Differentiation (Part-38) | S N De

In the previous article, we discussed Very Short Answer Type Questions of the Relation and Mapping Chapter (Questions 1–21) from the S.N. Dey Math Exercise. In this post, we begin solving Short Answer Type Questions (Questions 1–12) as part of our step-by-step S.N. Dey Math solution series.

Question 1

A right circular cylinder is inscribed in a sphere of radius $r$. Express the volume $v$ of the cylinder as a function of its height $x$.

Solution

Let the radius of the circular cylinder be $AB=R$ and the center of the sphere be $O$.

The height of the cylinder is $x$ (given). Thus, $OB=\frac{x}{2}$.

Now, in right triangle $ABO$:

\[ AO^2 = AB^2 + OB^2 \] \[ \Rightarrow r^2 = R^2 + \left(\frac{x}{2}\right)^2 \] \[ \Rightarrow R^2 = r^2 - \frac{x^2}{4} \]

The volume of the cylinder is $v = \pi R^2 x$:

\[ \Rightarrow v = \pi \left(r^2 - \frac{x^2}{4}\right)x \] \[ \Rightarrow v = \frac{\pi x}{4}(4r^2 - x^2) \]

Hence, the volume $v$ of the cylinder expressed as a function of its height $x$ is: \[ v = \frac{\pi x}{4}(4r^2 - x^2) \]

Question 2

A right circular cone is inscribed in a sphere of radius $a$. Express the volume $v$ of the cone as a function of its slant height $x$.

Solution

Let $O$ be the center of the sphere, and let the base radius of the cone be $BD=r$.

For $\Delta ABD$, the height $h = AD = \sqrt{AB^2 - BD^2}$:

\[ AD = \sqrt{x^2 - r^2} \quad \dots (1) \]

For $\Delta BOD$, $OD = \sqrt{OB^2 - BD^2}$:

\[ OD = \sqrt{a^2 - r^2} \quad \dots (2) \]

Since $AD = AO + OD$:

\[ \sqrt{x^2 - r^2} = a + \sqrt{a^2 - r^2} \]

Squaring both sides:

\[ x^2 - r^2 = a^2 + 2a\sqrt{a^2 - r^2} + a^2 - r^2 \] \[ \Rightarrow 2a\sqrt{a^2 - r^2} = x^2 - 2a^2 \]

Squaring again:

\[ 4a^2(a^2 - r^2) = (x^2 - 2a^2)^2 \] \[ 4a^4 - 4a^2 r^2 = x^4 - 4a^2 x^2 + 4a^4 \] \[ \Rightarrow r^2 = \frac{4a^2 x^2 - x^4}{4a^2} \]

Now, evaluating $h$:

\[ h = AD = \sqrt{x^2 - r^2} = \sqrt{x^2 - \frac{4a^2 x^2 - x^4}{4a^2}} = \frac{x^2}{2a} \]

Therefore, the volume of the cone $v = \frac{1}{3}\pi r^2 h$ becomes:

\[ v = \frac{1}{3}\pi \left(\frac{4a^2 x^2 - x^4}{4a^2}\right)\left(\frac{x^2}{2a}\right) \] \[ \Rightarrow v = \frac{\pi x^4}{24a^3}(4a^2 - x^2) \]

Question 4

If $F(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$, show that $F(0) = 1$.

Solution

Substituting $x = 0$ into $F(x)$:

\[ F(0) = \frac{(0-b)(0-c)}{(a-b)(a-c)} + \frac{(0-c)(0-a)}{(b-c)(b-a)} + \frac{(0-a)(0-b)}{(c-a)(c-b)} \] \[ = \frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)} \]

Taking a common denominator $(a-b)(b-c)(c-a)$:

\[ = -\frac{bc(b-c) + ca(c-a) + ab(a-b)}{(a-b)(b-c)(c-a)} = -\frac{p}{(a-b)(b-c)(c-a)} \quad \dots (1) \]

where $p = bc(b-c) + ca(c-a) + ab(a-b)$.

Expanding and factoring $p$:

\[ p = bc(b-c) + c^2 a - ca^2 + a^2 b - ab^2 \] \[ = bc(b-c) - a(b^2 - c^2) + a^2(b-c) \] \[ = (b-c)\{bc - a(b+c) + a^2\} \] \[ = (b-c)(bc - ab - ac + a^2) \] \[ = (b-c)\{b(c-a) - a(c-a)\} \] \[ = -(b-c)(c-a)(a-b) \quad \dots (2) \]

From (1) and (2):

\[ F(0) = \frac{-( -(b-c)(c-a)(a-b) )}{(a-b)(b-c)(c-a)} = 1 \]

Hence proved.

Question 5 (i)

If $f(x) = \frac{x-1}{x+1}$, show that $\frac{f(a) - f(b)}{1 + f(a)f(b)} = \frac{a-b}{1+ab}$.

Solution

First, calculate $f(a) - f(b)$:

\[ f(a) - f(b) = \frac{a-1}{a+1} - \frac{b-1}{b+1} \] \[ = \frac{(a-1)(b+1) - (b-1)(a+1)}{(a+1)(b+1)} \] \[ = \frac{(ab + a - b - 1) - (ab + b - a - 1)}{(a+1)(b+1)} \] \[ = \frac{2(a-b)}{(a+1)(b+1)} \quad \dots (1) \]

Next, calculate $1 + f(a)f(b)$:

\[ 1 + f(a)f(b) = 1 + \left(\frac{a-1}{a+1}\right)\left(\frac{b-1}{b+1}\right) \] \[ = \frac{(a+1)(b+1) + (a-1)(b-1)}{(a+1)(b+1)} \] \[ = \frac{(ab + a + b + 1) + (ab - a - b + 1)}{(a+1)(b+1)} \] \[ = \frac{2(ab + 1)}{(a+1)(b+1)} \quad \dots (2) \]

Dividing (1) by (2):

\[ \frac{f(a) - f(b)}{1 + f(a)f(b)} = \frac{\frac{2(a-b)}{(a+1)(b+1)}}{\frac{2(ab+1)}{(a+1)(b+1)}} = \frac{a-b}{1+ab} \]

Hence proved.

Question 5 (ii)

If $f(x) = \frac{1}{1-x}$, show that $f[f\{f(x)\}] = x$.

Solution

Evaluating step-by-step:

\[ f(f(x)) = \frac{1}{1 - \frac{1}{1-x}} = \frac{1}{\frac{1-x-1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} \]

Now, substitute this into $f(x)$ once more:

\[ f[f\{f(x)\}] = f\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}} = \frac{1}{\frac{x - (x-1)}{x}} = \frac{x}{1} = x \]

Hence proved.

Question 6

If $y = f(x) = \frac{ax-b}{bx-a}$, prove that $f(x) \cdot f(1/x)$ is independent of $x$.

Solution

Evaluate $f(1/x)$:

\[ f(1/x) = \frac{a(1/x) - b}{b(1/x) - a} = \frac{\frac{a - bx}{x}}{\frac{b - ax}{x}} = \frac{a - bx}{b - ax} = \frac{bx - a}{ax - b} \]

Now, multiply $f(x)$ and $f(1/x)$:

\[ f(x) \cdot f(1/x) = \left(\frac{ax - b}{bx - a}\right) \times \left(\frac{bx - a}{ax - b}\right) = 1 \]

Since the result is $1$ (a constant), it is independent of $x$. Hence proved.

Question 7

If $y = f(x) = \frac{px+q}{rx-p}$, show that $x = f(y)$.

Solution

We are given:

\[ y = \frac{px+q}{rx-p} \] \[ \Rightarrow y(rx-p) = px+q \] \[ \Rightarrow rxy - py = px+q \] \[ \Rightarrow rxy - px = py+q \] \[ \Rightarrow x(ry-p) = py+q \] \[ \Rightarrow x = \frac{py+q}{ry-p} \]

Observing the expression for $f(x)$, we see that $f(y) = \frac{py+q}{ry-p}$. Therefore, $x = f(y)$. Hence proved.

Question 8

If $y = f(x) = \frac{3x-5}{2x-m}$ and $f(y) = x$, find $m$.

Solution

Given $y = f(x) = \frac{3x-5}{2x-m}$ and $f(y) = x$:

\[ \frac{3y-5}{2y-m} = x \]

Substitute $y = \frac{3x-5}{2x-m}$ into the equation:

\[ \frac{3\left(\frac{3x-5}{2x-m}\right) - 5}{2\left(\frac{3x-5}{2x-m}\right) - m} = x \] \[ \frac{9x - 15 - 5(2x-m)}{2(3x-5) - m(2x-m)} = x \] \[ \frac{9x - 15 - 10x + 5m}{6x - 10 - 2mx + m^2} = x \] \[ -x + 5m - 15 = x(6x - 10 - 2mx + m^2) \] \[ -x + 5m - 15 = (6 - 2m)x^2 + (m^2 - 10)x \] \[ \Rightarrow (6 - 2m)x^2 + (m^2 - 9)x + (15 - 5m) = 0 \quad \dots (1) \]

For equation (1) to hold true for all values of $x$, the coefficients must vanish:

\[ 6 - 2m = 0 \implies m = 3 \] \[ m^2 - 9 = 0 \implies m = \pm 3 \] \[ 15 - 5m = 0 \implies m = 3 \]

Therefore, $m = 3$.

Download the complete solution eBook for Relation & Mapping (Class 11)

Get eBook Now

Question 9

If $y = f(x) = \frac{x-3}{2x+1}$ and $z = f(y)$, express $z$ as a function of $x$.

Solution

Given $z = f(y) = \frac{y-3}{2y+1}$:

Substituting $y = \frac{x-3}{2x+1}$ into the equation:

\[ z = \frac{\frac{x-3}{2x+1} - 3}{2\left(\frac{x-3}{2x+1}\right) + 1} \] \[ = \frac{(x-3) - 3(2x+1)}{2(x-3) + (2x+1)} \] \[ = \frac{x - 3 - 6x - 3}{2x - 6 + 2x + 1} \] \[ = \frac{-5x - 6}{4x - 5} = -\frac{5x + 6}{4x - 5} \]

Question 10

If $F(x) = \frac{4x-5}{3x-4}$, prove that $F\{F(x)\} = x$.

Solution \[ F\{F(x)\} = F\left(\frac{4x-5}{3x-4}\right) \] \[ = \frac{4\left(\frac{4x-5}{3x-4}\right) - 5}{3\left(\frac{4x-5}{3x-4}\right) - 4} \] \[ = \frac{4(4x-5) - 5(3x-4)}{3(4x-5) - 4(3x-4)} \] \[ = \frac{16x - 20 - 15x + 20}{12x - 15 - 12x + 16} \] \[ = \frac{x}{1} = x \]

Hence proved.

Question 11

If $f(x) = (a - x^n)^{1/n}$, where $a > 0$ and $n$ is a positive integer, show that $f(f(x)) = x$.

Solution \[ f(f(x)) = f\left((a - x^n)^{1/n}\right) \] \[ = \left(a - \left((a - x^n)^{1/n}\right)^n\right)^{1/n} \] \[ = (a - (a - x^n))^{1/n} \] \[ = (a - a + x^n)^{1/n} \] \[ = (x^n)^{1/n} = x \]

Hence proved.

Question 12 (i)

If $\phi(x) = \frac{1-x}{1+x}$, prove that $\phi(\phi(x)) = x$.

Solution \[ \phi(\phi(x)) = \phi\left(\frac{1-x}{1+x}\right) \] \[ = \frac{1 - \frac{1-x}{1+x}}{1 + \frac{1-x}{1+x}} \] \[ = \frac{(1+x) - (1-x)}{(1+x) + (1-x)} \] \[ = \frac{2x}{2} = x \]

Hence proved.

Question 12 (ii)

If $\phi(x) = \frac{1-x}{1+x}$, prove that $\phi(\phi(\cot\theta)) = \cot\theta$.

Solution \[ \phi(\phi(\cot\theta)) = \phi\left(\frac{1-\cot\theta}{1+\cot\theta}\right) \] \[ = \frac{1 - \frac{1-\cot\theta}{1+\cot\theta}}{1 + \frac{1-\cot\theta}{1+\cot\theta}} \] \[ = \frac{(1+\cot\theta) - (1-\cot\theta)}{(1+\cot\theta) + (1-\cot\theta)} \] \[ = \frac{2\cot\theta}{2} = \cot\theta \]

Hence proved.

To continue practicing further solutions, click Next: Part 6 →