Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.4 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.4

Question 1.

For the curve y = x^{3} given in figure below, draw

(i) y = – x^{3}

(ii) y = x^{3} + 1

(iii) y = x^{3} – 1

(iv) y = (x + 1)^{3} with the same scale.

Answer:

(i) y = – x^{3}

The graph y = – x^{3} is the reflection of the graph y = x^{3} about x-axis.

The graph of y = – f( x) is the reflection of the graph of y = f( x) about x – axis.

(ii) y = x^{3} + 1

The graph of y = x^{3} + 1, causes the graph y = x^{3} a shift to the upward by 1 unit.

The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

(iii) y = x^{3} – 1

The graph of y = x^{3} – 1, causes the graph y = x^{3} a shift to the downward by 1 unit.

The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.

(iv) y = (x + 1)^{0033}

The graph y = (x + 1)^{3} causes the graph of y = x^{3} a shift to the left by 1 unit.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

Question 2.

For the given curve y = x^{1/3} given in figure draw

(i) y = – x^{1/3}

(ii) y = x^{1/3} + 1

(iii) y = x^{1/3} – 1

(iv) y = (x + 1)^{1/3
}Answer:

(i) y = – x^{1/3}

-y = x^{1/3}

(-y)^{3} = x

-y^{3} = x

When y = 0 ⇒ – 0^{3} ⇒ x = 0

y = 1 ⇒ – 1^{3} = x ⇒ x = – 1

y = 2 ⇒ – 2^{3} = x ⇒ x = – 8

y = 3 ⇒ – 3^{3} = x ⇒ x = – 27

y = -1 ⇒ – (-1)^{3} = x ⇒ x = 1

y = -2 ⇒ – (-2)^{3} = x ⇒ x = 8

y = -3 ⇒ – (- 3)^{3} = x ⇒ x = 27

The graph of y = – x^{1/3} is the reflection of the graph of y = x^{1/3} about the x-axis.

The graph of y = – f(x) is the reflection of the graph of y = f(x) about x – axis.

(ii) y = x^{1/3} + 1

y – 1 = x^{1/3}

⇒ (y – 1)^{3} = x

When

y = 0 ⇒ (0 – 1 )^{3} = x ⇒ x = – 1

y = 1 ⇒ (1 – 1)^{3} = x ⇒ x = 0

y = 2 ⇒ ( 2 – 1 )^{3} = x ⇒ x = 1

y = 3 ⇒ (3 – 1)^{3} = x ⇒ x = 8

y = -1 ⇒ (-1 – 1)^{3} = x ⇒ x = – 8

y = -2 ⇒ (-2 – 1)^{3} = x ⇒ x = – 27

The graph of y = x^{1/3} + 1 causes the graph y = x^{1/3} a shift to the upward by 1 unit.

The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

(iii) y = x^{1/3} – 1

y + 1 = x^{1/3}

( y + 1)^{3} = x

When

y = 0 ⇒ (0 + 1)^{3} = x ⇒ x = 1

y = 1 ⇒ ( 1 + 1)^{3} = x ⇒ x = 8

y = 2 ⇒ (2 + 1)^{3} = x ⇒ x = 27

y = – 1 ⇒ (-1 + 1)^{3} = x ⇒ x = 0

y = – 2 ⇒ (-2 + 1)^{3} = x ⇒ x = – 1

y = – 3 ⇒ (-3 + 1)^{3} = x ⇒ x = – 8

The graph of y = x^{1/3} – 1 causes the graph y = x^{1/3} a shift to the downward by 1 unit.

The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.

(iv) (x + 1)^{1/3}

y^{3} = x + 1

When

y = 0 ⇒ 0^{3} = x + 1 ⇒ x = -1

y = 1 ⇒ 1^{3} = x + 1 ⇒ x = 0

y = 2 ⇒ 2^{3} = x + 1 ⇒ x = 8 – 1 = 7

y = 3 ⇒ 3^{3}= x + 1 ⇒ x = 27 – 1 = 26

y = – 1 ⇒ (-1)^{3} = x + 1 ⇒ x = – 1 – 1 = – 2

y = -2 ⇒ (-2)^{3} = x + 1 ⇒ x = – 8 – 1 = – 9

The graph of y = (x + 1)^{3} causes the graph y = x^{1/3} a shift to the left by 1 unit.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

Question 3.

Graph the functions f(x) = x^{3} and g (x) = \(\sqrt[3]{x}\) on the same coordinate plane. Find fog and the graph it on the plane as well. Explain your results.

Answer:

Given functions are f(x) = x3 and g(x) = x^{1/3}

fog (x) = f(g(x))

= f\(\left(x^{\frac{1}{3}}\right)\)

= \(\left(x^{\frac{1}{3}}\right)^{3}\) = x

f(x) = x^{3}

g(x) = x^{1/3}

Graph of fog(x) = x

Since fog(x) = x is symmetric about the line y = x, g(x) is the inverse image of f(x).

∴ g(x) = f^{-1}(x)

Question 4.

Write steps to obtain the graph of the function y = 3 (x – 1 )^{2} + 5 from the graph y = x^{2}

Answer:

Step 1:

Draw the graph y = x^{2}

Step 2:

The graph of y = (x – 1)^{2} shifts to the right for one unit.

The graph of y = (x – 1 )^{2} shifts the graph

y = x^{2} to the right by 1 unit.

The graph of y = f(x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.

Step 3:

The graph of y = 3 (x – 1)^{2} compresses towards y – axis that is moves away from the x – axis since the multiplying factor is which is greater than 1.

The graph of y = 3 (x – 1)^{2} compresses the graph y = (x – 1)^{2} towards the y-axis that is moving away from the x-axis since the multiplying factor is greater than 1.

For the graph y = kf(x), If k is a positive constant greater than one, the graph moves away from the x-axis. If k is a positive constant less than one, the graph moves towards the x-axis.

Step 4:

The graph of y = 3(x – 1)^{2} + 5 causes the shift to the upward for 5 units.

The graph of y = 3(x – 1)^{2} + 5 causes the graph y = 3(x – 1)^{2} shifts to the upward for 5 units.

The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

Question 5.

From the curve y = sin x, graph the functions.

(i) y = sin (- x)

(ii) y = -sin(-x)

(iii) y = sin\(\left(\frac{\pi}{2}+x\right)\) which is cos x

(iv) y = sin\(\left(\frac{\pi}{2}-x\right)\) which is also cos x

(Refer Trigonometry )

Answer:

y = sin x

(i) y = sin(-x)

y = – sin x

The graph of y = sin (- x) is the reflection of the graph of y = sin x about y-axis.

The graph of y = f(- x) is the reflection of the graph of y = f(x) about y – axis.

(ii) y = – sin (-x)

y = sin x

y = – sin (-x) is the reflection of y = sin (-x) about the x – axis.

The graph of y = – f( x) is the reflection of the graph of y = f( x) about x – axis.

(iii) y = sin\(\left(\frac{\pi}{2}+x\right)\)

The graph of y = sin \(\left(\frac{\pi}{2}+x\right)\) causes y = sin x a shift to the left by \(\frac{\pi}{2}\) units.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

(iv) y = sin\(\left(\frac{\pi}{2}-x\right)\)

The graph of sin \(\left(\frac{\pi}{2}-x\right)\) causes the graph y = sin x a shift to the right by \(\frac{\pi}{2}\) unit.

The graph of y = f(x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.

Question 6.

From the curve y = x draw

(i) y = -x

(ii) y = 2x

(iii) y = x + 1

(iv) y = \(\frac{1}{2}\)x + 1

(v) 2x + y + 3 = 0

Answer:

(i) y = -x

Graph of y = – x is the reflection of the graph of y = x about the x – axis.

The graph of y = – f(x) is the reflection of the graph of y = f(x) about x – axis.

(ii) y = 2x

The graph of y = 2x compresses the graph y = x towards the y-axis that is moving away from the x-axis since the multiplying factor is 2 which is greater than 1.

The graph of y = k f(x), k > 0 moves away from the x-axis if k is greater than 1.

(iii) y = x + 1

The graph of y = x + 1 causes the graph y = x shift to upward by 1 unit.

The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

(iv) y = \(\frac{1}{2}\)x + 1

When

x = 0 ⇒ y = \(\frac{1}{2}\) × 0 + 1 = 1

x = 2 ⇒ y = \(\frac{1}{2}\) × 2 + 1 = 2

x = 4 ⇒ y = \(\frac{1}{2}\) × 4 + 1 = 2 + 1 = 3

x = 6 ⇒ y = \(\frac{1}{2}\) × 6 + 1 = 3 + 1 = 4

x = – 2 ⇒ y = \(\frac{1}{2}\) × – 2 +1= – 1 + 1 = 0

x = – 4 ⇒ y = \(\frac{1}{2}\) × – 4 + 1 = – 2 + 1 = – 1

x = – 6 ⇒ y = \(\frac{1}{2}\) × – 6 + 1 = – 3 + 1 = – 2

The graph of y = \(\frac{1}{2}\)x + 1 stretches the graph y = x towards the x – axis since the multiplying factor is \(\frac{1}{2}\) which is less than 1 and shifts to the upward by 1 unit.

The graph of y = kf(x), k > 0 moves towards the x-axis if k is less than 1.

The graph of y = f(x) + d, d >0 causes the graph y = f(x) a shift to the upward by d units.

(v) 2x + y + 3 = 0

y = -2x – 3

When

x = 0 ⇒ y = -2 × 0 – 3 = -3

x = 1 ⇒ y = -2 × 1 – 3 = -5

x = \(\frac{1}{2}\) ⇒ y = – 2 × \(\frac{1}{2}\) – 3 = – 1 – 3 = – 4

x = 2 ⇒ y = -2 × 2 – 3 = – 4 – 3 = – 7

x = – 1 ⇒ y = -2 × – 1 – 3 = 2 – 3 = – 1

x = – 2 ⇒ y = 2 × – 2 – 3 = 4 – 3 = 1

x = – 3 ⇒ y = -2 × -3 – 3 = 6 – 3 = 3

The graph of y = – 2x – 3 stretches the graph y = x towards the x-axis since the multiplying factor is – 2 which is less than 1 and causes the shift to the downward by 3 units.5

The graph of y = kf(x), k > 0 moves towards the x-axis if k is less than 1.

The graph of y = f(x) – d, d >0 causes the graph y = f(x) a shift to the downward by d units.

Question 7.

From the curve y = |x| draw

(i) y = |x – 1| + 1

(ii) y = |x + 1| – 1

(iii) y = |x + 2| – 3

Answer:

y = |x|

(i) y = |x – 1| + 1

(a) Consider y = |x – 1|

x = 0 ⇒ y = – x + 1 ⇒ y = 1

x = 1 ⇒ y = x – 1 ⇒ y = 0

x = 2 ⇒ y = x – 1 ⇒ y = 1

x = 3 ⇒ y = x – 1 ⇒ y = 2

x = – 1 ⇒ y = – x + 1 ⇒ y = 2

x = – 2 ⇒ y = – x + 1 ⇒ y = 3

The graph of y = |x – 1| causes the graph y = |x| a shift to the right by 1 unit.

The graph of y = f(x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.

(b) Consider y = |x – 1| + 1

x = 0 ⇒ y = – x + 2 ⇒ y = 2

x = 1 ⇒ y = x ⇒ y = 1

x = 2 ⇒ y = x ⇒ y = 2

x = 3 ⇒ y = x ⇒ y = 3

x = – 1 ⇒ y = – x + 2 ⇒ y = 3

x = – 2 ⇒ y = – x + 2 ⇒ y = 4

The graph of y = |x – 1| + 1 shift the graph y = |x| to the right by 1 unit and causes a shift to the upward by 1 unit.

The graph of y = f( x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.

The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

(ii) y = |x + 1| – 1

(a) Consider y = |x + 1|

x = 0 ⇒ y = x + 1 ⇒ y = 1

x = 1 ⇒ y = x + 1 ⇒ y = 2

x = 2 ⇒ y = x + 1 ⇒ y = 3

x = 3 ⇒ y = x + 1 ⇒ y = 4

x = – 1 ⇒ y = x + 1 ⇒ y = 0

x = – 2 ⇒ y = – (x + 1) ⇒ y = 1

x = – 3 ⇒ y = – (x + 1) ⇒ y = 2

The graph of y = |x + 1| shifts the graph y = |x| to the left by 1 unit.

The graph of y = f( x + c), c > 0 causes the graph y = f(x) a shift to the left by e units.

(b) Consider y = |x + 1| – 1

x = 0 ⇒ y = x ⇒ y = 0

x = 1 ⇒ y = x ⇒ y = 1

x = 2 ⇒ y = x ⇒ y = 2

x = 3 ⇒ y = x ⇒ y = 3

x = – 1 ⇒ y = x ⇒ y = – 1

x = – 2 ⇒ y = – x – 2 ⇒ y = 0

x = – 3 ⇒ y = – x – 2 ⇒ y = 1

x = – 4 ⇒ y = – x – 5 ⇒ y = -1

The Graph of y = |x + 1 | – 1 shift the graph y = |x| to the left by 1 unit and causes a shift downward by 1 unit.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.

(iii) y = |x + 2| – 3

(a) Consider the curve y = |x + 2|

x = 0 ⇒ y = x + 2 ⇒ y = 2

x = 1 ⇒ y = x + 2 ⇒ y = 3

x = 2 ⇒ y = x + 2 ⇒ y = 4

x = 3 ⇒ y = x + 2 ⇒ y = 5

x = – 1 ⇒ y = x + 2 ⇒ y = 1

x = – 2 ⇒ y = x + 2 ⇒ y = 0

x = – 3 ⇒ y = – x – 2 ⇒ y = 1

The graph of y = |x + 2| shifts the graph y = |x| to the left by 2 units.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

(b) Consider the curve y = |x + 2| – 3

x = 0 ⇒ y = x – 1 ⇒ y = – 1

x = 1 ⇒ y = x – 1 ⇒ y = 0

x = 2 ⇒ y = x – 1 ⇒ y = 1

x = 3 ⇒ y = x – 1 ⇒ y = 2

x = – 1 ⇒ y = x – 1 ⇒ y = – 2

x = – 2 ⇒ y = x – 1 ⇒ y = – 3

x = – 3 ⇒ y = – x – 5 ⇒ y = – 2

The graph of y = |x + 2| – 3 shifts the graph y = |x| to the left by 2 units and causes a shift downward by 3 units.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the down ward by d units.

Question 8.

From the curve y = sin x, draw y = sin |x| (Hint: sin(- x) = – sin x)

Answer:

y = sin |x|

(a) y = sin x

(b) Consider y = sin |x|