Differentiation (Part-38) | S N De

In the previous three parts, we discussed Relation and Mapping as part of our comprehensive S.N. Dey Math Solution Series. We have already solved Exercise 2A, 2B, and 2C.

In this article, we will solve problems covering Very Short Answer Type Questions for Exercise 2D. Let us dive in!

EXERCISE: 2D (Very Short Answer Questions)

1.

Define a real valued function. What do you mean by domain of definition and range of a real function?

Sol.

Definition: Let $X$ and $Y$ be two non-empty sets of real variables. A definite rule $f$ associating every $x \in X$ to a unique $f(x) \in Y$ is called a real valued function.

The set of values over which $x$ varies is the domain, and the set of all resulting $f(x)$ values is the range.

4(i)

Given $f(x) = 3x - 9$; find $f(-1), f(3), f(x+h), f(x^2-1)$.

Sol.

Given $f(x) = 3x - 9$:

• $f(-1) = 3(-1) - 9 = -12$
• $f(3) = 3(3) - 9 = 0$
• $f(x+h) = 3(x+h) - 9$
• $f(x^2-1) = 3(x^2-1) - 9 = 3x^2 - 12$

4(ii)

If $f(x-1) = 7x - 5$, find $f(x)$ and $f(x+2)$.

Sol.

To find $f(x)$, replace $x$ with $(x+1)$:

$$f((x+1)-1) = 7(x+1) - 5 \implies f(x) = 7x + 2$$

To find $f(x+2)$, replace $x$ with $(x+3)$:

$$f((x+3)-1) = 7(x+3) - 5 \implies f(x+2) = 7x + 16$$

4(iii)

If $f(x+3) = 2x^2 - 3x - 1$, find $f(x+1)$.

Sol.

Replace $x$ with $(x-2)$:

$$f(x+1) = 2(x-2)^2 - 3(x-2) - 1$$

$$= 2(x^2 - 4x + 4) - 3x + 6 - 1 = 2x^2 - 11x + 13$$

4(iv)

Given $f(x+2) = x^2 - 6x + 2$, find $f(0), f(2), f(-2), f(x)$.

Sol.

• Putting $x = -2$: $f(0) = (-2)^2 - 6(-2) + 2 = 18$
• Putting $x = 0$: $f(2) = (0)^2 - 6(0) + 2 = 2$
• Putting $x = -4$: $f(-2) = (-4)^2 - 6(-4) + 2 = 42$
• Replacing $x$ with $(x-2)$:
  $$f(x) = (x-2)^2 - 6(x-2) + 2$$
  $$= x^2 - 4x + 4 - 6x + 12 + 2$$
  $$= x^2 - 10x + 18$$

5(i)

If $2f(x) + 3f(-x) = 15 - 4x$, prove that $f(x) = 3 + 4x$.

Sol.

Given: $2f(x) + 3f(-x) = 15 - 4x \quad \dots(1)$

Replace $x$ with $-x$: $2f(-x) + 3f(x) = 15 + 4x \quad \dots(2)$

Perform $2\times(1) - 3\times(2)$:

$$4f(x) - 9f(x) = 2(15-4x) - 3(15+4x)$$

$$-5f(x) = -15 - 20x \implies f(x) = 3 + 4x$$

5(ii)

If $3f(x) + 2f(-x) = 5(x-2)$, show that $f(1) = 3$.

Sol.

Given: $3f(x) + 2f(-x) = 5x - 10 \quad \dots(1)$

Replace $x$ with $-x$: $3f(-x) + 2f(x) = -5x - 10 \quad \dots(2)$

Perform $3\times(1) - 2\times(2)$:

$$9f(x) - 4f(x) = 3(5x-10) - 2(-5x-10)$$

$$5f(x) = 25x - 10 \implies f(x) = 5x - 2$$

Thus, $f(1) = 5(1) - 2 = 3$.

6.

If $f(x) = 2x^2 - 3x + 5$, find $f(a+h)$ and $\frac{f(a+h) - f(a)}{h}$.

Sol.

$$f(a+h) = 2(a+h)^2 - 3(a+h) + 5 = 2a^2 + 4ah + 2h^2 - 3a - 3h + 5$$

$$f(a+h) - f(a) = h(4a + 2h - 3)$$

$$\frac{f(a+h) - f(a)}{h} = 4a + 2h - 3$$

7.

If $\phi(x) = \frac{1-x}{1+x}$, show that $\phi(\cos 2\theta) = \tan^2 \theta$.

Sol.

$$\phi(\cos 2\theta) = \frac{1-\cos 2\theta}{1+\cos 2\theta} = \frac{2\sin^2 \theta}{2\cos^2 \theta} = \tan^2 \theta$$

8.

If $f(x) = 5^x$, prove that:
(i) $f(x+2) = 25f(x)$ | (ii) $f(x+y) = f(x)f(y)$
(iii) $\frac{f(x+1)}{f(x-1)} = 25$ | (iv) $f(\log_5 x) = x$

Sol.

(i) $f(x+2) = 5^{x+2} = 5^x \cdot 5^2 = 25f(x)$

(ii) $f(x+y) = 5^{x+y} = 5^x \cdot 5^y = f(x)f(y)$

(iii) $\frac{f(x+1)}{f(x-1)} = \frac{5^{x+1}}{5^{x-1}} = 5^{(x+1)-(x-1)} = 5^2 = 25$

(iv) $f(\log_5 x) = 5^{\log_5 x} = x$

9.

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

Sol.

Direct substitution gives $f(a) = a$ and $f(b) = b$, so $f(a)+f(b)=a+b \quad \dots(1)$

$$f(a+b) = \frac{a^2}{a-b} + \frac{b^2}{b-a} = \frac{a^2-b^2}{a-b} = a+b \quad \dots(2)$$

From (1) and (2), proved.

10.

If $\phi(x) = \log_e x$, show that:
(i) $\phi(e^x)=x$ | (ii) $\phi(x^m)=m\phi(x)$
(iii) $\phi(xy)=\phi(x)+\phi(y)$ | (iv) $\phi(x/y)=\phi(x)-\phi(y)$

Sol.

(i) $\log_e(e^x) = x$

(ii) $\log_e(x^m) = m\log_e x = m\phi(x)$

(iii) $\log_e(xy) = \log_e x + \log_e y = \phi(x)+\phi(y)$

(iv) $\log_e(x/y) = \log_e x - \log_e y = \phi(x)-\phi(y)$

11.

If $f(x) = e^{px+q}$, show that $f(a)f(b)f(c) = f(a+b+c)e^{2q}$.

Sol.

$$f(a)f(b)f(c) = e^{ap+q}e^{bp+q}e^{cp+q} = e^{p(a+b+c)+3q}$$

$$f(a+b+c)e^{2q} = e^{p(a+b+c)+q} \cdot e^{2q} = e^{p(a+b+c)+3q}$$

Hence proved.

12.

If $f(x) = |x| - 2x$, find $f(-1), f(1), f(-h)$.

Sol.

• $f(-1) = |-1| - 2(-1) = 1 + 2 = 3$
• $f(1) = |1| - 2(1) = 1 - 2 = -1$
• $f(-h) = |-h| - 2(-h) = |h| + 2h$

13.

If $g(x) = \frac{x-a}{x} + \frac{x}{x-b}$, prove that $g\left(\frac{a+b}{2}\right) = \frac{4ab}{a^2 - b^2}$.

Sol.

Let $x = \frac{a+b}{2}$. Then $x-a = \frac{b-a}{2}$ and $x-b = \frac{a-b}{2}$.

$$g\left(\frac{a+b}{2}\right) = \frac{b-a}{a+b} + \frac{a+b}{a-b} = \frac{-(a-b)^2+(a+b)^2}{a^2-b^2} = \frac{4ab}{a^2-b^2}$$

14.

If $f(x) = \frac{1}{x^2}$, show that $f(x+h) - f(x-h) = -\frac{4xh}{(x^2 - h^2)^2}$.

Sol.

$$f(x+h)-f(x-h) = \frac{1}{(x+h)^2} - \frac{1}{(x-h)^2} = \frac{(x-h)^2-(x+h)^2}{(x^2-h^2)^2} = -\frac{4xh}{(x^2-h^2)^2}$$

15.

If $g(\theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$, find $g(\pi/4 - \theta)$.

Sol.

Note $\tan(\pi/4-\theta) = \frac{1-\tan\theta}{1+\tan\theta}$.

$$g(\pi/4-\theta) = \frac{1-\frac{1-\tan\theta}{1+\tan\theta}}{1+\frac{1-\tan\theta}{1+\tan\theta}} = \frac{2\tan\theta}{2} = \tan\theta$$

16.

If $f(x) = \log_e \frac{1+x}{1-x}$, show that $f\left(\frac{2x}{1+x^2}\right) = 2f(x)$.

Sol.

$$1+\frac{2x}{1+x^2}=\frac{(1+x)^2}{1+x^2}, \quad 1-\frac{2x}{1+x^2}=\frac{(1-x)^2}{1+x^2}$$

$$f\left(\frac{2x}{1+x^2}\right) = \log_e\left(\frac{(1+x)^2}{(1-x)^2}\right) = 2\log_e\frac{1+x}{1-x} = 2f(x)$$

17.

If $f(x)=\log_e x$ and $g(x)=e^x$, prove $f(g(x))=g(f(x))$.

Sol.

$$f(g(x)) = \log_e(e^x) = x$$

$$g(f(x)) = e^{\log_e x} = x$$

Hence proved.

18.

If $f(x)=10x^2-13x+13$, solve $f(x)=16$.

Sol.

$$10x^2-13x-3=0 \implies (2x-3)(5x+1)=0$$

$\therefore x = \frac{3}{2}, -\frac{1}{5}$

19.

If $f(x)=4[x]-3|x|$, find $f(3.5)$ and $f(-3.5)$.

Sol.

$$f(3.5)=4[3.5]-3|3.5|=4(3)-3(3.5)=1.5$$

$$f(-3.5)=4[-3.5]-3|-3.5|=4(-4)-3(3.5)=-26.5$$

20.

If $f(x)=|x|, g(x)=x$, find product function $fg$.

Sol.

$$(fg)(x)=|x|\cdot x = \begin{cases} x^2 & x \ge 0 \\ -x^2 & x < 0 \end{cases}$$

21.

Find value of $[-3]+[-3.6]-|2.6+|-3||$.

Sol.

$$[-3]=-3, [-3.6]=-4, |-3|=3$$

$$|2.6+3|=5.6$$

$$\text{Value} = -3+(-4)-5.6 = -12.6$$

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