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

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.

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.

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.

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).

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.

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.

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

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.

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.

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

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

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

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

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.

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

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

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-.

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

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

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

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.

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.

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.

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.

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.

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.

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

p''(x)=-36<0 — maximum.

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

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.

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.

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.)

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.

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.

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.

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.

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.

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.

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.

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

Expanding: (18-2x)^2=324-72x+4x^2, so V(x)=324x-72x^2+4x^3\quad\ldots\text{(1)}.

Differentiating (1): V'(x)=324-144x+12x^2=12(x^2-12x+27)=12(x-3)(x-9)=0 \Rightarrow x=3 (rejecting x=9, a degenerate box).

V''(x)=24x-144, so V''(3)=72-144=-72<0 — maximum.

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

Expanding: (45-2x)(24-2x)=1080-138x+4x^2, so V(x)=1080x-138x^2+4x^3\quad\ldots\text{(1)}.

Differentiating (1): V'(x)=1080-276x+12x^2=12(x^2-23x+90)=12(x-5)(x-18)=0 \Rightarrow x=5 (rejecting x=18, outside domain).

V''(x)=24x-276, so V''(5)=120-276=-156<0 — maximum.

Let the circle have (fixed) radius r, and let the inscribed rectangle have sides x and y. Since the diagonal of the rectangle passes through the centre, it equals the diameter:

x^2+y^2=4r^2\quad\ldots\text{(1)}

From (1), y=\sqrt{4r^2-x^2}. Area:

A(x)=xy=x\sqrt{4r^2-x^2}\quad\ldots\text{(2)}, defined for 0<x<2r.

Differentiating (2): A'(x)=\sqrt{4r^2-x^2}+x\cdot\dfrac{-x}{\sqrt{4r^2-x^2}}=\dfrac{4r^2-2x^2}{\sqrt{4r^2-x^2}}.

Setting A'(x)=0: since the denominator is never zero on the domain, 4r^2-2x^2=0 \Rightarrow x^2=2r^2 \Rightarrow x=r\sqrt2.

The denominator \sqrt{4r^2-x^2} is always positive, 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. So A' changes sign + to - at x=r\sqrt2, confirming a maximum there.

From (1), y=\sqrt{4r^2-2r^2}=r\sqrt2=x.

Let the cylinder have radius r and height h. The (fixed) total surface area is

S=2\pi r^2+2\pi rh\quad\ldots\text{(1)}

From (1), h=\dfrac{S-2\pi r^2}{2\pi r}\quad\ldots\text{(2)} — this writes h in terms of r and the given constant S, so volume can be treated as a function of r alone.

Volume: V=\pi r^2h. Substituting (2):

V(r)=\pi r^2\cdot\dfrac{S-2\pi r^2}{2\pi r}=\dfrac{r(S-2\pi r^2)}{2}=\dfrac{S}{2}r-\pi r^3\quad\ldots\text{(3)}

Differentiating (3): V'(r)=\dfrac{S}{2}-3\pi r^2=0 \Rightarrow r^2=\dfrac{S}{6\pi}, i.e. S=6\pi r^2\quad\ldots\text{(4)}.

Substituting (4) into (2): h=\dfrac{6\pi r^2-2\pi r^2}{2\pi r}=\dfrac{4\pi r^2}{2\pi r}=2r.

V''(r)=-6\pi r<0 for r>0 — maximum.

Let the can have radius r and height h. The (fixed) volume is

V=\pi r^2h=100\quad\ldots\text{(1)}

From (1), h=\dfrac{100}{\pi r^2}\quad\ldots\text{(2)} — this writes h in terms of r, so surface area can be treated as a function of r alone.

Total surface area: S=2\pi r^2+2\pi rh. Substituting (2):

S(r)=2\pi r^2+2\pi r\cdot\dfrac{100}{\pi r^2}=2\pi r^2+\dfrac{200}{r}\quad\ldots\text{(3)}

Differentiating (3): S'(r)=4\pi r-\dfrac{200}{r^2}=0 \Rightarrow 4\pi r^3=200 \Rightarrow r^3=\dfrac{50}{\pi}\quad\ldots\text{(4)}.

Substituting (4) into (2) — multiply numerator and denominator by r: h=\dfrac{100r}{\pi r^3}=\dfrac{100r}{\pi\cdot\frac{50}{\pi}}=\dfrac{100r}{50}=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.

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}\quad\ldots\text{(1)}. Differentiating (1): A'(x)=\dfrac{x}{8}-\dfrac{28-x}{2\pi}=0 \Rightarrow x=\dfrac{112}{\pi+4}.

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

Let the cone have apex A on the sphere, height h, and base radius r, with its axis through the sphere's centre O. Let M be the centre of the base and C a point on the rim of the base.

Since OA=R and AM=h, the distance from O to the base plane is OM=|R-h|. Also MC=r, and OC=R (a radius of the sphere, since C lies on it). Triangle OMC is right-angled at M, so by Pythagoras:

OM^2+MC^2=OC^2 \Rightarrow (R-h)^2+r^2=R^2\quad\ldots\text{(1)}

Expanding (1): R^2-2Rh+h^2+r^2=R^2 \Rightarrow r^2=2Rh-h^2=h(2R-h)\quad\ldots\text{(2)}

Volume: V=\tfrac13\pi r^2h. Substituting (2):

V(h)=\tfrac13\pi h^2(2R-h)\quad\ldots\text{(3)}

Differentiating (3): V'(h)=\tfrac13\pi\left(4Rh-3h^2\right)=\tfrac13\pi h(4R-3h)=0 \Rightarrow h=\tfrac{4R}{3} (rejecting h=0, a degenerate cone).

V''(h)=\tfrac13\pi(4R-6h), so V''\left(\tfrac{4R}{3}\right)=\tfrac13\pi\left(4R-8R\right)=-\tfrac{4\pi R}{3}<0 — maximum.

Substituting h=\tfrac{4R}{3} into (3): V_{max}=\tfrac13\pi\left(\tfrac{4R}{3}\right)^2\left(2R-\tfrac{4R}{3}\right)=\tfrac13\pi\cdot\tfrac{16R^2}{9}\cdot\tfrac{2R}{3}=\tfrac{32}{81}\pi R^3, while the sphere's volume is \tfrac43\pi R^3.

Let the cone have radius r, height h, and slant height l. The (fixed) volume is

V=\tfrac13\pi r^2h \Rightarrow h=\dfrac{3V}{\pi r^2}\quad\ldots\text{(1)}

The curved (lateral) surface area is S=\pi rl, where l=\sqrt{r^2+h^2}. Squaring to avoid the square root:

S^2=\pi^2r^2l^2=\pi^2r^2(r^2+h^2)=\pi^2r^4+\pi^2r^2h^2\quad\ldots\text{(2)}

Substituting (1) into (2): \pi^2r^2h^2=\pi^2r^2\left(\dfrac{3V}{\pi r^2}\right)^2=\pi^2r^2\cdot\dfrac{9V^2}{\pi^2r^4}=\dfrac{9V^2}{r^2}, so

f(r)=S^2=\pi^2r^4+\dfrac{9V^2}{r^2}\quad\ldots\text{(3)} (a function of r alone, since V is fixed).

Differentiating (3): f'(r)=4\pi^2r^3-\dfrac{18V^2}{r^3}=0 \Rightarrow r^6=\dfrac{9V^2}{2\pi^2}\quad\ldots\text{(4)}.

To confirm this is the same as h^2=2r^2: from (1), h^2=\dfrac{9V^2}{\pi^2r^4}. Using (4), 9V^2=2\pi^2r^6, so h^2=\dfrac{2\pi^2r^6}{\pi^2r^4}=2r^2, i.e. h=\sqrt2\,r.

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).

With slant height l fixed and semi-vertical angle \alpha, from the right-angled triangle formed by r, h, and l:

r=l\sin\alpha,\qquad h=l\cos\alpha\quad\ldots\text{(1)}

Volume: V=\tfrac13\pi r^2h. Substituting (1):

V(\alpha)=\tfrac13\pi(l\sin\alpha)^2(l\cos\alpha)=\tfrac{\pi l^3}{3}\sin^2\alpha\cos\alpha\quad\ldots\text{(2)}

Differentiating (2) using the product rule: \dfrac{dV}{d\alpha}=\tfrac{\pi l^3}{3}\left[2\sin\alpha\cos\alpha\cdot\cos\alpha+\sin^2\alpha\cdot(-\sin\alpha)\right]=\tfrac{\pi l^3}{3}\sin\alpha\left(2\cos^2\alpha-\sin^2\alpha\right).

Setting this to zero (rejecting \sin\alpha=0, a degenerate cone): 2\cos^2\alpha-\sin^2\alpha=0 \Rightarrow \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 \alpha just below \tan^{-1}\sqrt2, \sin^2\alpha<\tfrac23 so 2-3\sin^2\alpha>0, giving \dfrac{dV}{d\alpha}>0 (V increasing); for \alpha just above it, \sin^2\alpha>\tfrac23 so 2-3\sin^2\alpha<0, giving \dfrac{dV}{d\alpha}<0 (V decreasing).

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

Let the cone have radius r, height h, and slant height l, with semi-vertical angle \alpha so that \sin\alpha=\dfrac{r}{l}. The (fixed) total surface area (base + curved surface) is

S=\pi r^2+\pi rl\quad\ldots\text{(1)}

From (1), l=\dfrac{S-\pi r^2}{\pi r}=\dfrac{S}{\pi r}-r\quad\ldots\text{(2)}.

Since l^2=r^2+h^2, we have h^2=l^2-r^2. Squaring (2):

l^2=\left(\dfrac{S}{\pi r}\right)^2-2\cdot\dfrac{S}{\pi r}\cdot r+r^2=\dfrac{S^2}{\pi^2r^2}-\dfrac{2S}{\pi}+r^2

so h^2=l^2-r^2=\dfrac{S^2}{\pi^2r^2}-\dfrac{2S}{\pi}\quad\ldots\text{(3)}.

Volume: V=\tfrac13\pi r^2h, so V^2=\tfrac19\pi^2r^4h^2. Substituting (3):

V^2=\tfrac19\pi^2r^4\left(\dfrac{S^2}{\pi^2r^2}-\dfrac{2S}{\pi}\right)=\tfrac19\left(S^2r^2-2\pi Sr^4\right)\quad\ldots\text{(4)}

Since S is fixed, maximising V is the same as maximising V^2, i.e. maximising g(r)=S^2r^2-2\pi Sr^4 (dropping the constant factor \tfrac19). Differentiating:

g'(r)=2S^2r-8\pi Sr^3=2Sr\left(S-4\pi r^2\right)=0

Rejecting r=0 (a degenerate cone): r^2=\dfrac{S}{4\pi}\quad\ldots\text{(5)}.

Substituting (5) into (2): since S=4\pi r^2, l=\dfrac{4\pi r^2}{\pi r}-r=4r-r=3r.

g''(r)=2S^2-24\pi Sr^2. At r^2=\tfrac{S}{4\pi}: g''=2S^2-24\pi S\cdot\dfrac{S}{4\pi}=2S^2-6S^2=-4S^2<0 — this confirms a maximum of V.

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

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.

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].

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.