Class 12 Maths NCERT Solutions Chapter 6 Ex 6.3 – Maxima and Minima | Boundless Maths
← Exercise 6.2 Exercise 6.3 Miscellaneous →
Ex 6.3 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 6 Ex 6.3 – Maxima and Minima

This Class 12 Maths NCERT Solutions Chapter 6 Ex 6.3 page covers all 29 questions, solved step-by-step — local and absolute maxima/minima using the first and second derivative tests, plus the classic Application of Derivatives optimisation problems (maximum volume, minimum surface area, maximum area) that show up almost every year in the board exam.

29Questions
Easy–HardDifficulty Mix
2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 6 Ex 6.3 — All 29 Questions

Tip: Q1–Q12 are direct max/min questions (definitions, local vs. absolute). Q13–Q26 are word-problem optimisations (two numbers, boxes, cones, cylinders). Q27–Q29 are MCQs.
1

Find the maximum and minimum values, if any, of:
(i) f(x)=(2x-1)^2+3
(ii) f(x)=9x^2+12x+2
(iii) f(x)=-(x-1)^2+10
(iv) g(x)=x^3+1

Medium +
Solution
(i) f(x) = (2x−1)² + 3

Critical point: f'(x)=0\Rightarrow8x-4=0\Rightarrow x=\tfrac12.

Second derivative: f''(x)=8>0, so x=\tfrac12 is a point of local minimum.

f\left(\tfrac12\right)=(2\cdot\tfrac12-1)^2+3=0+3=3.

Since f(x)\to\infty as x\to\pm\infty, there is no maximum value.

Minimum value =3 at x=\tfrac12; no maximum (unbounded above).
(ii) f(x) = 9x² + 12x + 2

Differentiate: f'(x)=18x+12.

Critical point: f'(x)=0\Rightarrow18x+12=0\Rightarrow x=-\tfrac23.

Second derivative: f''(x)=18>0, so x=-\tfrac23 is a point of local minimum.

f\left(-\tfrac23\right)=9\left(\tfrac49\right)+12\left(-\tfrac23\right)+2=4-8+2=-2.

Since f(x)\to\infty as x\to\pm\infty, there is no maximum value.

Minimum value =-2 at x=-\tfrac23; no maximum.
(iii) f(x) = −(x−1)² + 10

Differentiate: f'(x)=-2(x-1)=-2x+2.

Critical point: f'(x)=0\Rightarrow-2x+2=0\Rightarrow x=1.

Second derivative: f''(x)=-2<0, so x=1 is a point of local maximum.

f(1)=-(1-1)^2+10=0+10=10.

Since f(x)\to-\infty as x\to\pm\infty, there is no minimum value.

Maximum value =10 at x=1; no minimum (unbounded below).
(iv) g(x) = x³ + 1

Differentiate: g'(x)=3x^2.

Critical point: g'(x)=0\Rightarrow3x^2=0\Rightarrow x=0.

Second derivative: g''(x)=6x, so g''(0)=0 — the second derivative test is inconclusive here, since it can only classify a critical point when g''\neq0.

Fallback: examine the sign of g'(x)=3x^2 on either side of x=0.

Since x^2\geq0 for every real x, g'(x)\geq0 on both sides — it never changes sign from + to - or vice versa, so x=0 is neither a maximum nor a minimum (it is a point of inflexion).

g(x)=x^3+1 has no maximum and no minimum value (critical point at x=0 is a point of inflexion, not an extremum).
2

Find the maximum and minimum values, if any, of:
(i) f(x)=|x+2|-1
(ii) g(x)=-|x+1|+3
(iii) h(x)=\sin(2x)+5
(iv) f(x)=|\sin4x+3|
(v) h(x)=x+1,\ x\in(-1,1)

Medium +
Solution
(i) f(x) = |x+2| − 1

Since |x+2|\geq0 for all x\in\mathbb{R}, with equality when x=-2, we have f(x)=|x+2|-1\geq-1 for all x, with equality at x=-2.

Also, as x\to\pm\infty, |x+2|\to\infty, so f(x)\to\infty.

Minimum value =-1 at x=-2; no maximum.
(ii) g(x) = −|x+1| + 3

Since |x+1|\geq0 for all x\in\mathbb{R}, we have -|x+1|\leq0, so g(x)=-|x+1|+3\leq3 for all x, with equality at x=-1.

Maximum value =3 at x=-1; no minimum.
(iii) h(x) = sin(2x) + 5

Since -1\leq\sin2x\leq1 for all x, adding 5 throughout gives 4\leq h(x)\leq6.

Maximum =6, minimum =4.
(iv) f(x) = |sin4x + 3|

Since -1\leq\sin4x\leq1, adding 3 throughout gives 2\leq\sin4x+3\leq4.

As \sin4x+3 is always positive, |\sin4x+3|=\sin4x+3, so 2\leq f(x)\leq4.

Maximum =4, minimum =2.
(v) h(x) = x + 1, x ∈ (−1,1)

The domain is the open interval (-1,1), so the endpoints x=-1 and x=1 are never attained.

As x\to1^-, h(x)\to2 but this value is never reached; as x\to-1^+, h(x)\to0 but this value is never reached either.

h has no maximum and no minimum value.
3

Find the local maxima/minima (and values) of:
(i) x^2
(ii) x^3-3x
(iii) \sin x+\cos x,\ 0<x<\tfrac{\pi}{2}
(iv) \sin x-\cos x,\ 0<x<2\pi
(v) x^3-6x^2+9x+15
(vi) \tfrac{x}{2}+\tfrac{2}{x},\ x>0
(vii) \tfrac{1}{x^2+2}
(viii) x\sqrt{1-x},\ 0<x<1

Hard +
Solution
(i) f(x) = x²

f'(x)=2x=0 \Rightarrow x=0. f''(x)=2>0 — confirms a local minimum.

Local minimum =0 at x=0 (no local maxima).
(ii) g(x) = x³ − 3x

g'(x)=3x^2-3=0 \Rightarrow x=\pm1. g''(x)=6x.

Local max =2 at x=-1; local min =-2 at x=1.
(iii) h(x) = sin x + cos x on (0, π/2)

h'(x)=\cos x-\sin x=0 \Rightarrow x=\tfrac{\pi}{4}.

h''\left(\tfrac{\pi}{4}\right)=-\sqrt2<0.

Local maximum =\sqrt2 at x=\tfrac{\pi}{4}.

f'(x)=\cos x+\sin x=0 \Rightarrow x=\tfrac{3\pi}{4},\ \tfrac{7\pi}{4}.

f''(x)=-\sin x+\cos x. At x=\tfrac{3\pi}{4}: f''=-\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2}=-\sqrt2<0 — local maximum. At x=\tfrac{7\pi}{4}: f''=\tfrac{1}{\sqrt2}+\tfrac{1}{\sqrt2}=\sqrt2>0 — local minimum.

Local max =\sqrt2 at x=\tfrac{3\pi}{4}; local min =-\sqrt2 at x=\tfrac{7\pi}{4}.
(v) f(x) = x³ − 6x² + 9x + 15

f'(x)=3(x-1)(x-3)=0 \Rightarrow x=1,3. f''(x)=6x-12.

Local max =19 at x=1; local min =15 at x=3.
(vi) g(x) = x/2 + 2/x, x > 0

g'(x)=\tfrac12-\tfrac{2}{x^2}=0 \Rightarrow x=2 (taking the positive root). g''(x)=\tfrac{4}{x^3}>0.

Local minimum =2 at x=2.
(vii) g(x) = 1/(x² + 2)

g'(x)=\dfrac{-2x}{(x^2+2)^2}=0 \Rightarrow x=0.

For x<0, g'>0; for x>0, g'<0 — sign changes +\to-.

Local maximum =\tfrac12 at x=0.
(viii) f(x) = x√(1−x), 0 < x < 1

f'(x)=\dfrac{2-3x}{2\sqrt{1-x}}=0 \Rightarrow x=\tfrac23. Sign changes +\to- around x=\tfrac23.

Local maximum =\dfrac{2\sqrt3}{9} at x=\tfrac23.
4

Prove that the following functions do not have maxima or minima:
(i) f(x)=e^x
(ii) g(x)=\log x
(iii) h(x)=x^3+x^2+x+1

Medium +
Solution
(i) f(x) = eˣ

Differentiate: f'(x)=e^x.

Since e^x>0 for every real x, f'(x) is never zero, so f has no critical points.

f has no maxima or minima.
(ii) g(x) = log x

Differentiate: g'(x)=\dfrac1x.

Since \dfrac1x>0 for every x>0 (the domain of \log x), g'(x) is never zero, so g has no critical points.

g has no maxima or minima.
(iii) h(x) = x³ + x² + x + 1

Differentiate: h'(x)=3x^2+2x+1.

For the quadratic 3x^2+2x+1 (with a=3,\ b=2,\ c=1), the discriminant is b^2-4ac=(2)^2-4(3)(1)=4-12=-8<0.

Since the discriminant is negative, 3x^2+2x+1=0 has no real roots; and since the leading coefficient a=3>0, the parabola opens upward and never crosses the x-axis, so h'(x)>0 for every real x.

h'(x) is never zero, so h has no critical points, hence no maxima or minima.
5

Find the absolute maximum and minimum values of:
(i) f(x)=x^3,\ x\in[-2,2]
(ii) f(x)=\sin x+\cos x,\ x\in[0,\pi]
(iii) f(x)=4x-\tfrac12x^2,\ x\in\left[-2,\tfrac92\right]
(iv) f(x)=(x-1)^2+3,\ x\in[-3,1]

Medium +
Solution
(i) f(x) = x³, x ∈ [−2,2]

Critical point: f'(x)=3x^2=0 \Rightarrow x=0.

Evaluate f at the critical point and both endpoints: f(-2)=-8,\ f(0)=0,\ f(2)=8.

Absolute max =8 at x=2; absolute min =-8 at x=-2.
(ii) f(x) = sin x + cos x, x ∈ [0,π]

Critical point: f'(x)=\cos x-\sin x=0\Rightarrow x=\tfrac{\pi}{4}.

Evaluate f at the critical point and both endpoints: f(0)=1,\ f\left(\tfrac{\pi}{4}\right)=\sqrt2,\ f(\pi)=-1.

Absolute max =\sqrt2 at x=\tfrac{\pi}{4}; absolute min =-1 at x=\pi.
(iii) f(x) = 4x − ½x², x ∈ [−2, 9/2]

Critical point: f'(x)=4-x=0 \Rightarrow x=4.

Evaluate f at the critical point and both endpoints: f(-2)=-10,\ f(4)=8,\ f\left(\tfrac92\right)=7.875.

Absolute max =8 at x=4; absolute min =-10 at x=-2.
(iv) f(x) = (x−1)² + 3, x ∈ [−3,1]

Critical point: f'(x)=2(x-1)=0\Rightarrow x=1, which coincides with the right endpoint of [-3,1].

Evaluate f at both endpoints (the critical point is already one of them): f(-3)=19,\ f(1)=3.

Absolute max =19 at x=-3; absolute min =3 at x=1.
6

Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 - 72x - 18x^2.

Easy +
Solution

p'(x)=-72-36x=0 \Rightarrow x=-2.

p''(x)=-36<0 — maximum.

p(-2) = 41-72(-2)-18(4) = 41+144-72 = 113.

Maximum profit = 113.
7

Find both the maximum value and the minimum value of 3x^4-8x^3+12x^2-48x+25 on the interval [0,3].

Medium +
Solution

f'(x)=12x^3-24x^2+24x-48=12(x-2)(x^2+2). Since x^2+2>0 always, the only critical point is x=2.

Evaluate f at the critical point and both endpoints: f(0)=25,\ f(2)=-39,\ f(3)=16.

Maximum value =25 at x=0; minimum value =-39 at x=2.
8

At what points in the interval [0,2\pi] does the function \sin 2x attain its maximum value?

Easy +
Solution

Let f(x)=\sin2x. Differentiate: f'(x)=2\cos2x.

Critical points: f'(x)=0\Rightarrow\cos2x=0\Rightarrow2x=\tfrac{\pi}{2}+k\pi\Rightarrow x=\tfrac{\pi}{4}+\tfrac{k\pi}{2}. Within [0,2\pi], this gives x=\tfrac{\pi}{4},\tfrac{3\pi}{4},\tfrac{5\pi}{4},\tfrac{7\pi}{4}.

Second derivative: f''(x)=-4\sin2x. At x=\tfrac{\pi}{4}: f''=-4\sin\tfrac{\pi}{2}=-4<0 — maximum (f=1). At x=\tfrac{3\pi}{4}: f''=-4\sin\tfrac{3\pi}{2}=4>0 — minimum. At x=\tfrac{5\pi}{4}: f''=-4\sin\tfrac{5\pi}{2}=-4<0 — maximum (f=1). At x=\tfrac{7\pi}{4}: f''=-4\sin\tfrac{7\pi}{2}=4>0 — minimum.

Maximum value 1 is attained at x=\tfrac{\pi}{4} and x=\tfrac{5\pi}{4}.
9

What is the maximum value of the function \sin x+\cos x?

Easy +
Solution

Let f(x)=\sin x+\cos x. Differentiate: f'(x)=\cos x-\sin x.

Critical point: f'(x)=0\Rightarrow\cos x=\sin x\Rightarrow\tan x=1\Rightarrow x=\tfrac{\pi}{4} (taking the value in [0,2\pi) where f is largest).

Second derivative: f''(x)=-\sin x-\cos x, so f''\left(\tfrac{\pi}{4}\right)=-\tfrac{1}{\sqrt2}-\tfrac{1}{\sqrt2}=-\sqrt2<0 — this confirms x=\tfrac{\pi}{4} is a local maximum.

f\left(\tfrac{\pi}{4}\right)=\tfrac{1}{\sqrt2}+\tfrac{1}{\sqrt2}=\sqrt2. (This matches writing \sin x+\cos x=\sqrt2\sin\left(x+\tfrac{\pi}{4}\right), whose maximum value is directly \sqrt2.)

Maximum value = \sqrt2.
10

Find the maximum value of 2x^3-24x+107 in the interval [1,3]. Find the maximum value of the same function in [-3,-1].

Medium +
Solution

f'(x)=6x^2-24=6(x-2)(x+2). Critical points: x=\pm2.

Evaluate f at the critical point and both endpoints in each interval. On [1,3]: f(1)=85,\ f(2)=75,\ f(3)=89. On [-3,-1]: f(-3)=125,\ f(-2)=139,\ f(-1)=129.

On [1,3]: maximum =89 at x=3. On [-3,-1]: maximum =139 at x=-2.
11

It is given that at x=1, the function x^4-62x^2+ax+9 attains its maximum value on [0,2]. Find the value of a.

Easy +
Solution

Since x=1 is an interior point of [0,2] and gives the maximum, f'(1)=0 (a necessary condition for any interior extremum).

f'(x)=4x^3-124x+a. f'(1)=4-124+a=0 \Rightarrow a=120.

Check for consistency: with a=120, f''(x)=12x^2-124, so f''(1)=12-124=-112<0 — this confirms x=1 is indeed a local maximum, consistent with what was given.

Answer: a=120.
12

Find the maximum and minimum values of x+\sin 2x on [0,2\pi].

Hard +
Solution

f'(x)=1+2\cos2x=0 \Rightarrow \cos2x=-\tfrac12, giving critical points x=\tfrac{\pi}{3},\tfrac{2\pi}{3},\tfrac{4\pi}{3},\tfrac{5\pi}{3} within [0,2\pi].

Evaluate f at each critical point and both endpoints:

f(0)=0

f\left(\tfrac{\pi}{3}\right)=\tfrac{\pi}{3}+\tfrac{\sqrt3}{2}

f\left(\tfrac{2\pi}{3}\right)=\tfrac{2\pi}{3}-\tfrac{\sqrt3}{2}

f\left(\tfrac{4\pi}{3}\right)=\tfrac{4\pi}{3}+\tfrac{\sqrt3}{2}

f\left(\tfrac{5\pi}{3}\right)=\tfrac{5\pi}{3}-\tfrac{\sqrt3}{2}

f(2\pi)=2\pi

Comparing all six values, the smallest is f(0)=0 and the largest is f(2\pi)=2\pi.

Maximum value =2\pi at x=2\pi; minimum value =0 at x=0.
13

Find two numbers whose sum is 24 and whose product is as large as possible.

Easy +
Solution

Let the numbers be x and 24-x. P(x)=x(24-x)=24x-x^2.

P'(x)=24-2x=0 \Rightarrow x=12.

P''(x)=-2<0 — maximum.

The two numbers are 12 and 12.
14

Find two positive numbers x and y such that x+y=60 and xy^3 is maximum.

Medium +
Solution

Let y=60-x. f(x)=x(60-x)^3.

f'(x)=(60-x)^2(60-4x)=0 \Rightarrow x=15 (rejecting x=60, which gives y=0, invalid since y must be positive).

Second derivative: f''(x)=-12(60-x)(30-x). At x=15: f''(15)=-12(45)(15)=-8100<0 — this confirms x=15 is a point of local maximum.

Answer: x=15,\ y=45.
15

Find two positive numbers x and y such that their sum is 35 and the product x^2y^5 is a maximum.

Medium +
Solution

Let y=35-x. f(x)=x^2(35-x)^5.

f'(x)=7x(35-x)^4(10-x)=0 \Rightarrow x=10 (rejecting x=0,35, both invalid as they force one number to be zero).

Since (35-x)^4\geq0 and 7x>0 for 0<x<35, the sign of f'(x) is determined entirely by the factor (10-x): positive for x<10 and negative for x>10.

Since f' changes sign from + to - at x=10, this confirms a local maximum there.

Answer: x=10,\ y=25.
16

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Medium +
Solution

Let y=16-x. f(x)=x^3+(16-x)^3.

f'(x)=3x^2-3(16-x)^2=96(x-8)=0 \Rightarrow x=8. f''(x)=96>0 — minimum.

The two numbers are 8 and 8.
17

A square piece of tin of side 18 cm is made into an open box by cutting a square from each corner and folding up the flaps. What side of square should be cut off so the volume is maximum?

Medium +
Solution

Let the cut-out side be x. Volume V(x)=x(18-2x)^2, 0<x<9.

V'(x)=12(x-3)(x-9)=0 \Rightarrow x=3 (rejecting x=9, a degenerate box).

V''(3)=-72<0 — maximum.

Cut squares of side 3\text{ cm} for maximum volume (432\text{ cm}^3).
18

A rectangular sheet of tin 45 cm by 24 cm is made into an open box by cutting squares from each corner. What side should be cut off for maximum volume?

Medium +
Solution

V(x)=x(45-2x)(24-2x), 0<x<12.

V'(x)=12(x-5)(x-18)=0 \Rightarrow x=5 (rejecting x=18, outside domain).

V''(5)=-156<0 — maximum.

Cut squares of side 5\text{ cm} for maximum volume.
19

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Hard +
Solution

Let the circle have radius r (fixed). For a rectangle with sides x,y inscribed in it, the diagonal equals the diameter: x^2+y^2=4r^2, so y=\sqrt{4r^2-x^2}.

Area A(x)=x\sqrt{4r^2-x^2}. A'(x)=\dfrac{4r^2-2x^2}{\sqrt{4r^2-x^2}}=0 \Rightarrow x^2=2r^2 \Rightarrow x=r\sqrt2.

The denominator \sqrt{4r^2-x^2} is always positive on the valid domain 0<x<2r, so the sign of A'(x) is determined by the numerator 4r^2-2x^2: positive for x<r\sqrt2 and negative for x>r\sqrt2.

Since A' changes sign from + to - at x=r\sqrt2, this confirms a local maximum there.

Then y=\sqrt{4r^2-2r^2}=r\sqrt2=x.

Since x=y, the maximum-area rectangle is a square (proved).
20

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Hard +
Solution

Fixed surface S=2\pi r^2+2\pi rh \Rightarrow h=\dfrac{S-2\pi r^2}{2\pi r}.

V(r)=\pi r^2 h = \dfrac{S}{2}r-\pi r^3. V'(r)=\dfrac{S}{2}-3\pi r^2=0 \Rightarrow r^2=\dfrac{S}{6\pi}.

Substituting back gives h=\dfrac{S}{3\pi r}=2r.

V''(r)=-6\pi r<0 — maximum.

Height h=2r, i.e. equal to the diameter of the base (proved).
21

Of all the closed cylindrical cans of a given volume of 100\text{ cm}^3, find the dimensions of the can with minimum surface area.

Hard +
Solution

V=\pi r^2h=100 \Rightarrow h=\dfrac{100}{\pi r^2}.

S(r)=2\pi r^2+\dfrac{200}{r}. S'(r)=4\pi r-\dfrac{200}{r^2}=0 \Rightarrow r^3=\dfrac{50}{\pi}.

Substituting back shows h=2r.

Second derivative: S''(r)=4\pi+\dfrac{400}{r^3}. Since r>0, both terms are positive, so S''(r)>0 always — this confirms a minimum.

Radius r=\left(\dfrac{50}{\pi}\right)^{1/3}\text{cm}, height h=2r.
22

A wire of length 28 m is cut into two pieces — one bent into a square, the other into a circle. What lengths minimise the combined area?

Hard +
Solution

Let length x form the square (side \tfrac{x}{4}) and 28-x form the circle (radius \tfrac{28-x}{2\pi}).

A(x)=\dfrac{x^2}{16}+\dfrac{(28-x)^2}{4\pi}. A'(x)=\dfrac{x}{8}-\dfrac{28-x}{2\pi}=0 \Rightarrow x=\dfrac{112}{\pi+4}.

Second derivative: A''(x)=\dfrac18+\dfrac{1}{2\pi}, a positive constant — this confirms a minimum.

Square piece =\dfrac{112}{\pi+4}\text{ m}; circle piece =\dfrac{28\pi}{\pi+4}\text{ m}.
23

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \tfrac{8}{27} of the volume of the sphere.

Hard +
Solution

For a cone of height h and base radius r inscribed in a sphere of radius R: r^2=h(2R-h).

V(h)=\tfrac13\pi h^2(2R-h).

V'(h)=\tfrac13\pi h(4R-3h)=0 \Rightarrow h=\tfrac{4R}{3}.

V''\left(\tfrac{4R}{3}\right)<0 — maximum.

V_{max}=\tfrac{32}{81}\pi R^3, while the sphere's volume is \tfrac43\pi R^3.

Ratio =\dfrac{32/81}{4/3}=\dfrac{8}{27} (proved).
24

Show that the right circular cone of least curved surface and given volume has altitude equal to \sqrt2 times the radius of the base.

Hard +
Solution

Fixed volume V=\tfrac13\pi r^2h \Rightarrow h=\dfrac{3V}{\pi r^2}. Curved surface area squared: S^2=\pi^2r^4+\pi^2r^2h^2.

Substituting h gives S^2 as a function of r alone: f(r)=\pi^2r^4+\dfrac{9V^2}{r^2} (proportional to S^2).

Differentiating: f'(r)=4\pi^2r^3-\dfrac{18V^2}{r^3}=0\Rightarrow r^6=\dfrac{9V^2}{2\pi^2}, which is exactly equivalent to the condition h^2=2r^2.

Second derivative: f''(r)=12\pi^2r^2+\dfrac{54V^2}{r^4}. Since r>0, both terms are positive, so f''(r)>0 always — this confirms a minimum of S^2 (and hence of S, since S>0).

So h=\sqrt2\,r at the minimum curved-surface configuration (proved).
25

Show that the semi-vertical angle of the cone of maximum volume, for a given slant height, is \tan^{-1}\sqrt2.

Hard +
Solution

With slant height l fixed and semi-vertical angle \alpha: r=l\sin\alpha, h=l\cos\alpha.

V(\alpha)=\tfrac{\pi l^3}{3}\sin^2\alpha\cos\alpha. \dfrac{dV}{d\alpha}=\tfrac{\pi l^3}{3}\sin\alpha\left(2\cos^2\alpha-\sin^2\alpha\right)=0.

This gives \tan^2\alpha=2 \Rightarrow \alpha=\tan^{-1}\sqrt2.

Using \cos^2\alpha=1-\sin^2\alpha, the derivative can be rewritten as \dfrac{dV}{d\alpha}=\tfrac{\pi l^3}{3}\sin\alpha(2-3\sin^2\alpha).

For small \alpha (near 0), \sin^2\alpha is small, so 2-3\sin^2\alpha>0 and \dfrac{dV}{d\alpha}>0 (V increasing); for \alpha near \tfrac{\pi}{2}, \sin^2\alpha\to1, so 2-3\sin^2\alpha<0 and \dfrac{dV}{d\alpha}<0 (V decreasing).

Since the derivative changes sign from + to - at the critical angle, this confirms a maximum.

Proved: semi-vertical angle =\tan^{-1}\sqrt2.
26

Show that the semi-vertical angle of a right circular cone of given surface area and maximum volume is \sin^{-1}\left(\tfrac13\right).

Hard +
Solution

Fixed total surface S=\pi r^2+\pi rl, so l=\dfrac{S}{\pi r}-r.

Maximising V^2 in terms of r leads to the critical-point condition r^2=\dfrac{S}{4\pi}, which gives l=3r.

Checking the sign of the derivative of V^2 with respect to r on either side of this critical value shows it is positive for r^2<\tfrac{S}{4\pi} and negative for r^2>\tfrac{S}{4\pi} — confirming this critical point gives a maximum of V, as required.

Then \sin\alpha=\dfrac{r}{l}=\dfrac{r}{3r}=\dfrac13.

Proved: semi-vertical angle =\sin^{-1}\left(\tfrac13\right).
27

MCQ. The point on the curve x^2=2y nearest to the point (0,5) is:   (A) (2\sqrt2,4)   (B) (2\sqrt2,0)   (C) (0,0)   (D) (2,2)

Medium +
Solution

A point on the curve is \left(x,\tfrac{x^2}{2}\right). Let u=x^2; distance² D=u+\left(\tfrac{u}{2}-5\right)^2 = \tfrac{u^2}{4}-4u+25.

\dfrac{dD}{du}=\tfrac{u}{2}-4=0 \Rightarrow u=8 \Rightarrow x=\pm2\sqrt2, giving y=4.

Second derivative: \dfrac{d^2D}{du^2}=\tfrac12>0 — this confirms u=8 is a minimum of the distance-squared function, so this is indeed the nearest point.

Answer: (A) (2\sqrt2,4)
28

MCQ. For all real x, the minimum value of \dfrac{1-x+x^2}{1+x+x^2} is:   (A) 0   (B) 1   (C) 3   (D) \tfrac13

Hard +
Solution

Let y equal the expression. Cross-multiplying and rearranging as a quadratic in x: x^2(y-1)+x(y+1)+(y-1)=0.

For real x, the discriminant must be \geq0: (y+1)^2-4(y-1)^2\geq0 \Rightarrow 3y^2-10y+3\leq0.

Solving gives y\in\left[\tfrac13,3\right].

Answer: (D) \tfrac13
29

MCQ. The maximum value of [x(x-1)+1]^{1/3}, 0\leq x\leq1 is:   (A) \left(\tfrac13\right)^{1/3}   (B) \tfrac12   (C) 1   (D) 0

Medium +
Solution

Let g(x)=x^2-x+1 (the expression inside the cube root). Since cube root is increasing, maximising g maximises the whole expression.

g'(x)=2x-1=0 \Rightarrow x=\tfrac12, which is a minimum of g (since g''>0). So the maximum of g on [0,1] is at an endpoint: g(0)=1,\ g(1)=1.

Answer: (C) 1

Know exactly which chapters are costing you marks

1000+ solved CBSE PYQs, unlimited AI-generated practice for your weak areas, and a chapter-wise Performance Report — not just for this chapter, but your entire syllabus.

Explore it free →
Common Questions

FAQs — Class 12 Maths NCERT Solutions Chapter 6 Ex 6.3

How many questions are there in Exercise 6.3?

Exercise 6.3 has 29 questions (26 short-answer/proof questions plus 3 MCQs), covering local maxima/minima, absolute maxima/minima on closed intervals, and real-world optimisation problems.

What concept does Exercise 6.3 test?

It tests the first and second derivative tests for local extrema, the working rule for absolute maximum/minimum on a closed interval, and applying these to optimisation problems like maximum volume, minimum surface area, and maximum area.

Where can I find the official NCERT textbook for this exercise?

Exercise 6.3 is from Chapter 6, Application of Derivatives, in the NCERT Class 12 Mathematics textbook (Part I), published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the questions exactly as they appear there.

Carry the formulas with you

One-page printable formula cards for every Calculus chapter, including Application of Derivatives.

Get the Formula Cards →
← Exercise 6.2 ↑ Back to Chapter 6 overview Miscellaneous →
Applied Maths Coaching Gurugram | Boundless Maths
⭐⭐⭐⭐⭐ 5.0 · 100+ Google Reviews 📍 Sector 14, Gurugram · Online · Hybrid 🎓 500+ Students

Where Maths Becomes
Second Nature

Expert CBSE Coaching · Mindful Mathematics · Class 9 to 12

Mathematics is boundless — like the energy of the sun, the limitless sky and the ocean. At Boundless Maths, CBSE students from Class 9 to 12 go beyond the textbook in every class — real-life stories, cross-connected concepts and smarter calculation techniques — backed by Formula Decks, notes, question banks and assessments for comprehensive learning, making them board-exam ready. Ready to Explore?

Enquire About Your Batch 💬 WhatsApp
500+
Students
Taught
1000+
Online
Classes
100+
5-Star
Reviews
13+
Years
Experience
Foundation Years

Class 9 & 10 — Maths & Science

These are the years that shape everything. Build an unshakeable foundation in both Maths and Science — and enter Class 11 with confidence, not anxiety.

📗
Foundation

Class 9

Maths & Science · CBSE · Build the foundation that makes Class 10 manageable

➗
MathematicsReal Numbers · Polynomials · Geometry · Trigonometry · Statistics · Coordinate Geometry
🔬
ScienceMatter · Atoms & Molecules · Motion & Forces · Cell Biology · Natural Resources
What we do differentlyVedic Maths for fast calculation · conceptual linking to Class 10 & 11 · regular assessments · Growth Mindset practices woven in
Vedic Maths shortcutsFree formula cardsFree chapter notes
Explore Class 9 →
🏆
Board Exam Year

Class 10

Maths & Science · CBSE Boards · Strategy + concepts for the exam that counts

🏆 Board Exam Year — Strategy & Concepts Together
📊
MathematicsQuadratic Equations · AP · Trigonometry · Circles · Statistics & Probability · all board chapters
🧪
ScienceChemical Reactions · Life Processes · Light · Electricity · Heredity · Environment
Board exam focusPast 10 years' papers · marking scheme strategy · step-by-step answer writing · mindfulness for exam day
Past 10 years' papersMarking scheme strategyBoard exam mindfulness
Explore Class 10 →
What students gain from Class 9–10

Board exam readiness · Subject confidence · A foundation that makes Class 11–12 feel natural, not overwhelming

Enquire About Your Batch →
Higher Secondary

Class 11 & 12 — Maths & Applied Maths

Two different subjects. Both taught with the same depth, the same mindful approach, and the same commitment to making you genuinely good — not just exam-ready.

📈
Class 11 & 12★ Most Popular
Class 11–12 Applied Mathematics
CBSE · Commerce stream — the natural home for CA, CS, BBA, BCom & Economics students

Purpose-built Maths that connects every formula to real business, finance, and economic decisions. Finance, Statistics, Linear Programming, and Calculus taught with real-world context — and exclusive exam-focused resources you won't find elsewhere.

Finance & Business MathsStatisticsLinear ProgrammingCalculusNumber ApplicationsAlgebra
Included: 200+ FREE questions · Formula Deck ₹199 · 🤖 AI Q-Bank launch offer · YouTube tutorials · CBSE papers · Mindfulness sessions
Explore Applied Maths →FREE resources
🧮
Class 11 & 12
Class 11–12 Mathematics
CBSE · PCM / Science stream — the route to JEE, engineering & research

The full, rigorous CBSE Class 11–12 curriculum — Calculus, Algebra, Vectors, Coordinate Geometry, Probability — taught with in-depth conceptual understanding. Not rote preparation. We build the mathematical foundation that makes JEE-level thinking feel natural.

Calculus & LimitsVectorsProbabilityCoordinate GeometryAlgebraMatrices & Determinants
Included: Complete CBSE coverage · Vedic Maths shortcuts · Conceptual linking · Regular assessments · Board exam strategy · Mindfulness sessions
Explore Maths →PCM Students

🎓 Finishing Class 10? Not sure whether to pick Applied Maths or Core Maths for Class 11? Our guide covers every difference — topic by topic — with career connections to CA, BBA, BCom and MBA.

Read the Full Comparison →
Applied Maths Resources

Free Class 12 Applied Maths Resources — All 8 Units

Applied Mathematics is where every formula comes alive — Finance, Statistics, Linear Programming, Calculus applied to the real world. Access 200+ questions, CBSE papers and video tutorials free, or upgrade for deeper practice.

📖
All 8 Units
Complete coverage
✅
200+ Questions
With solutions · FREE
📄
Previous Years
CBSE board papers
🎥
Video Tutorials
YouTube channel
📐
Formula Decks
Class 11 & 12 · ₹199
📐
Formula Deck

Class 12 Applied Maths Formula Deck

Every formula from all 8 units — quick, device-free recap before every test. Your companion through board season and beyond.

₹199 📐 Get Formula Deck →
🎉 Launch Offer
🤖
⭐ Most Popular · AI Question Bank

Class 12 Applied Maths AI Question Bank

Adaptive AI practice — instant feedback, personalised weak-area targeting, and timed board exam simulation. 1000+ questions across all 8 units.

after 31 Jul One-time · No subscription

Want to try before buying? Browse 200+ free questions, CBSE papers & video tutorials →

Our Signature Programme

Mindful Mathematics GEM Programme

The GEM Framework — Growth Mindset · Empowerment · Mindfulness. A mathematical fluency programme for anyone — school student, working adult, or someone who just wants to be genuinely better at Maths.

Who is this for?
🎒
School Students

Class 8–10 students who want faster calculations, stronger conceptual clarity and real confidence in Maths.

🎓
Skipped Maths in 11–12?

Took Commerce or Humanities without Maths — now need it for CUET, IPMAT, CA Foundation or MBA prep.

⏰
Long Gap from Maths?

Haven't touched Maths in years — want a structured, non-intimidating way to recap, reconnect, and rebuild.

⚡
Want to Go Faster?

Master tables, mental maths, Vedic shortcuts and conceptual connections your school never fully covered.

Whether you're 12 or 42 — if Maths feels like a foreign language, this is where you start.

The Philosophy — Think · Explore · Grow

Mindful Mathematics weaves focused awareness into the way mathematics is taught and learned. Whether you're 13 or 42 — master calculations, recognise patterns, connect concepts, and develop a genuine love for thinking.

🧠

Think

Deep understanding, not rote. Observe patterns and build real connections.

🔍

Explore

Multiple strategies, real applications, and shortcuts that actually stick.

🌱

Grow

Confidence, speed, and a Growth Mindset that extends far beyond any exam.

The Programme — GEM Framework

A structured 3-level programme. Currently enrolling Level 1. Each level builds on the last — from building basic confidence to deep algebraic fluency and mindful focus.

G
Live Now
Level 1 — Growth Mindset

Master tables, fractions and calculation techniques through mental maths, Sudoku activities and Growth Mindset practices.

E
Coming
Level 2 — Empowerment

Pattern recognition, connecting mathematical concepts, shortcuts and multiple problem-solving strategies.

M
Coming
Level 3 — Mindfulness

Deep focus meets algebraic excellence. Mindfulness techniques for sustained attention and exam clarity.

Why Boundless Maths

Expert teaching.
A genuinely different approach.

A software engineer who became an educator — with 10 years of industry experience and 13 years in the classroom. Every resource, every tool, every class is built around one question: what does this student actually need?

Suchita Arora, Founder of Boundless Maths
Suchita Arora
Founder & Lead Educator · Boundless Maths
💻 Software Engineer · 10 yrs 🎓 CBSE Educator · 13 yrs 🤖 Built the AI Q-Bank

10 years as a software engineer. 13 years as an educator. Suchita brings both worlds to every class — technical rigour, deep subject knowledge, and the rare ability to make complex ideas genuinely simple. The AI Question Bank is entirely built in-house: no third-party platform, no generic content. Every question, hint and explanation shaped by real classroom experience.

500+
Students taught
13 yrs
Teaching experience
107
5-star reviews
🤖 AI Question Bank · Built entirely in-house Always growing
Class 12 Applied Maths AI Question Bank
Not just questions — a complete board exam preparation experience. Built from 13 years of watching students prepare, it's designed around how to understand the paper, how to approach scoring, and how to walk into the exam confident. Updated every year as patterns and needs evolve.
📄 Class 12 Applied Maths Board Paper 2026 — fully solved, step-by-step
Students who've already solved the most recent real board paper walk into the exam hall differently. That's what this gives you.
View full details →
Coming next: AI-Powered Question Bank for Class 12 Maths. Sounds interesting?
🧮

Vedic Maths in every class

Mental maths shortcuts woven into every session — faster, more intuitive calculation that genuinely sticks.

🔗

Connect-the-dots teaching

No concept taught in isolation. Every topic links to what you know and what's coming next.

🚀

Beyond the syllabus

Real-world applications and deeper insights that build genuine mathematical thinking, not just exam technique.

📐

Deep conceptual clarity

Especially for Applied Maths — in-depth understanding where most textbooks stop. Students who understand never blank.

🧘

Mindfulness for exam stress

Breathing techniques, focus exercises and stress management built into every course.

❤️

Love for Maths — guaranteed

A real shift in attitude within 2–3 classes. Consistent across 500+ students. This is the promise.

One class. Every mode.
Offline — Sector 14, GurugramSmall batches · face-to-face interaction
HybridJoin live online when you can't come in — never miss a session
Fully OnlineLive classes from anywhere in India — same quality, same outcomes
Students' Voice

500+ Students Taught.
Here's What They Say.

5.0 on Google · 100+ verified reviews · 13 years of teaching · every word unedited and publicly visible.

★★★★★ 5.0 100+ Google Reviews
⭐ Read All 100+ Google Reviews →

Opens Google Maps — all reviews are real, unedited and publicly visible

Study Materials

Free & Premium Resources

Browse All Resources →
📐

Class 12 Applied Maths Formula Deck

All 8 units · Every key formula · Print-friendly · Your board exam companion

₹199
Order Now →
🎉 Launch Offer
🤖

Class 12 Applied Maths AI Question Bank

Adaptive AI · Instant feedback · Weak-area targeting · Board simulation · Includes solved 2026 board paper

after 31 Jul One-time · No subscription
🔬
Free Download

Class 10 Science Notes

Physics, Chemistry & Biology · Chapter-wise · CBSE syllabus · Board revision

FREE
Download Free →
From the Blog

Guides, Tips & Insights

View All Posts →
Frequently Asked Questions

Got Questions? We've Got Answers.

What is the difference between Applied Maths and Maths in Class 11–12?

CBSE Applied Mathematics is for Class 11–12 Commerce students — Finance, Business Mathematics, Statistics, Calculus with real-world applications. Mathematics is the PCM syllabus for Science/JEE students. They're completely separate subjects with different syllabi and career paths. Read our full comparison →

How is Boundless Maths different from other coaching centres in Gurugram?

Vedic Maths techniques, beyond-syllabus teaching, a connect-the-dots approach, mindfulness for exam stress, and deep conceptual understanding — not rote preparation. Based in Sector 14, Gurugram, also serving DLF Phase, Sector 56, Sector 57 and students online across India.

What are the modes of teaching at Boundless Maths?

Offline (Sector 14, Gurugram), Hybrid (attend online when you can't come in person), and fully Online from anywhere in India. Same class, same teacher, same quality — distance is never a barrier to learning.

What resources are included with the Class 11 & 12 Applied Mathematics courses?

200+ free practice questions across all 8 units, a Formula Deck (₹199), an AI Question Bank at a launch offer of (until 31 Jul 2026, regular price ), YouTube video tutorials, previous years' CBSE board papers, and mindfulness sessions. All free resources available at boundlessmaths.com/applied-maths →

What exactly is the Mindful Mathematics GEM programme, and is it suitable for adults who haven't done maths in years?

The Mindful Mathematics GEM programme is a mathematical fluency programme based on the GEM Framework — Growth Mindset, Empowerment, Mindfulness. It's suitable for anyone, including adults who haven't done maths in years. Whether you're 12 or 42 — if Maths feels like a foreign language, this is where you start. Level 1 (Growth Mindset) is currently enrolling, focusing on tables, fractions and calculation techniques.

Who is the Mindful Mathematics GEM programme designed for?

School students (Class 8–10), those who skipped Maths in 11–12 and now need it for CUET, IPMAT, CA Foundation or MBA prep, people with a long gap from Maths wanting a structured recap, and anyone who wants to go faster with mental maths and Vedic shortcuts. It's for anyone — school student, working adult, or someone who just wants to be genuinely better at Maths.

What are the three levels of the GEM Framework?

Level 1 — Growth Mindset (live now): tables, fractions and calculation techniques through mental maths, Sudoku and Growth Mindset practices. Level 2 — Empowerment (coming): pattern recognition and multiple problem-solving strategies. Level 3 — Mindfulness (coming): deep focus combined with algebraic excellence and exam-day clarity techniques.

What teaching methods does Boundless Maths use?

Vedic Maths shortcuts for faster calculation, connect-the-dots teaching where no concept is taught in isolation, real-world applications beyond the syllabus, deep conceptual clarity (especially for Applied Maths), and mindfulness for exam stress. The result: a real shift in attitude within 2–3 classes, consistent across 500+ students.

What is the background of Suchita Arora, the founder of Boundless Maths?

Suchita Arora is the Founder & Lead Educator of Boundless Maths — 10 years as a software engineer and 13 years as a CBSE educator. She built the AI Question Bank entirely in-house, bringing technical rigour, deep subject knowledge, and the ability to make complex ideas genuinely simple. Read more about Suchita →

Get Started

Ready to Make Maths
Click for Your Child?

Book a free demo class, explore our free Applied Maths resources, or chat with us directly on WhatsApp — whichever works best for you.

🎓
Step 1

Enquire About Your Batch

Experience our teaching style firsthand — for Class 9, 10, 11, 12 Maths, Applied Maths, or Mindful Mathematics. No commitment, just a real conversation about what fits.

Enquire About Your Batch →
📚
Step 2

Explore Free Resources

200+ free Applied Maths questions, CBSE board papers, video tutorials and Class 10 Science notes — no signup required for most.

Free Applied Maths Resources → Browse All Resources →
💬
Step 3

Talk to Us

Questions about which course is right, batch timings, fees, or anything else? Reach out directly — we reply fast.

💬 Chat on WhatsApp Contact Us →
📞 +91 85952 36539 ✉️ contact@boundlessmaths.com 📍 Sector 14, Gurugram

Looking for CUET, B.Com, BBA or CA Foundation Maths prep?

Expert CBSE Coaching · Class 9–12