Miscellaneous Exercise Class 12 Maths Application of Derivatives

Question 01

Show that the function given by $f(x) = \frac{\log x}{x}$ has maximum at $x = e$.

$$f'(x) = \frac{x(\frac{1}{x}) – \log x(1)}{x^2} = \frac{1 – \log x}{x^2}$$ For critical points, set $f'(x) = 0 \implies 1 – \log x = 0 \implies \log x = 1 \implies x = e$. $$f”(x) = \frac{x^2(-\frac{1}{x}) – (1-\log x)(2x)}{x^4} = \frac{-x – 2x(1-\log x)}{x^4}$$ At $x=e$: $f”(e) = \frac{-e – 0}{e^4} = \frac{-1}{e^3} < 0$.

Since $f”(e) < 0$, Maxima at $x = e$.

Question 02

The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Let equal sides be $x$. Given $\frac{dx}{dt} = -3$ cm/s.

Area $A = \frac{1}{2} \cdot \text{base} \cdot \text{height}$. Height $h = \sqrt{x^2 – (b/2)^2}$. $$A = \frac{b}{2}\sqrt{x^2 – \frac{b^2}{4}} = \frac{b}{4}\sqrt{4x^2 – b^2}$$ $$\frac{dA}{dt} = \frac{b}{4} \cdot \frac{1}{2\sqrt{4x^2 – b^2}} \cdot 8x \frac{dx}{dt} = \frac{bx}{\sqrt{4x^2 – b^2}} \frac{dx}{dt}$$ When $x = b$: $$\frac{dA}{dt} = \frac{b(b)}{\sqrt{4b^2 – b^2}} (-3) = \frac{-3b^2}{\sqrt{3b^2}} = -\sqrt{3}b$$

Area decreases at $\sqrt{3}b$ cm²/s.

Question 03

Find the intervals in which the function $f$ given by $f(x) = \frac{4\sin x – 2x – x\cos x}{2 + \cos x}$ is (i) increasing (ii) decreasing.

Simplify $f(x) = \frac{4\sin x}{2+\cos x} – x$.

$$f'(x) = \frac{(2+\cos x)(4\cos x) – 4\sin x(-\sin x)}{(2+\cos x)^2} – 1$$ $$f'(x) = \frac{8\cos x + 4\cos^2 x + 4\sin^2 x}{(2+\cos x)^2} – 1 = \frac{8\cos x + 4}{(2+\cos x)^2} – 1$$ $$f'(x) = \frac{8\cos x + 4 – (4 + \cos^2 x + 4\cos x)}{(2+\cos x)^2} = \frac{\cos x(4 – \cos x)}{(2+\cos x)^2}$$

Since $(2+\cos x)^2 > 0$ and $4-\cos x > 0$, the sign depends on $\cos x$.

Increasing in $(0, \pi/2) \cup (3\pi/2, 2\pi)$; Decreasing in $(\pi/2, 3\pi/2)$.

Question 04

Find intervals in which $f(x) = x^3 + \frac{1}{x^3}, x \neq 0$ is (i) increasing (ii) decreasing.

$$f'(x) = 3x^2 – \frac{3}{x^4} = \frac{3(x^6 – 1)}{x^4}$$ For critical points: $x^6 – 1 = 0 \implies x = \pm 1$. Since $x^4 > 0$, sign depends on $x^6 – 1$.

Increasing ($f'(x)>0$): $(-\infty, -1) \cup (1, \infty)$.

Decreasing ($f'(x)<0$): $(-1, 1) - \{0\}$.

Question 05

Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with its vertex at one end of the major axis.

Let vertex be $(-a, 0)$ and other points $(a\cos\theta, \pm b\sin\theta)$.

$$Area = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2}(2b\sin\theta)(a + a\cos\theta) = ab(\sin\theta + \sin\theta\cos\theta)$$ $$A'(\theta) = ab(\cos\theta + \cos 2\theta) = 0 \implies 2\cos^2\theta + \cos\theta – 1 = 0$$ $$(2\cos\theta – 1)(\cos\theta + 1) = 0 \implies \cos\theta = 1/2 \implies \theta = \pi/3$$ $$A(\pi/3) = ab(\frac{\sqrt{3}}{2})(1 + \frac{1}{2}) = \frac{3\sqrt{3}}{4}ab$$

Max Area: $\frac{3\sqrt{3}}{4}ab$ sq units.

Question 06

A tank with rectangular base and sides, open at top, depth 2m and volume 8m³. Building costs Rs 70/m² for base and Rs 45/m² for sides. Minimize cost.

Let base be $x$ and $y$. Depth $h=2$. Volume $2xy = 8 \implies xy = 4 \implies y = 4/x$.

Area Base = $xy = 4$. Area Sides = $2(2x + 2y) = 4(x+y)$. Cost $C = 70(4) + 45(4(x + \frac{4}{x})) = 280 + 180(x + \frac{4}{x})$ $$\frac{dC}{dx} = 180(1 – \frac{4}{x^2}) = 0 \implies x^2 = 4 \implies x = 2$$ Min Cost $C(2) = 280 + 180(2 + 2) = 280 + 720 = 1000$.

Least Cost: Rs 1000

Question 07

Sum of perimeter of circle and square is $k$. Prove sum of areas is least when side of square is double the radius.

Let circle radius $r$, square side $x$. Constraint: $2\pi r + 4x = k \implies x = \frac{k – 2\pi r}{4}$.

Sum of Areas $A = \pi r^2 + x^2 = \pi r^2 + (\frac{k-2\pi r}{4})^2$ $$\frac{dA}{dr} = 2\pi r + 2(\frac{k-2\pi r}{4})(\frac{-2\pi}{4}) = 2\pi r – \frac{\pi(k-2\pi r)}{4}$$ Set $\frac{dA}{dr} = 0 \implies 8r = k – 2\pi r \implies k = 2r(4+\pi)$. Substitute back into perimeter eq: $2\pi r + 4x = 2\pi r + 8r \implies 4x = 8r \implies x = 2r$.

Side of square = 2 × Radius. Verified.

Question 08

A window is a rectangle surmounted by a semicircle. Total perimeter 10m. Find dimensions for maximum light.

Let width $2x$ (radius $x$) and height $y$.

Perimeter: $2x + 2y + \pi x = 10 \implies y = \frac{10 – 2x – \pi x}{2}$. Light $\propto$ Area $A = 2xy + \frac{1}{2}\pi x^2 = x(10 – 2x – \pi x) + \frac{\pi x^2}{2} = 10x – 2x^2 – \frac{\pi x^2}{2}$. $$A'(x) = 10 – 4x – \pi x = 0 \implies x(4+\pi) = 10 \implies x = \frac{10}{4+\pi}$$

Width: $\frac{20}{4+\pi}$m, Height: $\frac{10}{4+\pi}$m.

Question 09

A point on the hypotenuse of a triangle is at distance $a$ and $b$ from the sides. Show minimum length of hypotenuse is $(a^{2/3} + b^{2/3})^{3/2}$.

Let hypotenuse length be $L$. From geometry, $L = a \sec\theta + b \csc\theta$.

$$\frac{dL}{d\theta} = a \sec\theta \tan\theta – b \csc\theta \cot\theta = 0$$ $$\implies \frac{a \sin\theta}{\cos^2\theta} = \frac{b \cos\theta}{\sin^2\theta} \implies \tan^3\theta = \frac{b}{a} \implies \tan\theta = (b/a)^{1/3}$$ Substituting back: $L = a \sqrt{1+\tan^2\theta} + b \sqrt{1+\cot^2\theta}$ After simplifying algebraically, $L = (a^{2/3} + b^{2/3})^{3/2}$.

Question 10

Find points of local maxima, minima, inflexion for $f(x) = (x-2)^4(x+1)^3$.

$$f'(x) = 4(x-2)^3(x+1)^3 + 3(x-2)^4(x+1)^2 = (x-2)^3(x+1)^2 [4(x+1) + 3(x-2)]$$ $$f'(x) = (x-2)^3(x+1)^2(7x-2)$$ Critical points: $x = 2, -1, 2/7$.

At $x=-1$: Sign does not change ($+ \to +$). Point of Inflexion.
At $x=2/7$: Sign changes $+ \to -$. Local Maxima.
At $x=2$: Sign changes $- \to +$. Local Minima.

Question 11

Find absolute max and min of $f(x) = \cos^2 x + \sin x, x \in [0, \pi]$.

$f(x) = 1 – \sin^2 x + \sin x$. Let $\sin x = t, t \in [0, 1]$. $g(t) = -t^2 + t + 1$. $g'(t) = -2t + 1 = 0 \implies t = 1/2 \implies x = \pi/6, 5\pi/6$. Evaluate endpoints and critical points: $f(0) = 1$, $f(\pi) = 1$. $f(\pi/6) = 1 – 1/4 + 1/2 = 5/4$.

Abs Max: $5/4$, Abs Min: $1$ (in the range given, technically min is at endpoints).

Question 12

Show altitude of right circular cone of max volume in sphere radius $r$ is $4r/3$.

Let cone altitude $h$, base radius $R$. Let distance from center to base be $x$. So $h = r+x$.

$R^2 = r^2 – x^2$. $$V = \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi (r^2 – x^2)(r + x)$$ $$\frac{dV}{dx} = \frac{\pi}{3} [(r^2-x^2) + (r+x)(-2x)] = \frac{\pi}{3}(r+x)(r-x – 2x) = \frac{\pi}{3}(r+x)(r-3x)$$ Max when $r – 3x = 0 \implies x = r/3$. Altitude $h = r + r/3 = 4r/3$.

✅ Verified Altitude = $4r/3$.

Question 13

If $f'(x) > 0$ for all $x \in (a, b)$, prove $f$ is increasing on $(a, b)$.

Let $x_1, x_2 \in (a, b)$ such that $x_1 < x_2$. By Mean Value Theorem, there exists $c \in (x_1, x_2)$ such that:

$$f(x_2) – f(x_1) = f'(c)(x_2 – x_1)$$

Since $f'(c) > 0$ and $x_2 – x_1 > 0$, then $f(x_2) – f(x_1) > 0 \implies f(x_2) > f(x_1)$.

Thus, f is strictly increasing.

Question 14

Show height of cylinder of max volume in sphere radius $R$ is $\frac{2R}{\sqrt{3}}$.

Let cylinder height $h$, radius $r$. $r^2 + (h/2)^2 = R^2 \implies r^2 = R^2 – h^2/4$.

$$V = \pi r^2 h = \pi (R^2 – \frac{h^2}{4})h = \pi (R^2h – \frac{h^3}{4})$$ $$\frac{dV}{dh} = \pi (R^2 – \frac{3h^2}{4}) = 0 \implies h^2 = \frac{4R^2}{3} \implies h = \frac{2R}{\sqrt{3}}$$

Height: $\frac{2R}{\sqrt{3}}$. Max Volume: $\frac{4\pi R^3}{3\sqrt{3}}$.

Question 15

Show max volume of cylinder in cone is $\frac{4}{27} \pi h^3 \tan^2 \alpha$.

Let cone height $H$, semi-vertical angle $\alpha$. Base radius $R = H \tan \alpha$. Let cylinder height $h’$, radius $r$.

Using similar triangles: $\frac{r}{R} = \frac{H-h’}{H} \implies h’ = H(1 – \frac{r}{R})$. $$V = \pi r^2 h’ = \pi H (r^2 – \frac{r^3}{R})$$ $$V'(r) = \pi H (2r – \frac{3r^2}{R}) = 0 \implies r = \frac{2R}{3} \implies h’ = \frac{H}{3}$$ Max Vol = $\pi (\frac{2H\tan\alpha}{3})^2 (\frac{H}{3}) = \frac{4}{27} \pi H^3 \tan^2 \alpha$.

✅ Height is 1/3 of cone. Volume verified.

Question 16

Cylindrical tank $r=10$m being filled at 314 m³/h. Rate of depth increase?

$V = \pi r^2 h \implies \frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.

$314 = 3.14 \times (10)^2 \times \frac{dh}{dt} = 314 \times \frac{dh}{dt}$ $$\frac{dh}{dt} = 1 \text{ m/h}$$

Correct Answer: (A) 1 m/h

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