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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 ### 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, ### 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, My Personal Notes arrow_drop_up