Students appreciate clear and concise OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(e) that guide them through exercises.
S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(e)
Draw the graph of each of the following quadratic functions.
Question 1.
y = x² – 5x + 6 ; 0 ≤ x ≤ 5.
Solution:
Given y = x² – 5x + 6 ; 0 ≤ x ≤ 5
The table of values is given as under :
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 6 | 2 | 0 | 0 | 2 | 6 |
Question 2.
y = – x² + 2x + 3 ; – 3 ≤ x ≤ 5.
Solution:
Given y = – x² + 2x + 3 where – 3 ≤ x ≤ 5
The table of values is given as under;
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | -12 | -5 | 0 | 3 | 4 | 3 | 0 | -5 | -12 |
Question 3.
y = x² – 4x + 4 ; – 1 ≤ x ≤ 5.
Solution:
Given y = x² – 4x + 4 ; – 1 ≤ x ≤ 5
The table of values is given as under
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
Question 4.
Solve graphically and compare your answer with algebraic solution either by factorization or formula method :
(i) y = x² – 5x + 6
(ii) y = – x² + 2x + 3
(iii) y = x² – 4x + 4
(iv) y = x² – x – 6
(v) y = x² – 6x + 9
(vii) y = x² – 4x + 5 = 0
(viii) y = x² + 2x + 2 = 0
Solution:
(i) y = x² – 5x + 6
The table of values is given as under:
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 6 | 2 | 0 | 0 | 2 | 6 |
Clearly the graph of curve intersects x-axis at two points at
x = 2 and x = 3
∴ x = 2, 3
Verification : y = x² – 2x – 3x + 6 = x (x – 2) – 3 (x – 3)
⇒ y = (x – 2) (x – 3)
Clearly x = 2 and x = 3 are the two solutions of given curve.
(ii) Given y = – x² + 2x + 3
The table of values is given as under :
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | -12 | -5 | 0 | 3 | 4 | 3 | 0 | -5 | -12 |
Clearly the graph of curve intersects x-axis at two points at x = – 1 and x = 3
Verification : y = – x³ + 2x + 3 = – x² + 3x – x + 3
= – x (x – 3) – 1(x – 3)
= – (x – 3) (x + 1)
Clearly x = – 1 and x = 3 are two solutions of given curve.
(iii) Given y = x² – 4x + 4
The table of values is given as under :
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
Clearly the graph of curve intersects x-axis at one point i.e. x = 2
Verification : y = x² – 4x + 4 = (x – 2)²
Clearly x = 2 be the solution of given curve.
(iv) Given y = x² – x – 6
The table of values is given as under
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 0 | -4 | -6 | -6 | -4 | 0 | 6 |
Clearly the graph meets x-axis at (- 2, 0) and (3, 0).
∴ x = – 2, 3 are the real solutions of given curve
Verification : y = x² – x – 6 = x² – 3x + 2x – 6
= x (x – 3) + 2 (x – 3) = (x + 2) (x – 3)
Clearly – 2 and 3 are the solutions of y = x² – x – 6 = 0
(v) Given eqn. of curve be y = x² – 6x + 9
The tale of values is given as under :
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 9 | 4 | 1 | 0 | 1 | 4 |
Clearly the graph of curve intersects x-axis at (3, 0).
∴ x = 3, 3 be the solution of given curve.
Verification : y = x² – 6x + 9 = x² – 3x – 3x + 9
= x (x – 3) – 3 (x – 3) = (x – 3)²
Clearly x = 3 be the solution of given curve.
(vi) Given eqn. of curve bey = – x² + x + 12 ⇒ y = – (x² + x – 12)
The table of values is given as under :
| x | -4 | -2 | 0 | 1 | 2 | 3 |
| y | 0 | 10 | 12 | 10 | 6 | 0 |
Clearly the graph of the curve intersects x-axis at x = – 4 and x = 3.
Hence x = – 4, 3 are two real solutions of given curve.
Verification : y = – (x² – 4x + 3x – 12) = – [x (x – 4) + 3 (x – 4)]
= – (x + 3)(x – 4)
Clearly x = – 3 and 4 are the solutions of y = – x² + x + 12 = 0
(vii) Given curve be y = x² – 4x + 5 = 0
The table of values is given as under
| x | 0 | 1 | 2 | 3 | 4 |
| y | 5 | 2 | 1 | 2 | 5 |
Clearly the curve does not meets x-axis at any point so it has no real solution.
Verification : x² – 4x + 5 = 0 ⇒ x = \(\frac{4 \pm \sqrt{16-4 \times 1 \times 5}}{2}=\frac{4 \pm 2 i}{2}=2 \pm i\)
(viii) Given curve be y = x² + 2x + 2 = 0
The table of values is given as under :
| x | 0 | 1 | -1 | -2 | -3 |
| y | 2 | 5 | 1 | 2 | 5 |
Clearly the graph of curve does not meet x-axis
∴ given curve has no real solution.
Verification : y = x² + 2x + 2 = 0
⇒ x = \(\frac{-2 \pm \sqrt{4-8}}{2}\)
⇒ x = \(\frac{-2 \pm 2 i}{2}\)
⇒ x = – 1 ± i