Class 12 Maths NCERT Solutions Chapter 6 Miscellaneous Exercise

f'(x) = \dfrac{1-\log x}{x^2} (quotient rule). Set f'(x)=0 \Rightarrow \log x = 1 \Rightarrow x=e.

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}}.

\dfrac{dA}{dt} = \dfrac{ab}{2\sqrt{a^2-b^2/4}}\cdot\dfrac{da}{dt} (Chain Rule).

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

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

Differentiating using the quotient rule and simplifying (the x\sin x\cos x and 2x\sin x terms cancel) gives a strikingly clean result: f'(x) = \dfrac{\cos x(4-\cos x)}{(2+\cos x)^2}.

Since 4-\cos x>0 always and the denominator is always positive, 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}. Since x^4>0 always, the sign 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, area A=y(a-x), where y=\tfrac{b}{a}\sqrt{a^2-x^2}.

Maximising A^2 gives the critical point x=-\tfrac{a}{2}, at which y=\tfrac{\sqrt3}{2}b.

A = \tfrac{\sqrt3 b}{2}\left(a+\tfrac{a}{2}\right) = \tfrac{\sqrt3 b}{2}\cdot\tfrac{3a}{2}.

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). C'(x)=180\left(1-\dfrac{4}{x^2}\right)=0 \Rightarrow x=2, giving y=2. C''(x)>0 — minimum.

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}.

Combined area A(r)=\pi r^2+s^2. \dfrac{dA}{dr}=2\pi r+2s\left(-\dfrac{\pi}{2}\right)=2\pi r-\pi s.

Setting \dfrac{dA}{dr}=0 gives directly s=2r. \dfrac{d^2A}{dr^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).

Total area (light) A(x)=2xy+\tfrac12\pi x^2 = 10x - x^2\left(2+\tfrac{\pi}{2}\right).

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

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

Let \theta be the angle the hypotenuse makes with one leg. Writing the intercept form of the hypotenuse through the given point gives hypotenuse length L(\theta)=a\sec\theta+b\csc\theta.

\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) 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} back into L 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.

For a cone of height h inscribed in a sphere of radius r, the base radius satisfies \rho^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.

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.

For cylinder radius x, height y inscribed in sphere of radius R: x^2+\tfrac{y^2}{4}=R^2.

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

V''(y)<0 — maximum.

Substituting back: V_{max}=\dfrac{4\pi R^3}{3\sqrt3}.

By similar triangles, cylinder radius x and height y relate as y=h-x\cot\alpha (equivalently x=(h-y)\tan\alpha).

V(x)=\pi x^2(h-x\cot\alpha). V'(x)=\pi x(2h-3x\cot\alpha)=0 \Rightarrow x=\dfrac{2h}{3}\tan\alpha.

Then y=h-x\cot\alpha=h-\dfrac{2h}{3}=\dfrac{h}{3}.

V''(x)<0 — maximum.

Substituting: 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.