Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.2

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 

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-x³        —(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, 

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,

      —(i)

                —(ii)                         

Adding equation (i) and (ii), we get 

                   —(iii)

Subtracting equation (ii) from equation (i)

  —(iv)

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

Therefore,