Class 12 Maths NCERT Solutions Chapter 6 Miscellaneous Exercise

f'(x) = \dfrac{1-\log x}{x^2}\quad\ldots\text{(1)} (quotient rule). Setting f'(x)=0 in (1): \log x = 1 \Rightarrow x=e.

From (1): for x<e, \log x<1 \Rightarrow f'(x)>0.

For x>e: \log x>1 \Rightarrow f'(x)<0. Sign change +\to-.

Let equal side =a. Height h=\sqrt{a^2-\tfrac{b^2}{4}}, so area A=\tfrac{b}{2}\sqrt{a^2-\tfrac{b^2}{4}}\quad\ldots\text{(1)}.

Differentiating (1) by the chain rule: \dfrac{dA}{dt} = \dfrac{ab}{2\sqrt{a^2-b^2/4}}\cdot\dfrac{da}{dt}\quad\ldots\text{(2)}.

Given \dfrac{da}{dt}=-3; at a=b: \sqrt{a^2-b^2/4} = \dfrac{b\sqrt3}{2}.

Substituting into (2): \dfrac{dA}{dt} = \dfrac{b\cdot b}{2\cdot(b\sqrt3/2)}\cdot(-3) = \dfrac{b}{\sqrt3}\cdot(-3) = -\sqrt3\,b.

Let N=4\sin x-2x-x\cos x and D=2+\cos x. By the product rule, N'=4\cos x-2-(\cos x-x\sin x)=3\cos x-2+x\sin x, and D'=-\sin x.

By the quotient rule, the numerator of f'(x) is N'D-ND'=N'D+N\sin x:

N'D=(3\cos x-2+x\sin x)(2+\cos x)=4\cos x+3\cos^2x-4+2x\sin x+x\sin x\cos x

N\sin x=(4\sin x-2x-x\cos x)\sin x=4\sin^2x-2x\sin x-x\sin x\cos x

Adding these, the x\sin x\cos x terms cancel directly, and the 2x\sin x and -2x\sin x terms also cancel, leaving:

N'D+N\sin x=4\cos x+3\cos^2x-4+4\sin^2x=4\cos x+3\cos^2x-4+4(1-\cos^2x)=4\cos x-\cos^2x=\cos x(4-\cos x)

So f'(x)=\dfrac{\cos x(4-\cos x)}{(2+\cos x)^2}\quad\ldots\text{(1)}.

In (1), 4-\cos x>0 always and the denominator is always positive, so the sign of f'(x) follows the sign of \cos x.

f'(x)=3x^2-\dfrac{3}{x^4} = \dfrac{3(x^6-1)}{x^4}\quad\ldots\text{(1)}. Since x^4>0 always, the sign of (1) follows x^6-1.

x^6-1>0 when |x|>1; x^6-1<0 when |x|<1 (excluding 0).

With vertex at (a,0) and the other two vertices at (x,\pm y) on the ellipse, the base of the triangle has length 2y and the height (horizontal distance from the base to the vertex) is a-x. So area A=\tfrac12(2y)(a-x)=y(a-x)\quad\ldots\text{(1)}, where y=\tfrac{b}{a}\sqrt{a^2-x^2}\quad\ldots\text{(2)} (from the ellipse equation).

It's easier to maximise A^2 (equivalent, since A>0). Using (1) and (2):

A^2=y^2(a-x)^2=\dfrac{b^2}{a^2}(a^2-x^2)(a-x)^2=\dfrac{b^2}{a^2}(a-x)^3(a+x)\quad\ldots\text{(3)}

Let f(x)=(a-x)^3(a+x) (dropping the constant b^2/a^2 factor in (3)). By the product rule:

f'(x)=-3(a-x)^2(a+x)+(a-x)^3=(a-x)^2\big[-3(a+x)+(a-x)\big]=(a-x)^2(-2a-4x)=-2(a-x)^2(a+2x)\quad\ldots\text{(4)}

Setting f'(x)=0 in (4): either a-x=0 (rejected — this puts all three vertices at the same point, a degenerate triangle), or a+2x=0 \Rightarrow x=-\dfrac{a}{2}.

From (4), just left of x=-\tfrac{a}{2}, (a+2x)<0 so f'(x)>0; just right of it, (a+2x)>0 so f'(x)<0. Sign change +\to- confirms a maximum.

Substituting x=-\tfrac{a}{2} into (2): y=\dfrac{b}{a}\sqrt{a^2-\tfrac{a^2}{4}}=\dfrac{b}{a}\cdot\dfrac{a\sqrt3}{2}=\dfrac{\sqrt3}{2}b.

Substituting into (1): A=y(a-x)=\dfrac{\sqrt3}{2}b\left(a+\dfrac{a}{2}\right)=\dfrac{\sqrt3}{2}b\cdot\dfrac{3a}{2}=\dfrac{3\sqrt3}{4}ab.

Let base dimensions be x, y. Volume =2xy=8 \Rightarrow xy=4 — so the base area is fixed at 4\text{ m}^2, costing 70\times4=280 regardless of shape.

Total side area =2(2x)+2(2y)=4(x+y), costing 45\times4(x+y)=180(x+y).

Cost C(x) = 280+180\left(x+\dfrac{4}{x}\right)\quad\ldots\text{(1)}. Differentiating (1): C'(x)=180\left(1-\dfrac{4}{x^2}\right)=0 \Rightarrow x=2, giving y=2. C''(x)>0 — minimum.

Substituting x=2 into (1): C(2)=280+180(2+2)=280+720=1000.

Let circle radius r, square side s. Constraint: 2\pi r+4s=k \Rightarrow s=\dfrac{k-2\pi r}{4}\quad\ldots\text{(1)}, so \dfrac{ds}{dr}=-\dfrac{2\pi}{4}=-\dfrac{\pi}{2}.

Combined area A(r)=\pi r^2+s^2\quad\ldots\text{(2)}. By the chain rule, differentiating (2): \dfrac{dA}{dr}=2\pi r+2s\dfrac{ds}{dr}=2\pi r+2s\left(-\dfrac{\pi}{2}\right)=2\pi r-\pi s.

Setting \dfrac{dA}{dr}=0: 2\pi r-\pi s=0 \Rightarrow 2\pi r=\pi s \Rightarrow s=2r.

\dfrac{d^2A}{dr^2}=2\pi-\pi\dfrac{ds}{dr}=2\pi-\pi\left(-\dfrac{\pi}{2}\right)=2\pi+\dfrac{\pi^2}{2}>0 — minimum.

Let rectangle width =2x (so semicircle radius =x), height =y. Perimeter: 2x+2y+\pi x=10 \Rightarrow y=5-x\left(1+\tfrac{\pi}{2}\right)\quad\ldots\text{(1)}.

Total area (light) A(x)=2xy+\tfrac12\pi x^2\quad\ldots\text{(2)}. Substituting (1) into (2): A(x) = 10x - x^2\left(2+\tfrac{\pi}{2}\right)\quad\ldots\text{(3)}.

Differentiating (3): A'(x)=10-x(4+\pi)=0 \Rightarrow x=\dfrac{10}{4+\pi}.

A''(x)<0 — maximum. Substituting x=\dfrac{10}{4+\pi} into (1) gives y=\dfrac{10}{4+\pi} too.

Let the right angle be at O, with legs along the horizontal and vertical directions, and let P be the given point, at perpendicular distance a from the vertical leg and b from the horizontal leg. Let the hypotenuse through P meet the horizontal leg at X and the vertical leg at Y, making angle \theta with the horizontal leg at X.

Drop perpendiculars from P to each leg. The small triangle formed with foot M on the horizontal leg (so PM=b) is similar to the full triangle OXY, sharing the angle \theta at X. In this small right triangle, PM is opposite \theta, so \sin\theta=\dfrac{PM}{XP}=\dfrac{b}{XP} \Rightarrow XP=b\csc\theta.

Similarly, the small triangle with foot N on the vertical leg (so PN=a) shares the angle (90^\circ-\theta) at Y, giving \cos\theta=\dfrac{PN}{PY}=\dfrac{a}{PY} \Rightarrow PY=a\sec\theta.

The hypotenuse length is therefore

L(\theta)=XY=XP+PY=a\sec\theta+b\csc\theta\quad\ldots\text{(1)}

Differentiating (1): \dfrac{dL}{d\theta}=a\sec\theta\tan\theta-b\csc\theta\cot\theta=0 \Rightarrow \tan^3\theta=\dfrac{b}{a}.

To confirm this critical point gives a minimum (not a maximum), examine the boundary behaviour of L(\theta) in (1) on the valid domain \left(0,\tfrac{\pi}{2}\right).

As \theta\to0^+, \csc\theta\to\infty, so L\to\infty; as \theta\to\tfrac{\pi}{2}^-, \sec\theta\to\infty, so L\to\infty as well.

Since L is large at both ends of the interval and there is only one critical point in between, that single critical point must give the minimum value of L.

Substituting t=\tan\theta=(b/a)^{1/3} into (1) and simplifying using 1+t^2=\dfrac{a^{2/3}+b^{2/3}}{a^{2/3}} gives L=a(1+t^2)^{3/2}.

Using the product rule: f'(x)=(x-2)^3(x+1)^2(7x-2). Critical points: x=-1,\ \tfrac27,\ 2.

Testing the sign of f'(x) in each interval: positive on (-\infty,-1) and (-1,\tfrac27) — no sign change at x=-1 since (x+1)^2\geq0.

Negative on (\tfrac27,2), positive on (2,\infty).

f'(x)=-2\sin x\cos x+\cos x=\cos x(1-2\sin x)=0 \Rightarrow x=\tfrac{\pi}{6},\tfrac{\pi}{2},\tfrac{5\pi}{6} (within [0,\pi]).

Evaluate: f(0)=1,\ f\left(\tfrac{\pi}{6}\right)=\tfrac54,\ f\left(\tfrac{\pi}{2}\right)=1,\ f\left(\tfrac{5\pi}{6}\right)=\tfrac54,\ f(\pi)=1.

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

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

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

Expanding (1): r^2-2rh+h^2+\rho^2=r^2 \Rightarrow \rho^2=2rh-h^2=h(2r-h)\quad\ldots\text{(2)}.

Substituting (2) into the volume formula: V(h)=\tfrac13\pi\rho^2h=\tfrac13\pi h^2(2r-h)\quad\ldots\text{(3)}.

Differentiating (3): V'(h)=\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(4r-8r)=-\tfrac{4\pi r}{3}<0 — maximum.

This is the Increasing/Decreasing Test established earlier in this chapter (Exercise 6.2): if f is differentiable on an interval with f'(x)>0 at every point of it, f is increasing on that interval.

Here f'(x)>0 for every x\in(a,b) — exactly the hypothesis of that test — so applying it directly to (a,b) gives the result.

Consider the rectangular cross-section of the cylinder through the sphere's centre: width 2x (the cylinder's diameter) and height y. Since all four corners of this rectangle lie on the sphere's great circle, its diagonal equals the sphere's diameter 2R:

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

From (1), x^2=R^2-\tfrac{y^2}{4}. Substituting into V=\pi x^2y: V(y)=\pi\left(R^2-\tfrac{y^2}{4}\right)y\quad\ldots\text{(2)}.

Differentiating (2): V'(y)=\pi R^2-\tfrac{3\pi}{4}y^2=0 \Rightarrow y=\dfrac{2R}{\sqrt3}.

V''(y)=-\tfrac{3\pi}{2}y<0 for y>0 — maximum.

Substituting y=\tfrac{2R}{\sqrt3} into (1): x^2=R^2-\dfrac{1}{4}\cdot\dfrac{4R^2}{3}=R^2-\dfrac{R^2}{3}=\dfrac{2R^2}{3}, so V_{max}=\pi\cdot\dfrac{2R^2}{3}\cdot\dfrac{2R}{\sqrt3}=\dfrac{4\pi R^3}{3\sqrt3}.

Let the cylinder have radius x and height y, inscribed in the cone with its top rim touching the slant surface. The cone has total height h and semi-vertical angle \alpha, so its base radius is h\tan\alpha.

The top rim of the cylinder sits at height y above the base, i.e. at distance (h-y) below the apex. By similar triangles, the cone's radius at that level is proportional to its distance from the apex: \dfrac{\text{radius at distance }(h-y)}{h-y}=\dfrac{\text{base radius}}{h}=\dfrac{h\tan\alpha}{h}=\tan\alpha.

Since this radius equals the cylinder's radius x: x=(h-y)\tan\alpha, equivalently y=h-x\cot\alpha\quad\ldots\text{(1)}.

Substituting (1) into V=\pi x^2y: V(x)=\pi x^2(h-x\cot\alpha)\quad\ldots\text{(2)}.

Differentiating (2): V'(x)=\pi x(2h-3x\cot\alpha)=0 \Rightarrow x=\dfrac{2h}{3}\tan\alpha (rejecting x=0, a degenerate cylinder).

Substituting into (1): y=h-x\cot\alpha=h-\dfrac{2h}{3}=\dfrac{h}{3}.

V''(x)=\pi(2h-6x\cot\alpha), so at x=\tfrac{2h}{3}\tan\alpha: V''=\pi\left(2h-6\cdot\tfrac{2h}{3}\right)=\pi(2h-4h)=-2\pi h<0 — maximum.

Substituting into (2): V_{max}=\pi\left(\dfrac{2h}{3}\tan\alpha\right)^2\cdot\dfrac{h}{3} = \dfrac{4}{27}\pi h^3\tan^2\alpha.

V=\pi r^2 h, with r=10 fixed: \dfrac{dV}{dt}=100\pi\dfrac{dh}{dt}.

314 = 100\pi\dfrac{dh}{dt}. Using \pi\approx3.14: 314=314\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt}=1.