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

Complete the square: 9x^2+12x+2 = 9\left(x+\tfrac23\right)^2-2.

Since \sin(2x) \in [-1,1], h(x) \in [4,6].

Since \sin4x \in [-1,1], \sin4x+3 \in [2,4], always positive, so |\sin4x+3| = \sin4x+3.

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)=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.

h'(x)=3x^2+2x+1. Its discriminant = 4-12=-8<0, and the leading coefficient is positive, so h'(x)>0 for every real x.

Critical point: f'(x)=3x^2=0 \Rightarrow x=0. Evaluate: f(-2)=-8,\ f(0)=0,\ f(2)=8.

Critical point: x=\tfrac{\pi}{4}. Evaluate: 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(-2)=-10,\ f(4)=8,\ f\left(\tfrac92\right)=7.875.

Critical point (also the right endpoint): x=1. Evaluate: 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(0)=25,\ f(2)=-39,\ f(3)=16.

\sin2x attains its maximum value 1 when 2x=\tfrac{\pi}{2}+2k\pi, i.e. x=\tfrac{\pi}{4}+k\pi.

Within [0,2\pi]: x=\tfrac{\pi}{4} and x=\tfrac{5\pi}{4}.

\sin x+\cos x = \sqrt2\sin\left(x+\tfrac{\pi}{4}\right), and \sin(\cdot) has maximum value 1.

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

On [1,3] (critical point x=2 inside): f(1)=85,\ f(2)=75,\ f(3)=89.

On [-3,-1] (critical point x=-2 inside): 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.

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

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].

Evaluating f at all critical points and the endpoints x=0,\ 2\pi shows the extreme values actually occur at the endpoints.

f(0)=0; f(2\pi)=2\pi, and every interior critical value lies strictly between these two.

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). Sign changes +\to- — 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). This gives the maximum.

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.

V'(x)=12(x-3)(x-9)=0 \Rightarrow x=3 (rejecting x=9, a degenerate box). V''(3)=-72<0 — maximum.

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.

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.

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

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.

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. S''(r)>0 — 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}. A'(x)=\dfrac{x}{8}-\dfrac{28-x}{2\pi}=0 \Rightarrow x=\dfrac{112}{\pi+4}.

A''(x)>0 — minimum.

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.

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 and minimising w.r.t. r leads to the critical-point condition h^2=2r^2.

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 (a maximum on checking sign change).

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.

Then \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.

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.