Differentiation (Part-38) | S N De

 

   In the previous three parts, we have discussed RELATION AND MAPPING as a part of our full S.N.Dey Math Solution series. We have solved Exercise : 2A, 2B and 2C. In this chapter , we will discuss Very Short Answer Type Questions, Short Answer Type Questions  and Long Answer Type Questions .  So,  Let's start. 

EXERCISE: 2D

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

Definition of a real valued function:

Let $\,X\,$ and $\,Y\,$ be two non-empty sets of real variables. Then a definite rule $\,f\,$, which associates every $\, x \in X\,$ to a unique $\,f(x)\,$ or $\,\,y\in Y,\,\,$ is called a real valued function on the set of real variables.

The set of values over which the variable $\,x\,$ varies is called the domain of definition of the real function (when $\,f(x)\,$ possesses finite value for every $\,x\,$ over the set) and the set of all values of $\,f(x)\,$ is called the range of the function.

4 (i) Given $\,\,f(x)=3x-9;\,\,$ Find $\,\,f(-1), \,f(3),\, f(x+h), \,f(x^2-1).$

Sol. $\,\,f(x)=3x-9 \\ \Rightarrow f(-1)=3(-1)-9=-12,\\ f(3)=3(3)-9=0\\ f(x+h)=3(x+h)-9\\ f(x^2-1)=3(x^2-1)-9=3x^2-12.$

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

Sol. Since $\,f(x-1)=7x-5,\,\,$ to find $\,f(x)\,$ we replace $\,\,x\,\,$ with $\,\,x+1\,\,$   

 so that $f(\overline{x+1}-1)=7(x+1)-5=7x+2.$

Since $\,f(x-1)=7x-5,\,\,$ to find $\,f(x+2)\,$ we replace $\,\,x\,\,$ with $\,\,x+3\,\,$   

 so that $f(\overline{x+3}-1)=7(x+3)-5=7x+16.$

4(iii) If $\,\,f(x+3)=2x^2-3x-1,\,\,$ find the value of $\,\,f(x+1).$

Sol. If $\,\,f(x+3)=2x^2-3x-1,\,\,$  to find the value of $\,\,f(x+1),\,\,$ we need to replace $\,\,x\,$ by $\,\,(x-2)\,$   

so that we get $\,\,f(\overline{x-2}+3) \\=f(x+1)\\=2(x-2)^2-3(x-2)-1\\=2(x^2-4x+4)-3x+6-1\\=2x^2-11x+13$

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

Sol.  $\,\,f(x+2)=x^2-6x+2, \\ f(-2+2)=(-2)^2-6(-2)+2 \\ [\quad \text{by putting,}\,\,x=-2]\\ \Rightarrow f(0)=4+12+2=18$

Now,   $\,\,f(x+2)=x^2-6x+2, \\ f(0+2)=(0)^2-6(0)+2 \\ [\quad \text{by putting,}\,\,x=0]\\ \Rightarrow f(2)=2$

 $\,\,f(x+2)=x^2-6x+2, \\ f(-4+2)=(-4)^2-6(-4)+2 \\ [\quad \text{by putting,}\,\,x=0]\\ \Rightarrow f(-2)=16+24+2=42$

$\,\,f(x+2)=x^2-6x+2, \\ f(\overline{x-2}+2)=(x-2)^2-6(x-2)+2\\ [\quad \text{by putting,}\,\,x=x-2]\\ \Rightarrow f(x)=x^2-4x+4-6x+12+2\\ \Rightarrow f(x)=x^2-10x+18$

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

Since, $\,\,2f(x)+3f(-x)=15-4x,\cdots(1) \\ 2f(-x)+3f(x)=15+4x\cdots (2) \\ [\,\,\text{Replacing x with (-x)}]$

Now , $\,\,(1) \times 2-(2) \times 3$ gives   

 $\,4f(x)+6f(-x)-6f(-x)-9f(x)=2(15-4x)-3(15+4x) \\ \Rightarrow -5f(x)=-15-20x \\ \Rightarrow f(x)=\frac{-15-20x}{-5}\\ \Rightarrow f(x)=3+4x $

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

Since, $\,\,3f(x)+2f(-x)=5(x-2)\cdots(1) \\ 3f(-x)+2f(x)=5(-x-2)\cdots (2) \\ [\,\,\text{Replacing x with (-x)}]$

Now , $\,\,(1) \times 3-(2) \times 2$ gives   

 $\,9f(x)+6f(-x)-6f(-x)-4f(x)\\=3(5x-10)-2(-5x-10) \\ \Rightarrow 5f(x)=15x-30+10x+20\\ \Rightarrow f(x)=\frac{25x-10}{5}\\ \Rightarrow f(x)=5x-2 \\ \Rightarrow 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. $\,\,f(x)=2x^2-3x+5, \\  \therefore f(a+h)=2(a+h)^2-3(a+h)+5 \\ \Rightarrow f(a+h)=2a^2+4ah+2h^2-3a-3h+5 \\ \therefore f(a+h)-f(a)\\=2a^2+4ah+2h^2-3a-3h+5\\-(2a^2-3a+5)\\=h(4a+2h-3) \\ \Rightarrow \frac{f(a+h)-f(a)}{h}=4a+2h-3$

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

Sol. $\,\,\phi(x)=\frac{1-x}{1+x} \\ \therefore \phi(\cos{2 \theta})\\=\frac{1-\cos{2 \theta}}{1+\cos{2 \theta}} \\ =\frac{2 \sin^2\theta}{2 \cos^2\theta}\\= \tan^2\theta$

8(i) If $\,\,f(x)=5^x,\,\,$ prove that $\, f(x+2)=25f(x)$

Sol. $\,\,f(x+2)=5^{x+2}\\=5^x.5^2\\=25 \times 5^x\\=25f(x)$

8(ii) If $\,\,f(x)=5^x,\,\,$ prove that $\, f(x+y)=f(x).f(y)$

Sol. $\,\,f(x+y)=5^{x+y}\\=5^x.5^y\\=f(x).f(y)$

8(iii) If $\,\,f(x)=5^x,\,\,$ prove that $\, \frac{f(x+1)}{f(x-1)}=25$

Sol. $\,\,f(x+1)=5^{x+1}=5^x.5=5f(x)\\ f(x-1)=5^{x-1}\\=5^x.5^{-1}\\=f(x).5^{-1}$

So, $\, \frac{f(x+1)}{f(x-1)}\\=\frac{5f(x)}{f(x).5^{-1}}\\=\frac{5}{5^{-1}}\\=5 \times 5=25$

8(iv) If $\,\,f(x)=5^x,\,\,$ prove that $\, f(\log_5 x)=x$

Sol. $\,\,f(\log_5 x)=5^{\log_5 x}=x$

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

$f(a)+f(b)=f(a+b)$

Sol. By the given problem, we get  

$\,\,f(a)=a, \,\,f(b)=b \cdots(1)\\f(a+b)=a.\frac{a+b-b}{a-b}+b.\frac{a+b-a}{b-a} \\=\frac{a^2}{a-b}+\frac{b^2}{b-a}\\=\frac{a^2}{a-b}-\frac{b^2}{a-b}\\=\frac{a^2-b^2}{a-b}\\=\frac{(a+b)(a-b)}{a-b}\\=a+b \cdots(2)$

So, from (1) and (2) we get, $\,\,f(a+b)=f(a)+f(b)$  

Hence  completes the result.

10(i). If $\,\,\phi(x)=\log_e x,\,\,$ show that  $\,\,\phi(e^x)=x.$

Sol. $\,\,\phi(x)=\log_e x \\ \implies \phi(e^x)=\log_e (e^x)\\ \implies \phi(e^x)=x\log_e e=x.1=x$

10(ii). If $\,\,\phi(x)=\log_e x,\,\,$ show that  $\,\,\phi(x^m)=m\phi(x).$

Sol. $\,\,\phi(x)=\log_e x \\ \implies \phi(x^m)=\log_e (x^m)\\ \implies \phi(x^m)=m\log_e x=m.\phi(x)$

10(iii). If $\,\,\phi(x)=\log_e x,\,\,$ show that  $\,\,\phi(xy)=\phi(x)+\phi(y).$

Sol. $\,\,\phi(x)=\log_e x \\ \Rightarrow \phi(xy)=\log_e(xy) \\ \Rightarrow \phi(xy)=\log_e(x)+\log_e(y)=\phi(x)+\phi(y) $

10(iv). If $\,\,\phi(x)=\log_e x,\,\,$ show that  $\,\,\phi(x/y)=\phi(x)-\phi(y).$

Sol. $\,\,\phi(x)=\log_e x \\ \Rightarrow \phi(x/y)=\log_e(x/y) \\ \Rightarrow \phi(x/y)=\log_e(x)-\log_e(y)=\phi(x)-\phi(y) $

11. If $\,\,f(x)=e^{px+q},\,\,p,q \,$ are constants, Show that $\,\,f(a).f(b).f(c)=f(a+b+c).e^{2q}$

Sol. Since $\,\,f(x)=e^{px+q},\,\,p,q \,$ are constants,  

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

Now , $f(a+b+c).e^{2q}\\=e^{p(a+b+c)+q}.e^{2q}\\=e^{ap}.e^{bp}.e^{cp}.e^q.e^{q+q}\\=e^{ap+q}.e^{bp+q}.e^{cp+q}\\=f(a).f(b).f(c)$

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

Sol. $\,\,f(x)=|x|-2x \\ f(-1)=|-1|-2(-1)=1+2=3 \\ f(1)=|1|-2(1)=1-2=-1 \\ f(-h)=|-h|-2(-h)=|h|+2h$

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

Sol. First we calculate $\,\,\,x-a,\,x-b. $   

We see $\,\,x-a=\frac{a+b}{2}-a=\frac{a+b-2a}{2}=\frac{b-a}{2} \\ x-b=\frac{a+b}{2}-b=\frac{a+b-2b}{2}=\frac{a-b}{2}$

$\,\,g(x)=\frac{x-a}{x}+\frac{x}{x-b}, \\ \Rightarrow g(\frac{a+b}{2})=\frac{b-a}{2x}+\frac{2x}{a-b}=\frac{b-a}{a+b}+\frac{a+b}{a-b} \\ [x=\frac{a+b}{2} \Rightarrow 2x=a+b] \\  \Rightarrow g(\frac{a+b}{2})=\frac{-(a-b)^2+(a+b)^2}{a^2-b^2}=\frac{4ab}{a^2-b^2}$

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

Sol. $\,\,f(x)=\frac{1}{x^2}, \\ \Rightarrow f(x+h)=\frac{1}{(x+h)^2}, \cdots (1)\\ f(x-h)=\frac{1}{(x-h)^2}\cdots (2)$  

Hence, from (1) and (2), we get ,   

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

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

Sol. We know, $\,\tan(\pi/4-\theta)\\=\frac{1-\tan \theta}{1+\tan \theta}\\ g(\pi/4-\theta)=\frac{1-\tan(\pi/4-\theta)}{1+\tan(\pi/4-\theta)}\\=\frac{1-\frac{1-\tan \theta}{1+\tan \theta}}{1+\frac{1-\tan \theta}{1+\tan \theta}}\\=\frac{(1+\tan \theta)-(1-\tan \theta)}{(1+\tan \theta)+(1-\tan \theta)} \\ [\text{multiplying numerator and denominator by} \\ (1+\tan \theta)]\\=\frac{2 \tan \theta}{2}\\=\tan \theta$

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

Sol.  We first calculate , $\,\, 1+2x/(1+x^2)=\frac{1+x^2+2x}{1+x^2}, \cdots (1)\\ 1-2x/(1+x^2)=\frac{1+x^2-2x}{1+x^2}\cdots (2)$

$\,f(2x/(1+x^2))\\=\log_e \frac{1+x^2+2x}{1+x^2-2x},\\ [\text{Using (1),(2)}]\\=\log_e \frac{(1+x)^2}{(1-x)^2}\\=2 \log_e\frac{1+x}{1-x}\\=2 f(x)$ 

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

Sol. $\,\,f(g(x))\\=\log_e(g(x))\\=\log_e(e^x)\\=x\log_e e\\=x \cdots (1)$

Again, $\,\,g(f(x))\\=e^{f(x)}\\=e^{\log_e x}\\= x\cdots (2)$ 

Hence , from (1) and (2), the result follows.

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

Sol. $\, f(x)=16 \\ \Rightarrow 10x^2-13x+13=16 \\ \Rightarrow 10x^2-13x+13-16=0 \\ \Rightarrow 10x^2-13x-3=0 \\ \Rightarrow 10x^2-15x+2x-3=0 \\ \Rightarrow 5x(2x-3)+1(2x-3)=0 \\ \Rightarrow (2x-3)(5x+1)=0 \\ \Rightarrow x= \frac32,-\frac15. $

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

Sol. $\,\,f(3.5)=4[3.5]-3|3.5|\\=4 \times 3 -3 \times 3.5\\=12-10.5\\=1.5$

Again, $\,\,f(-3.5)=4[-3.5]-3|-3.5|\\=4 \times (-4) -3 \times 3.5\\=-16-10.5\\=-26.5$

20. If $\,\,\mathbb R\,$ be the set of real numbers and $\,f(x)=|x|,\,g(x)=x,\,\,$ find the product function $\,\,fg.$

Sol. We know, $\,|x|=x,\,\,x \geq 0, \\ ~~~~~= -x, x<0.$ 

So, $\,\,f(x).g(x)=|x|.x \\~~~~~~~~~~~~~~~~~~=x^2, x\geq 0 \\~~~~~~~~~~~~~~~~~~=-x^2, x<0.$

21. Find the value of $\,\,[-3]+[-3.6]-|2.6+|-3|$

Sol. $\,\,[-3]+[-3.6]-|2.6+|-3|\\=-3-4-2.6+3\\=-6.6$

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