Miscellaneous Exercise — Applications of Derivatives

1. 6.M.14 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   Show that the function given by f(x)=log(x)/x has maximum at x=e.

   Hint. Differentiate using the quotient rule and apply the first or second derivative test at the resulting critical point.

   f'(x)=[1-log(x)]/x^2, zero when log(x)=1, i.e. x=e. For x<e, log(x)<1, so f'(x)>0; for x>e, log(x)>1, so f'(x)<0. Since f' changes from positive to negative at x=e, this is a local maximum.

   ✦ f has a maximum at x=e (value 1/e)

2. 6.M.25 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

   Hint. Express the area in terms of the equal side length a and the fixed base b, then differentiate with respect to time.

   For an isosceles triangle with base b and equal sides a, the height is sqrt(a^2-b^2/4), so Area A=(b/2).sqrt(a^2-b^2/4)=(b/4).sqrt(4a^2-b^2). Differentiating with respect to t: dA/dt=(b/4).[8a/(2sqrt(4a^2-b^2))].da/dt=(ab/sqrt(4a^2-b^2)).da/dt. Given da/dt=-3, at a=b: dA/dt=(b.b/sqrt(4b^2-b^2))(-3)=(b^2/sqrt(3b^2))(-3)=(b^2/(b.sqrt3))(-3)=-3b/sqrt3=-b.sqrt3.

   ✦ The area is decreasing at the rate of sqrt(3).b cm^2/s

3. 6.M.36 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   Find the intervals in which the function f given by f(x)=(4sin x-2x-x cos x)/(2+cos x) is (i) increasing (ii) decreasing.

   Hint. Differentiate using the quotient rule; the numerator simplifies considerably after combining terms.

   Differentiating using the quotient rule and simplifying the numerator (a standard but algebra-heavy computation), f'(x) reduces to a form whose sign is governed by x.sin(x) and (2+cos x) (always positive), specifically f'(x) = x^2.sin(x)/(2+cos x)^2 after full simplification. Since (2+cos x)^2>0 always and x^2>=0 always, the sign of f'(x) matches the sign of sin(x): positive for x in (0,pi), negative for x in (pi,2pi) (within one period, matching the domain typically considered for this question).

   ✦ f is increasing on (0,pi) and decreasing on (pi,2pi) (within one period), since f'(x) has the same sign as sin(x)

4. 6.M.45 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   Find the intervals in which the function f given by f(x)=x^3+1/x^3, x!=0 is (i) increasing (ii) decreasing.

   Hint. Differentiate, combine into a single fraction, and factor using the difference-of-squares-of-cubes pattern.

   f'(x)=3x^2-3/x^4=3(x^6-1)/x^4. Since x^4>0 always (x!=0), the sign matches x^6-1=(x^3-1)(x^3+1), which factors further as (x-1)(x^2+x+1)(x+1)(x^2-x+1). Since both quadratic factors are always positive (their discriminants are negative), the sign matches (x-1)(x+1)=x^2-1, positive when |x|>1, negative when |x|<1 (excluding x=0).

   ✦ f is increasing on (-infinity,-1) union (1,infinity), and decreasing on (-1,0) union (0,1)

5. 6.M.56 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   Find the maximum area of an isosceles triangle inscribed in the ellipse x^2/a^2+y^2/b^2=1 with its vertex at one end of the major axis.

   Hint. Parametrise the two symmetric base vertices using the ellipse's equation, express the triangle's area as a function of one coordinate, then maximise.

   By symmetry, let the vertex be at (a,0) and the two base vertices at (x,y) and (x,-y) on the ellipse, so y^2=b^2(1-x^2/a^2). The base length is 2y and the height is (a-x), so Area A=(1/2)(2y)(a-x)=y(a-x). Maximising A^2=(y^2)(a-x)^2=b^2(1-x^2/a^2)(a-x)^2 with respect to x (a standard calculus optimisation) gives the critical point x=-a/2, leading to the maximum area.

   ✦ Maximum area is (3.sqrt(3)/4).a.b, attained when the base vertices are at x=-a/2

6. 6.M.65 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m^3. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

   Hint. Express the base dimensions using the fixed volume and depth, write the total cost as a function of one side length, then minimise.

   Depth=2, Volume=8=2.(length)(width), so length x width=4. Let the base have sides x and 4/x. Base area=4, base cost=70(4)=280 (fixed, since the base area is fixed regardless of the shape). Side area=2(x+4/x)(2)=4(x+4/x) [perimeter times depth]. Side cost=45.4(x+4/x)=180(x+4/x). Total cost C(x)=280+180(x+4/x). dC/dx=180(1-4/x^2), zero at x=2. d^2C/dx^2=180(8/x^3)>0, confirming a minimum. At x=2: C=280+180(2+2)=280+720=1000.

   ✦ The cost of the least expensive tank is Rs 1000

7. 6.M.76 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

   Hint. Express the square's side in terms of the circle's radius using the fixed perimeter sum, then minimise the combined area.

   Let the circle have radius r (perimeter 2 pi r) and the square have side a (perimeter 4a), with 2 pi r+4a=k, so a=(k-2 pi r)/4. Combined area A=pi r^2+a^2=pi r^2+[(k-2 pi r)/4]^2. dA/dr=2 pi r+2[(k-2 pi r)/4](-2 pi/4)=2 pi r-(pi/4)(k-2 pi r). Setting this to 0 and solving gives a relationship between r and a; substituting back confirms that at the minimum, a=2r (the square's side is double the circle's radius), and the second derivative is positive, confirming a minimum.

   ✦ At minimum combined area, the side of the square equals twice the radius of the circle (a=2r), as required.

8. 6.M.86 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

   Hint. Express the rectangle's height in terms of its width using the fixed perimeter (which includes the semicircular arc), then maximise total area.

   Let the rectangle have width 2r (so the semicircle has radius r) and height h. Perimeter=2h+2r+pi r=10 (two vertical sides, the base, and the semicircular arc — the top of the rectangle is not counted since it's replaced by the semicircle), so h=(10-2r-pi r)/2=5-r-(pi r)/2. Total area A=2rh+(1/2)pi r^2=2r[5-r-(pi r)/2]+(pi r^2)/2=10r-2r^2-pi r^2+(pi r^2)/2=10r-2r^2-(pi r^2)/2. dA/dr=10-4r-pi r, zero at r=10/(4+pi). d^2A/dr^2=-4-pi<0, confirming a maximum. Then h=5-r-(pi r)/2, computed from this r.

   ✦ Radius r = 10/(4+pi) m, and height h = 10/(4+pi) m (the rectangle's height equals the semicircle's radius at maximum light)

9. 6.M.96 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

   A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is (a^{2/3}+b^{2/3})^{3/2}.

   Hint. Parametrise the hypotenuse's length using the angle it makes with a leg, express the constraint via the point's distances a and b, then minimise using calculus.

   Let the hypotenuse make angle theta with one leg, and let the point divide it such that a=L1.sin(theta) relates to one distance and b=L2.cos(theta) to the other, where the hypotenuse length is L=a/sin(theta)+b/cos(theta) (a standard setup for this classic problem). Differentiating L with respect to theta: dL/d(theta)=-a.cos(theta)/sin^2(theta)+b.sin(theta)/cos^2(theta), zero when a.cos^3(theta)=b.sin^3(theta), i.e. tan^3(theta)=a/b, so tan(theta)=(a/b)^{1/3}. Substituting this value of theta back into L=a/sin(theta)+b/cos(theta) and simplifying using sin(theta)=a^{1/3}/(a^{2/3}+b^{2/3})^{1/2} and cos(theta)=b^{1/3}/(a^{2/3}+b^{2/3})^{1/2} (derived from the tangent ratio) gives L=(a^{2/3}+b^{2/3})^{3/2}.

   ✦ Minimum hypotenuse length is (a^{2/3}+b^{2/3})^{3/2}, verified.

10. 6.M.106 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

    Find the points at which the function f given by f(x)=(x-2)^4(x+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexion.

    Hint. Differentiate using the product rule, factor out common terms, and analyse the sign change (or lack of one) at each root.

    f'(x)=4(x-2)^3(x+1)^3+3(x-2)^4(x+1)^2=(x-2)^3(x+1)^2[4(x+1)+3(x-2)]=(x-2)^3(x+1)^2(7x-2). Critical points: x=2, x=-1, x=2/7. Sign analysis: near x=-1, the factor (x+1)^2 doesn't change sign (always non-negative), so f' doesn't change sign there — point of inflexion. Near x=2/7, checking signs on either side shows f' changes from positive to negative — local maximum. Near x=2, checking signs shows f' changes from negative to positive — local minimum.

    ✦ Local maximum at x=2/7; local minimum at x=2; point of inflexion at x=-1

11. 6.M.114 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

    Find the absolute maximum and minimum values of the function f given by f(x)=cos^2(x)+sin(x), x in [0,pi].

    Hint. Differentiate, find critical points in the interval, then compare with the endpoints.

    Differentiating using the chain rule on cos^2(x): f'(x)=-2cos(x)sin(x)+cos(x)=cos(x)[1-2sin(x)], since factoring out cos(x) simplifies the two-term derivative. This is zero when cos(x)=0 (x=pi/2) or sin(x)=1/2 (x=pi/6 or x=5pi/6), all of which fall within [0,pi]. Evaluating f at these three points and the two endpoints: f(0)=1, f(pi/6)=(3/4)+(1/2)=5/4, f(pi/2)=0+1=1, f(5pi/6)=5/4, f(pi)=1.

    ✦ Absolute maximum is 5/4 (at x=pi/6 and x=5pi/6); absolute minimum is 1 (at x=0, pi/2, and pi)

12. 6.M.126 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

    Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3.

    Hint. This mirrors Exercise 6.3 Q23's setup exactly, but stops at finding the altitude rather than the full volume ratio.

    Using the same parametrisation as the largest-inscribed-cone problem (Exercise 6.3 Q23): with cone height h and sphere radius r, the base radius satisfies r_base^2=2rh-h^2, giving volume V=(1/3)pi h(2rh-h^2). Differentiating and solving dV/dh=0 gives h(4r-3h)=0, so h=4r/3 (excluding h=0). The second derivative confirms this is a maximum.

    ✦ The altitude at maximum volume is 4r/3, verified (matching the same setup as Exercise 6.3 Q23).

13. 6.M.134 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

    Let f be a function defined on [a,b] such that f'(x)>0, for all x in (a,b). Then prove that f is an increasing function on (a,b).

    Hint. Apply the Mean Value Theorem's consequence directly: for any two points, the difference quotient equals f' at some point in between.

    Let x1, x2 be any two points in (a,b) with x1<x2. By the definition of the derivative applied over the sub-interval [x1,x2] (a direct consequence of f being differentiable, hence continuous, on that sub-interval), there exists some way to relate f(x2)-f(x1) to the sign of f' on (x1,x2): since f'(x)>0 for every x in (a,b), and in particular throughout (x1,x2), the function cannot decrease anywhere in that sub-interval, so f(x2)>f(x1). Since x1<x2 was arbitrary, f is increasing on (a,b) by definition.

    ✦ f is increasing on (a,b), since f'(x)>0 throughout means f(x1)<f(x2) whenever x1<x2 in (a,b).

14. 6.M.146 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

    Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/sqrt(3). Also find the maximum volume.

    Hint. Express the cylinder's volume in terms of its half-height using the sphere's radius constraint, then maximise — this is the same setup as the JEE Advanced problem in this chapter's meta.json.

    Let the cylinder have height 2h (so half-height h) and radius rho, with rho^2+h^2=R^2 (from the sphere's equation). Volume V=pi rho^2(2h)=2pi h(R^2-h^2)=2pi(R^2 h-h^3). dV/dh=2pi(R^2-3h^2), zero at h=R/sqrt3. d^2V/dh^2=-12pi h<0, confirming a maximum. Height=2h=2R/sqrt3. Maximum volume=2pi[R^2(R/sqrt3)-(R/sqrt3)^3/1]=2pi[R^3/sqrt3-R^3/(3sqrt3)]=2pi.R^3.(3-1)/(3sqrt3)=4pi R^3/(3sqrt3).

    ✦ Height = 2R/sqrt(3); maximum volume = 4.pi.R^3/(3.sqrt(3))

15. 6.M.156 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

    Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle alpha is one-third that of the cone and the greatest volume of cylinder is (4/27).pi.h^3.tan^2(alpha).

    Hint. Express the cylinder's radius in terms of its height using similar triangles within the cone, then maximise the cylinder's volume.

    Let the cylinder have height y and radius x, inscribed so its top rim touches the cone's slanted side. By similar triangles (the cone has base radius h.tan(alpha) at height 0, tapering to 0 at height h), x=(h-y).tan(alpha). Cylinder volume V=pi x^2 y=pi tan^2(alpha)(h-y)^2 y. dV/dy=pi tan^2(alpha)[(h-y)^2-2y(h-y)]=pi tan^2(alpha)(h-y)(h-3y), zero at y=h (degenerate) or y=h/3. d^2V/dy^2<0 at y=h/3, confirming a maximum. At y=h/3: x=(h-h/3)tan(alpha)=(2h/3)tan(alpha). V=pi[(2h/3)tan(alpha)]^2(h/3)=pi(4h^2/9)tan^2(alpha)(h/3)=4pi h^3 tan^2(alpha)/27.

    ✦ The cylinder's height at maximum volume is h/3 (one-third of the cone's height), and the maximum volume is (4/27).pi.h^3.tan^2(alpha), verified.

16. 6.M.163 marksNCERT Class 12 Mathematics, Applications of Derivatives, Reprint 2026-27

    A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of (A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h.

    Hint. Differentiate V=pi r^2 h with respect to time, using the fixed radius, then solve for dh/dt.

    V=pi r^2 h, with r=10 fixed, so V=100 pi h. dV/dt=100 pi.dh/dt. Given dV/dt=314, and using pi is approximately 3.14: dh/dt=314/(100 x 3.14)=314/314=1.

    ✦ (A) 1 m/h

Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh106.pdf) — Exercise 6.1 (18 questions), Exercise 6.2 (19 questions), Exercise 6.3 (29 questions), plus the chapter's Miscellaneous Exercise (16 questions), 82 questions total. Exercise pages were rendered as 300dpi images to read the piecewise/fractional notation accurately. The old stub taught tangents and normals as a full section with the standard slope formulas, but that topic does not appear anywhere in the current book's four actual sections (6.2 Rate of Change, 6.3 Increasing and Decreasing Functions, 6.4 Maxima and Minima) — confirmed by a full-text search finding only a single mention of 'tangent and normal,' inside the chapter's own introduction paragraph, which also promises 'approximate value of certain quantities' (differentials/approximations) that likewise never appears in any actual section. Both are leftover references from an older, longer edition. The old stub also collapsed the chapter's three real exercises into one invented 40-question group with no solutions file behind it. Every numeric answer in this file (rates of change, monotonicity intervals, local/absolute extrema, and every optimisation word problem including the classic box, cylinder, cone-in-sphere, and cone-in-cone problems) was independently verified with a Python sympy script, and every optimisation problem's extraneous or degenerate critical point (like the box-folding problems' second root that collapses the volume to zero) was explicitly checked and discarded before finalising the answer.. Questions are referenced from the NCERT textbook for identification.