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Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.2
  • Last Updated : 25 Jan, 2021

Question 1. If f(x)=x2-3x+4, then find the values of x satisfying the equation f(x)=f(2x+1).

Solution:

We have, 

f(x)=x2-3x+4 

Now,

f(2x+1)=(2x+1)2-3(2x+1)+4



f(2x+1)=4x2+1+4x-6x-3+4

f(2x+1)=4x2-2x+2

It is given that 

f(x)=f(2x+1)

x2-3x+4=4x2-2x+2

0=4x2-x2-2x+3x+2-4

3x2+x-2=0

3x2+3x-2x-2=0

3x(x+1)-2(x+1)=0

(x+1)(3x-2)=0

x+1=0 or 3x-2=0

x=-1 or x=2/3

Therefore, the value of x are -1 and 2/3.

Question 2. If f(x)=(x-a)2 (x-b)2 ,find f(a+b).

Solution: 

We have, 

f(x)=(x-a)2(x-b)2

Now, let us find f(a+b)

f(a+b)=(a+b-a)2(a+b-b)2



f(a+b)=b2a2

Therefore, f(a+b)=(ba)2

Question 3. If y=f(x)=(ax-b)/(bx-a), show that x=f(y).

Solution:

Given,

y=f(x)=(ax-b)/(bx-a)

As we know, f(y)=(ay-b)/(by-a)

Let us prove that x=f(y).

We have,

y=(ax-b)/(bx-a)

By cross multiplying,

y(bx-a)=ax-b

bxy-ay=ax-b

bxy-ax=ay-b

x(by-a)=ay-b

x=(ay-b)/(by-a)

Therefore, x=f(y)

Hence proved.

Question 4.  If f (x) = 1 / (1 – x), show that f [f {f (x)}] = x.

Solution:

Given:

f (x) = 1 / (1 – x)

Let us prove that f [f {f (x)}] = x.

Firstly, let us solve for f {f (x)}.

f {f (x)} = f {1/(1 – x)}

= 1 / 1 – (1/(1 – x))

= 1 / [(1 – x – 1)/(1 – x)]

= 1 / (-x/(1 – x))

= (1 – x) / -x

= (x – 1) / x

∴ f {f (x)} = (x – 1) / x

Now, we shall solve for f [f {f (x)}]



f [f {f (x)}] = f [(x-1)/x]

= 1 / [1 – (x-1)/x]

= 1 / [(x – (x-1))/x]

= 1 / [(x – x + 1)/x]

= 1 / (1/x)

∴ f [f {f (x)}] = x

Hence proved.

Question 5. If f (x) = (x + 1) / (x – 1), show that f [f (x)] = x.

Solution:

Given:

f (x) = (x + 1) / (x – 1)

We have to prove that f [f (x)] = x.

f [f (x)] = f [(x+1)/(x-1)]

= [(x+1)/(x-1) + 1] / [(x+1)/(x-1) – 1]

= [[(x+1) + (x-1)]/(x-1)] / [[(x+1) – (x-1)]/(x-1)]

= [(x+1) + (x-1)] / [(x+1) – (x-1)]

= (x+1+x-1)/(x+1-x+1)

= 2x/2

= x

∴ f [f (x)] = x

Hence proved.

Question 6. If 

f(x) =    \begin{cases}       x^2,\text{when } x<0\\      x, \text{when }0\leq x<1\\      \frac{1}{x},\text{when }x\geq1    \end{cases}

Find:

(i) f (1/2)

(ii) f (-2)

(iii) f (1)

(iv) f (√3)

(v) f (√-3)

Solution:

(i) f (1/2)

When, 0 ≤ x ≤ 1, f(x) = x

∴ f (1/2) = 1/2

(ii) f (-2)

When, x < 0, f(x) = x2

f (–2) = (–2)2

= 4

∴ f (–2) = 4

(iii) f (1)

When, x ≥ 1, f (x) = 1/x  

f (1) = 1/1

∴ f(1) = 1

(iv) f (√3)

We have √3 = 1.732 > 1  

When, x ≥ 1, f (x) = 1/x  

∴ f (√3) = 1/√3

(v) f (√-3)

We know √-3 is not a real number and the function f(x) is defined only when x ∈ R.

∴ f (√-3) does not exist.

Question 7. If f(x)=x3-(1/x3), show that f(x)+f(1/x)=0.

Solution: 

We have,

f(x)=x3-(1/x)3              —(i)

Now,

f(1/x)=(1/x)3-(1/(1/x)3)

f(1/x)=(1/x)3-x       —(ii)

Adding equation (i) and (ii), we get 

f(x)+f(1/x) = (x3-1/x3)+(1/x3-x3

f(x)+f(1/x)=x3-x3+1/x3-1/x3

f(x)+f(1/x)=0    

Hence, proved.

Question 8. If f(x)= 2x/(1+x2),show that f(tan θ)=sin 2θ.

Solution. 

We have,

f(x)=2x/(1+x2)

Now,

f(tan θ)=2(tan θ)/(1+tan2 θ)

f(tan θ)=sin 2θ    (Because, sin 2θ = 2(tan θ)/(1+tan2θ))

Hence, proved.

Question 9. If f(x)=(x-1)/(x+1), then show that 

i) f(1/x)=-f(x) 

ii) f(1/(-x))=-1/f(x)

Solution.

i) We have, 

f(x)=(x-1)/(x+1)

Now, 

f(1/x)=((1/x)-1)/((1/x)+1)

f(1/x)=((1-x)/x)/((1+x)/x)

f(1/x)=(1-x)/(1+x)=f(-x)

Hence, proved.

ii) We have, 

f(x)=(x-1)/(x+1)

Now, 

f(1/(-x))= ((1/(-x))-1)/((1/(-x))+1)

f(1/(-x))=((1+x)/(-x))/((1-x)/(-x))

f(1/(-x))=(1+x)/(1-x)

f(1/(-x))=(-1)/((x-1)/(x+1))

f(1/(-x))=-1/f(x)

Hence, proved.

Question 10. If f(x)=(a-xn)1/n ,a>0 and n ∈ N, then prove that f(f(x))=x for all x.

Solution.

We have, 

f(x)=(a-x^n)^\frac{1}{n}, a>0\\ \newline \\ Now, \\ f(f(x))=f(a-x^n)^\frac{1}{n}\\ f(f(x))=[a-\{(a-x^n)^\frac{1}{n}\}^n]^\frac{1}{n} \\ f(f(x))=[a-(a-x^n)]^\frac{1}{n}\\ f(f(x))=[a-a+x^n]^\frac{1}{n}\\ f(f(x))=(x^n)^\frac{1}{n}\\ f(f(x))=x\\ Hence, proved

Question 11. If for non-zero x, a f(x)+b f(1/x) = 1/x – 5, where a ≠ b, then find f(x).

Solution. 

We have,

af(x)+bf(\frac{1}{x})=\frac{1}{x}-5       —(i)

af(\frac{1}{x})+bf(x)=\frac{1}{\frac{1}{x}}-5\\ af(\frac{1}{x})+bf(x)=x-5

af(\frac{1}{x})+bf(x)=x-5                 —(ii)                         

Adding equation (i) and (ii), we get 

af(x)+bf(x)+bf(\frac{1}{x})+af(\frac{1}{x})=\frac{1}{x}-5+x-5\\ (a+b)f(x)+f(\frac{1}{x})(a+b)=\frac{1}{x}+x-10         

f(x)+f(\frac{1}{x})=\frac{1}{a+b}[\frac{1}{x}+x-10]                    —(iii)

Subtracting equation (ii) from equation (i)

af(x)-bf(x )+bf(\frac{1}{x})-af(\frac{1}{x})=\frac{1}{x}-5-x+5 \\ (a-b)f(x)-f(\frac{1}{x})(a-b)=\frac{1}{x}-x \\ f(x)-f(\frac{1}{x})=\frac{1}{a-b}[\frac{1}{x}-x]   —(iv)

Adding equations(iii) and (iv), we get 

2f(x)=\frac{1}{a+b}[\frac{1}{x}+x-10]+\frac{1}{a-b}[\frac{1}{x}-x]\\ 2f(x)=\frac{(a-b)[\frac{1}{x}+x-10]+(a+b)[\frac{1}{x}-x]}{(a+b)(a-b)} \\ 2f(x)=\frac{\frac{a}{x}+ax-10a-\frac{b}{x}-bx+10b+\frac{a}{x}-ax+\frac{b}{x}-bx}{a^2-b^2} \\ 2f(x)=\frac{\frac{2a}{x}-10a+10b-2bx}{a^2-b^2} \\ f(x)=\frac{1}{a^2-b^2}\times\frac{1}{2}[\frac{2a}{x}-10a+10b-2bx]\\ f(x)=\frac{1}{a^2-b^2}[\frac{a}{x}-5a+5b-bx]

Therefore, 

f(x)=\frac{1}{a^2-b^2}[\frac{a}{x}-5a+5b-bx]

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