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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.4

Question 1.

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

(i) r = -x^{3}

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

(iii) y = x^{3} – 1

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

Solution:

(i) It is the reflection on y axis

(ii) The graph of y = x^{3} + 1 is shifted upward to 1 unit.

(iii) The graph of y = x^{3} – 1 is shifted downward to 1 unit.

(iv) The graph of y = (x + 1)^{3} is shifted to the left for 1 unit.

Question 2.

Solution:

Question 3.

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

Solution:

Question 4.

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

Solution:

Draw the graph of y = x^{2}

To get y = (x – 1)^{2} we have to shift the curve 1 unit to the right.

Then we have to draw the curve y = 3(x – 1)^{2} and finally, we have to draw y = 3(x – 1)^{2} + 5

Question 5.

From the curve y = sin x, graph the functions

(i) y = sin(-x)

(ii) y = -sin(-x)

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

(iv) y = sin (\(\frac{\pi}{2}\) – x) which is also cos x (refer trigonometry)

Solution:

First we have to draw the curve y = sin x

(i) y = sin (-x) = – sin x = f(x)

(ii) y = -sin(-x) = -[-sin x] = sin x

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. 2

Solution:

y = x

(i) y = -x

(ii) y = 2x

y = 2x the graph moves away from the x-axis, as multiplying factor is 2 which is greater than one.

(iii) y = 2x + 1

(iv) y = 1/2x + 1

y = \(\frac{1}{2}\)x moves towards x – axis by a side factor \(\frac{1}{2}\) which is less than y = \(\frac{1}{2}\)x + 1 upwards by 1 unit.

(v) y = -2x – 3

Question 7.

From the curve y = |x|, draw

(i) y= |x – 1| + 1

(ii) y = |x + 1| – 1

(iii) y = |x + 2| – 3.

Solution:

Given, y = |x|

If y = x

If y = -x

(i) y = |x – 1| + 1

y = x – 1 + 1

y = -x + 1 + 1 = x

(ii) y = |x + 1| – 1

y = x + 1 – 1 = x

y = -x -1 – 1

y = -x – 2

y = – (x + 2)

(iii) y = |x + 2| – 3

y = x + 2 – 3 ⇒ x – 1

y = -x – 2 + 3 = 1 – x

y = -(x – 1)

Question 8.

From the curves = sin x, draw y = sin |x| (Hint: sin(-x) = -sin x.)

Solution:

y = sin |x|

∴ y = sin x

∴ y = sin (-x) = – sin x

y = – sin x