TN 11th, Maths Chapter 1, Exercise 1.4


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Class : XI
Subject : Maths
Chapter : 1
Exercise : 1.4

Que 1 :

For the curve y = x³ given in Figure, draw


(i) y = -x³
(ii) y = x³ + 1
(iii) y = x³–1
(iv) y = (x + 1)³, with the same scale.

Solution :

(i) y = -x³




y=-f(x) It is the reflection of the graph of About the x-axis





(ii) y = x³ + 1




The graph of y = f(x) + 1 is shifted upward to 1 unit.




(iii) y = x³ – 1




The graph of y = x³ – 1 is shifted downward to 1 unit




(iv) y = (x + 1)³





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





Que 2 :
                                   
For the curve, y = (x^⅓) given in figure draw.


(i) y = (-x^⅓)
(ii) y=(x^⅓) +1
(iii) y= (x^⅓) -1
(iv) y=(x+1)^⅓




Solution :

(i) y = (- x)^⅓




Then y = -x^⅓ is the reflection of the graph of y=x^⅓ about the x-axis.




(ii) y=(x^⅓) +1

 



Then y= (x^⅓)+1 is the x graph of y = x^⅓ shifts to the upwards for one unit.




(iii) y= (x^⅓) -1

     



Then y = (x^⅓)-1 is the graph of (x^⅓) shifts to the downward for one unit.




(iv) y=(x+1)^⅓





Then y = (x+1)^⅓, it causes the graph of x^⅓, shift to the left for one unit.


Que 3 :

Graph the functions f(x) = x³ And g(x) = (³√x) on the same coordinate plane. Find fog and graph it on the plane as well. Explain your results.

Solution :

Given functions are f(x) = x³ and g(x) = x⅓.

Now, fog(x) = f(g(x)) = f (x⅓) = (x³)^⅓ = x




Since fog(x) = x is symmetric about the line y=x, g(x) is the inverse of f(x) ∴ g(x) = f-¹(x).




Que 4 :

Write the steps to obtain the graph of the function y = 3(x – 1)²+5 from the graph y = x².

Solution:




Step 1: Draw The Graph Of y = x².

Step 2: The Graph Of y = ( x-1)², shifts to the right for 1 unit.

Step 3: The Graph Of y = 3(x-1)², Compresses towards the y-axis that is moves away from the x-axis since the multiplying factor is 3 Which is greater than 1.

Step 4:  The Graph Of y=3(x-1)²+5, causes the shift to the upwards for 5 units.




Que 5 :

From The Curve y = sin x, Graph the functions
(i) y = sin(-x)
(ii) y = -sin(-x)
(iii) y = sin ((Ï€/2) + x) which is cos x
(iv) y = sin ((Ï€/2) – x) which is also cos x (refer trigonometry)
Solution:

Solution :

(i) y = sin(-x)


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




(ii) y = -sin(-x)



y = -sin(-x) is the reflection of y = sin(-x) which is Same as y = sin x.





(iii) y = sin ((Ï€/2) + x) which is cos x



y = sin [(Ï€/2) + x] it causes the shift to the left for (Ï€/2) units.





(iv) y = sin ((Ï€/2) – x) which is also cos x 


y = sin [(Ï€/2) - x ] causes the shift to the right for (Ï€/2) unit to the sin(-x) curve.





Que 6 :

From the curve y = x, draw

(i) y = -x
(ii) y = 2x
(iii) y = x + 1
(iv) y = (1/2)x + 1
(v) 2x + y + 3 = 0.2

Solution :


(i) y = -x




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





(ii) y = 2x




Graph of y=2x Compresses towards the y-axis that is move away from the x-axis since the multiplying factor is 2, which is greater than 1.





(iii) y = x + 1




The Graph Of y=x+1, causes the shift to the upwards for one unit.





(iv) y = (1/2)x + 1




The graph of y= (1/2)x + 1, stretches towards the X-axis since the multiplying factor 1/2 is which is less than one and shifts to the upward for one unit.




(v) 2x + y + 3 = 0.2




The graph of y = -2x - 3, stretches towards the X-axis since the multiplying factor is - 2 which is less than one and causes the shifts to the downward for 3 units.



Que 7 :

From the curve y = |x|, draw

(i) y= |x – 1| + 1
(ii) y = |x + 1| – 1
(iii) y = |x + 2| – 3.

Solution :

(i) y= |x – 1| + 1




The graph of y=|x - 1|+1, shifts to the right for one unit and causes the shift to the upward for one unit.



(ii) y = |x + 1| – 1





The graph of y = |x + 1|-1, shifts to the left for one unit and causes the shift to the downward for one unit.



(iii) y = |x + 2| – 3



The graph of y =|x + 2|-3, shifts to the left for 2 units and causes the shift to the downward for 3 units.



Que 8 :

From the curve y = sin x, draw y=sin bax! (Hint: sin(x) =-sin x.)

Solution :




We know That 

|x| = x if x ≥ 0; -x if x < 0;

∴ sin|x| = sin x if x ≥ 0

and sin |x| = sin (-x) = - sin x if x <0.

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





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