c) to negative (right of c), then c is a point of local maximum. We will also use the quotient rule: (uv)' = u'v - uv'v².

1. Domain: x > 0 (so that log x is defined).
2. Differentiate f(x) = log xx by the quotient rule with u = log x, v = x: f'(x) = (log x)' · x - log x · (x)'x² = 1x· x - log x · 1x² = 1 - log xx².
3. Set f'(x) = 0. Since x² > 0, the equation reduces to 1 - log x = 0, i.e. log x = 1, giving x = e.
4. Sign of f' around x = e. The denominator x² > 0 throughout, so the sign of f' matches the sign of 1 - log x.
   • For x < e: log x < 1, so 1 - log x > 0, hence f'(x) > 0 (f increasing).
   • For x > e: log x > 1, so 1 - log x < 0, hence f'(x) < 0 (f decreasing).
   f' changes sign +→- at x = e. By the First Derivative Test, x = e is a local (and absolute, since f' > 0 everywhere left of e and f' < 0 everywhere right) maximum on (0,∞).
5. Maximum value: f(e) = log ee = 1e.

f(x) has its maximum at x = e, value 1e.

Concept used. Related rates. The area A depends on the equal-side length a; both A and a vary with time t. Use the chain rule: dAdt = dAda· dadt.

1. Let the two equal sides each have length a at time t. The base b is fixed. Drop a perpendicular from the apex to the midpoint of the base; its length (the altitude) is h = √a² - b²4.
2. Area: A = 12· b· h = b2√a² - b²4.
3. Differentiate A with respect to a: dAda = b2· 12√a² - b²4· 2a = ab2√a² - b²4.
4. Chain rule: dAdt = dAda· dadt = ab2√a² - b²4· dadt. Given: dadt = -3 cm/s (decreasing).
5. Substitute a = b (the moment we are asked about): √a² - b²4 = √b² - b²4 = √3b²4 = b32. Then dAdt = b· b2· b32· (-3) = b²b3· (-3) = b3· (-3) = -3 b.
6. The negative sign indicates the area is decreasing. The rate of decrease is 3 b cm²/s.

Area decreases at 3 b cm²/s when a = b.

Concept used. A function f is increasing on an interval I if f'(x) ≥ 0 for all x ∈ I, and decreasing if f'(x) ≤ 0. Compute f' via the quotient rule, simplify, and read off the sign.

1. Let N = 4sin x - 2x - xcos x and D = 2 + cos x. So f = ND.
2. Differentiate. N' = 4cos x - 2 - (cos x - xsin x) = 4cos x - 2 - cos x + xsin x = 3cos x - 2 + xsin x. D' = -sin x.
3. Quotient rule: f'(x) = N' D - N D'D². Let's expand the numerator. aligned N'D &= (3cos x - 2 + xsin x)(2 + cos x)
   &= 6cos x + 3cos² x - 4 - 2cos x + 2xsin x + xsin xcos x
   &= 4cos x + 3cos² x - 4 + 2xsin x + xsin xcos x. aligned aligned -N D' &= -(4sin x - 2x - xcos x)(-sin x) = (4sin x - 2x - xcos x)sin x
   &= 4sin² x - 2xsin x - xsin xcos x. aligned Sum (numerator of f'): aligned N'D - ND' &= 4cos x + 3cos² x - 4 + 2xsin x + xsin xcos x
   & + 4sin² x - 2xsin x - xsin xcos x
   &= 4cos x + 3cos² x + 4sin² x - 4
   &= 4cos x + 3cos² x + 4(1 - cos² x) - 4
   &= 4cos x - cos² x
   &= cos x (4 - cos x). aligned
4. So f'(x) = cos x (4 - cos x)(2 + cos x)².
5. Sign analysis. Note (2 + cos x)² > 0 always, and 4 - cos x ≥ 3 > 0 always (since cos x ≤ 1). So the sign of f'(x) matches the sign of cos x.
   • f'(x) > 0 when cos x > 0. On [0, 2π] this is the union [0,π2)∪(3π2, 2π].
   • f'(x) < 0 when cos x < 0, i.e. on (π2, 3π2).
6. Stated as the standard NCERT answer on [0, 2π]:
   • Increasing on [0,π2]∪[3π2, 2π].
   • Decreasing on [π2, 3π2].

Increasing on [0,π2]∪[3π2, 2π]; decreasing on [π2, 3π2].

Concept used. Compute f'(x) and determine where it is positive (increasing) or negative (decreasing).

1. Differentiate f(x) = x³ + x^-3: f'(x) = 3x² - 3x^-4 = 3(x² - 1x⁴) = 3(x⁶ - 1)x⁴.
2. x⁴ > 0 for all x≠ 0. So the sign of f'(x) is the sign of x⁶ - 1.
3. Factor: x⁶ - 1 = (x² - 1)(x⁴ + x² + 1). The factor x⁴ + x² + 1 > 0 for all real x (sum of positive squares plus 1). So the sign of f' is the sign of x² - 1.
4. x² - 1 > 0 ⇔ |x| > 1 ⇔ x < -1 or x > 1. x² - 1 < 0 ⇔ -1 < x < 1 (with x ≠ 0).
5. Conclusion:
   • f is increasing on (-∞, -1]∪[1, ∞).
   • f is decreasing on [-1, 0)∪(0, 1].

Increasing on (-∞, -1]∪[1,∞); decreasing on [-1, 0)∪(0, 1].

Concept used. Place the vertex of the isosceles triangle at (-a, 0) (one end of the major axis). By the isosceles symmetry, the opposite side is a vertical chord x = constant. Express the area as a function of the chord's x-coordinate and maximise.

1. Let the apex be A(-a, 0). Let the other two vertices be P(x, y) and Q(x, -y) on the ellipse (symmetric about the x-axis to keep the triangle isosceles). Since P is on the ellipse, x²a² + y²b² = 1 ⇒ y = b√1 - x²a². Allowed range: -a < x ≤ a. (When x = -a, the chord collapses to the point A.)
2. Base of the triangle (the chord PQ): length = 2y.
   Height from A to the base: horizontal distance = x - (-a) = x + a.
3. Area: S(x) = 12· 2y· (x + a) = y(x + a) = (x + a) b√1 - x²a².
4. Differentiate. Let u(x) = (x + a) and v(x) = b√1 - x²a². Then u' = 1 and v' = b· -x/a²√1 - x²/a² = -bxa²√1 - x²/a². Product rule: S'(x) = b√1 - x²a² + (x + a)· -bxa²√1 - x²/a². Multiply through by √1 - x²/a² to clear the denominator and set S'(x) = 0: b(1 - x²a²) - (x + a)bxa² = 0. Divide by ba²: (a² - x²) - x(x + a) = 0. Factor: a² - x² = (a - x)(a + x). So (a - x)(a + x) - x(x + a) = (a + x)(a - x - x) = (a + x)(a - 2x).
5. So S'(x) = 0 ⇒ (a + x)(a - 2x) = 0, giving x = -a (degenerate) or x = a2.
6. Check sign. For -a < x < a2: (a + x) > 0 and (a - 2x) > 0, so S' > 0 (rising). For a2 < x < a: (a + x) > 0 and (a - 2x) < 0, so S' < 0 (falling). Therefore x = a2 is a local (and absolute) maximum.
7. Maximum area. At x = a2, y = b√1 - 14 = b· 32. Then S_max = (x + a) y = 3a2· b32 = 33 ab4.

Maximum area =334 ab.

Concept used. Express the cost as a function of one variable using the volume constraint, then minimise. Volume = (base area)×(depth); cost = 70·(base area) + 45·(total side area).

1. Let the base of the tank have length x m and breadth y m, with depth 2 m. Volume constraint: 2xy = 8 ⇒ xy = 4 ⇒ y = 4x.
2. Areas:
   • Base: xy = 4 m² (already fixed by the volume constraint).
   • Sides (4 walls, two pairs of equal walls): two walls of x× 2 each (total 2· 2x = 4x m²) and two walls of y× 2 each (total 4y m²). Side total = 4x + 4y m².
3. Cost (in Rs): C = 70· 4 + 45· (4x + 4y) = 280 + 180(x + y) = 280 + 180(x + 4x).
4. Differentiate: dCdx = 180(1 - 4x²). Set to zero: 1 - 4x² = 0 ⇒ x² = 4 ⇒ x = 2 (positive root).
5. Second derivative test: d² Cdx² = 180· 8x³. At x = 2, d² Cdx² = 180· 1 = 180 > 0. Minimum.
6. Find y: y = 42 = 2 m. Both x = 2 and y = 2: the base is a 2× 2 square.
7. Minimum cost: C = 280 + 180· (2 + 2) = 280 + 720 = 1000 Rs.

Least cost = Rs 1000 (with x = y = 2 m, depth = 2 m).

Concept used. Same approach as the ``wire cut into circle + square'' problem from Ex 6.3: write the total area as a single-variable function and minimise.

1. Let the circle have radius r and the square have side s. Perimeters add to k: 2π r + 4s = k ⇒ s = k - 2π r4.
2. Total area: A = π r² + s² = π r² + (k - 2π r)²16.
3. Differentiate with respect to r: dAdr = 2π r + 2(k - 2π r)(-2π)16 = 2π r - π(k - 2π r)4. Set dAdr = 0: 2π r = π(k - 2π r)4 ⇒ 8π r = π(k - 2π r) ⇒ 8r = k - 2π r. Solve for r: 8r + 2π r = k ⇒ r(8 + 2π) = k ⇒ r = k2(π + 4).
4. Compute s: s = k - 2π r4 = k - 2π· k2(π + 4)4 = k - π kπ + 44 = k(π + 4) - π kπ + 44 = 4k4(π + 4) = kπ + 4.
5. Ratio: sr = k/(π+4)k/[2(π+4)] = 2. So s = 2r: the side of the square equals the diameter of the circle.
6. Second derivative: d² Adr² = 2π + π· 2π4 = 2π + π²2 > 0. Minimum confirmed.

s = 2r minimises the sum of areas.

Concept used. ``Maximum light'' = maximum area of the window. Express the area in terms of one variable using the perimeter constraint, then maximise.

1. Let the rectangle have width 2x (so the semicircle's diameter is 2x, radius x) and height y.
2. Perimeter of the window (outside path): two vertical sides + bottom + semicircle. P = 2y + 2x + π x = 10 ⇒ y = 10 - (2 + π)x2 = 5 - (2 + π)x2.
3. Area: A = (2x)y + 12π x² = 2xy + π x²2. Substitute y: A(x) = 2x(5 - (2 + π)x2) + π x²2 = 10x - (2 + π)x² + π x²2. Combine: -(2 + π)x² + π x²2 = -2x² - π x² + π x²2 = -2x² - π x²2. So A(x) = 10x - 2x² - π x²2 = 10x - (4 + π)x²2.
4. Differentiate: A'(x) = 10 - (4 + π)x. Set A'(x) = 0 ⇒ x = 10π + 4.
5. Second derivative: A''(x) = -(π + 4) < 0. Maximum.
6. Dimensions. With x = 10π + 4:
   • Rectangle width 2x = 20π + 4 m.
   • Rectangle height y = 5 - (2 + π)· 10/(π + 4)2 = 5 - 5(2 + π)π + 4 = 5(π + 4) - 5(2 + π)π + 4 = 10π + 4 m.
   Note that y = x = 10π + 4.

Width of window =20π + 4 m; height of rectangle =10π + 4 m.

Concept used. Place the right triangle in standard position with legs along the axes. The point on the hypotenuse at perpendicular distances a and b from the two legs is just the point (a, b) (the perpendicular distance from a point to the y-axis is its x-coordinate, and similarly for the x-axis). Parametrise the line by its angle of inclination and minimise the hypotenuse length.

1. Let the right triangle have its right angle at the origin, legs along the positive x- and y-axes, and hypotenuse from (p, 0) to (0, q). The fixed point (a, b) lies on the hypotenuse, so its coordinates satisfy the line's intercept form: ap + bq = 1.
2. Hypotenuse length: L = √p² + q². Parametrise via the angle θ the hypotenuse makes with the x-axis. Then the segment from (a, b) along the hypotenuse to the x-axis has length bsinθ (the rise b along a line of slope -tanθ), and from (a, b) along the hypotenuse to the y-axis has length acosθ. Hence L(θ) = acosθ + bsinθ = asecθ + bcscθ, 0 < θ < π2.
3. Differentiate: L'(θ) = asecθ - bcscθ = asinθcos²θ - bcosθsin²θ. Set L'(θ) = 0: asinθcos²θ = bcosθsin²θ ⇒ asin³θ = bcos³θ ⇒ tan³θ = ba ⇒ tanθ = (ba)^1/3.
4. At this θ: write tanθ = b^1/3a^1/3. Build the right triangle with opposite = b^1/3 and adjacent = a^1/3, hypotenuse = √a^2/3 + b^2/3. So sinθ = b^1/3√a^2/3 + b^2/3, cosθ = a^1/3√a^2/3 + b^2/3.
5. Substitute into L: L_min = acosθ + bsinθ = a· √a^2/3 + b^2/3a^1/3 + b· √a^2/3 + b^2/3b^1/3. Simplify each term: a· a^-1/3 = a^2/3 and b· b^-1/3 = b^2/3. L_min = (a^2/3 + b^2/3)√a^2/3 + b^2/3 = (a^2/3 + b^2/3)^3/2.

L_min = (a^2/3 + b^2/3)^3/2.

Concept used. First Derivative Test. Compute f'(x), factor, and look at where f' changes sign. At a sign change +→- we get a local max; -→+ a local min; no sign change means a point of inflexion.

1. Differentiate using the product rule. Let u = (x - 2)⁴ and v = (x + 1)³. u' = 4(x - 2)³, v' = 3(x + 1)². Then f'(x) = u'v + uv' = 4(x - 2)³ (x + 1)³ + (x - 2)⁴· 3(x + 1)².
2. Factor common terms (x - 2)³ (x + 1)²: f'(x) = (x - 2)³ (x + 1)² [4(x + 1) + 3(x - 2)]. Simplify the bracket: 4x + 4 + 3x - 6 = 7x - 2. So f'(x) = (x - 2)³ (x + 1)² (7x - 2).
3. Roots of f': x = 2, x = -1, x = 27.
4. Sign analysis on the real line (split into intervals by these roots, in order -1 < 27 < 2).
   • For x < -1: (x - 2)³ < 0, (x + 1)² > 0, (7x - 2) < 0. Product: (-)(+)(-) = +. So f' > 0.
   • For -1 < x < 27: (x - 2)³ < 0, (x + 1)² > 0 (still), (7x - 2) < 0. Product = +. So f' > 0. Sign across -1: +→+. No change.
   • For 27 < x < 2: (x - 2)³ < 0, (x + 1)² > 0, (7x - 2) > 0. Product = (-)(+)(+) = -. So f' < 0. Sign across 27: +→-. Local maximum at x = 27.
   • For x > 2: (x - 2)³ > 0, (x + 1)² > 0, (7x - 2) > 0. Product = +. So f' > 0. Sign across 2: -→+. Local minimum at x = 2.
5. At x = -1: f' does not change sign (+→+) because (x + 1)² keeps it non-negative on both sides. So x = -1 is a point of inflexion.

Local max at x = 27; local min at x = 2; point of inflexion at x = -1.

Concept used. Absolute extrema on a closed interval = compare f at the critical points inside (a, b) and at the endpoints.

1. Rewrite using cos² x = 1 - sin² x: f(x) = 1 - sin² x + sin x. Let t = sin x. For x ∈ [0, π], t ∈ [0, 1]. Then g(t) = 1 - t² + t = -t² + t + 1.
2. This is a downward parabola in t with vertex at t = 12. Vertex value: g(12) = -14 + 12 + 1 = 54. Endpoint values on t∈ [0, 1]: g(0) = 1, g(1) = -1 + 1 + 1 = 1.
3. So on t ∈ [0, 1], g ranges from 1 (at t = 0 and t = 1) up to 54 (at t = 12).
4. Translate back. t = 12 means sin x = 12, so x = π6 or x = 5π6 (both in [0, π]). t = 0 means sin x = 0, so x = 0 or π. t = 1 means sin x = 1, so x = π2.
5. Compile: aligned f(0) &= 1 + 0 = 1,
   f(π6) &= cos²(π6) + sin(π6) = 34 + 12 = 54,
   f(π2) &= 0 + 1 = 1,
   f(5π6) &= 34 + 12 = 54,
   f(π) &= 1 + 0 = 1. aligned Absolute max = 54 at x = π6 and 5π6. Absolute min = 1 at x = 0, π2, π.

Absolute max = 54; absolute min = 1.

Concept used. This is the cone-in-sphere problem from Q23 of Ex 6.3, restated. The cone's altitude h measured from apex to base, with base of radius a on the sphere, satisfies a² = 2rh - h².

1. Place the apex of the cone at the top of the sphere (the ``north pole''). The base lies at a depth h below the apex; its centre is at distance h - r below the sphere's centre O. Pythagoras on (centre, base-circle centre, base-circle edge): a² + (h - r)² = r² ⇒ a² = 2rh - h².
2. Volume of the cone: V = 13π a² h = π3(2rh² - h³).
3. Differentiate: dVdh = π3(4rh - 3h²) = π h3(4r - 3h). Set dVdh = 0: h = 0 (degenerate) or h = 4r3.
4. Second derivative: d² Vdh² = π3(4r - 6h); d² Vdh²|_{h = 4r/3} = π3(4r - 8r) = -4π r3 < 0. So h = 4r3 is a maximum.

Altitude of the maximum-volume inscribed cone = 4r3.

Concept used. Lagrange's Mean Value Theorem (MVT). If f is continuous on [c, d] and differentiable on (c, d), then there exists ξ ∈ (c, d) with f(d) - f(c) = f'(ξ)(d - c).

1. Pick any two points x₁, x₂ ∈ (a, b) with x₁ < x₂. We must show f(x₁) < f(x₂) (definition of strictly increasing).
2. f is differentiable on (a, b) ⊇ (x₁, x₂), so f is continuous on [x₁, x₂] (differentiability implies continuity) and differentiable on (x₁, x₂). The hypotheses of the MVT hold on [x₁, x₂].
3. By the MVT, there exists ξ ∈ (x₁, x₂) such that f(x₂) - f(x₁) = f'(ξ)(x₂ - x₁).
4. By hypothesis, f'(ξ) > 0 (since ξ ∈ (a, b)). Also x₂ - x₁ > 0 (since x₁ < x₂). Therefore the right-hand side is the product of two positive numbers, hence positive: f(x₂) - f(x₁) > 0 ⇒ f(x₁) < f(x₂).
5. Since x₁, x₂ ∈ (a, b) were arbitrary with x₁ < x₂, f is strictly increasing on (a, b).

Hence f is increasing on (a, b).

Concept used. Inscribe a right circular cylinder symmetrically in a sphere of radius R. With centre of the sphere at origin and cylinder axis along the vertical, the top and bottom rims are at ± h2 (where h is the cylinder's height). Pythagoras: r² + (h2)² = R², where r is the cylinder's radius.

1. Constraint: r² + h²4 = R² ⇒ r² = R² - h²4. Valid for 0 < h < 2R.
2. Volume: V = π r² h = π h(R² - h²4) = π R² h - π h³4.
3. Differentiate: dVdh = π R² - 3π h²4. Set to zero: π R² = 3π h²4 ⇒ h² = 4 R²3 ⇒ h = 2R3.
4. Second derivative: d² Vdh² = -3π h2 < 0 for h > 0. Maximum confirmed.
5. Maximum volume. With h = 2R3: h²4 = 4R²/34 = R²3, so r² = R² - R²3 = 2R²3. V_max = π r² h = π· 2R²3· 2R3 = 4π R³33.

h = 2R3; V_max = 4π R³33.

Concept used. Place the cone with apex pointing up. Let the inscribed cylinder have height y measured from the cone's base. By similar triangles, the cylinder's radius x relates to its height through the cone's geometry.

1. Set up. The cone has apex at (0, h) (with the base on the x-axis) and base radius htanα. Let the inscribed cylinder have height y (resting on the base) and radius x. The top rim of the cylinder is at height y; its radius equals the cross-section of the cone at that height.
2. At height y inside the cone (measured from the base), the remaining vertical distance to the apex is h - y. By similar triangles between the full cone (height h, base radius htanα) and the smaller cone at height y (height h - y, radius x): xh - y = tanα ⇒ x = (h - y)tanα.
3. Volume of the cylinder: V = π x² y = π (h - y)² tan²α · y = ²α · y(h - y)².
4. Differentiate f(y) = y(h - y)² (the ²α is a positive constant and doesn't affect the location of the optimum): f'(y) = (h - y)² + y· 2(h - y)(-1) = (h - y)[(h - y) - 2y] = (h - y)(h - 3y).
5. Set f'(y) = 0: y = h (degenerate; cylinder collapses to a point) or y = h3. Take y = h3.
6. Sign check: for 0 < y < h3, (h - y) > 0 and (h - 3y) > 0, so f' > 0. For h3 < y < h, (h - y) > 0 and (h - 3y) < 0, so f' < 0. Sign change +→-: local max.
7. Compute the maximum volume. At y = h3: h - y = 2h3, so V_max = ²α· h3· (2h3)² = ²α· h3· 4h²9 = 4π h³ tan²α27.

Cylinder height = h3 (one-third of the cone's height); V_max = 4π h³ tan²α27.

Concept used. Related rates. The volume V in a cylindrical tank of fixed radius R = 10 m and variable depth h is V = π R² h. Differentiate both sides with respect to time t.

1. Volume: V = π (10)² h = 100π h.
2. Differentiate with respect to t: dVdt = 100π· dhdt.
3. Given: dVdt = 314 m³/h. Solve for dhdt: dhdt = 314100π = 3.14π m/h.
4. Take π ≈ 3.14: dhdt ≈ 3.143.14 = 1 m/h.

(A) 1 m/h.