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S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(f)
Question 1.
Show that
(a) x² – 3x + 6 > 0 for all x
(b) 4x – x² – 6 < 0 for all x
(c) 2x² – 4x + 7 is always +ve
(d) – 2x² + 3x – 4 is always -ve
(e) – x² + 3x – 3 is always -ve.
Solution:
(a) Here D = b² – 4ac = (- 3)² – 4 x 1 x 6
= 9 – 24 = – 15 < 0
Thus the expression x² – 3x + 6 and coeff. of x² have same sign.
Here coeff. of x² = a = 1 > 0
∴ x² – 3x + 6 > 0 ∀ x
(b) Here a = – 1, b = + 4; c = – 6
and Discriminant D = b² – 4ac = 4² – 4 x ( – 1) x (- 6)
= 16 – 24 = – 8 < 0
Thus the given expression 4x – x² – 6 has same sign as that of a.
Here a = – 1 < 0
∴ 4x – x² – 6 < 0 for all x.
(c) Here a = 2 ; b = – 4 ; c = 1
and discriminant D = b² – 4 ac = (- 4)² – 4 x 2 × 7 = 16 – 56 = – 40 < 0
∴ roots are imaginary. Thus the given expression 2x² – 4x + 7 and a has same sign. Here a = 2 > 0
∴ 2x² – 4x + 7 > 0 for all x
(d) Here a = – 2 ; b = – 3 and c = – 4
and discriminant D = b² – 4ac = (- 3)² – 4 x (- 2) x (- 4) = 9 – 32 = – 23 < 0
∴ roots are imaginary
Hence the given expression – 2x² + 3x – 4 and a has same sign. Here a = – 2 < 0
∴ – 2x² + 3x – 4 < 0 ∀ x
(e) Here a = – 1 ; b = 3 ; c = -3
Here discriminant D = b² – 4ac = 32 – 4 x (- 1) x (- 3) = 9 – 12 = – 3 < 0
∴ roots are imaginary. Hence given expression and ‘a’ has same sign.
Here, a = – 1 < 0
∴ – x² + 3x – 3 < 0 ∀ x.
Question 2.
Explain why 3x² + kx – 1 is never always positive for any value of k.
Solution:
Here, a = 3 ; b = k; c = – 1
Here discriminant D = b² – 4ac = k² – 4 x 3 x (- 1) = k² + 12 > 0
∴ roots are real and unequal.
Let a and P be its two roots s.t a > P
Then when x > α or x < ß, the given expression 3x² + kx – 1 has same sign as that of ‘a’
Here a = 3 > 0
∴ 3x² + kx – 1 > 0
But when ß < x < α, then 3x² + kx – 1 and a has opposite sign.
∴ 3x² + kx – 1 < 0
Hence 3x² + kx – 1 is never always positive for any value of k.
Question 3.
Under what conditions is 2x² + kx + 2 always positive ?
Solution:
Comparing 2x² + kx + 2 with ax² + bx + c, we have a = 2 ; b = k; c = 2
Here discriminant D = b² – 4ac = k² – 4 x 2 x 2 = k² – 16
Here a = 2 > 0 if D < 0 then 2x² + kx + 2 > 0 ∀ x i.e. if – 16 < 0 if k² < 16
if | k |< 4 if – 4 < k < 4
Hence, 2x² + kx + 2 > 0 ∀x, where – 4 < k < 4
Question 4.
Find the values of a so that the expression x? – (a + 2) x + 4 is always positive.
Solution:
On comparing x² – (a + 2) x + 4 with Ax² + Bx + C
we have A = 1 ; B = – (a + 2) and C = 4
Here Discriminant D = b² – 4AC = (a + 2)² – 16
Also, A = 1 > 0 if D < 0 then given expression is always be positive.
i.e. if (a + 2)² – 16 < 0 if (a + 2)² < 16 if |a + 2| < 4
if – 4 < a + 2 < 4 [∵ |x| < l ⇒ – l < x < l]
if – 6 < a < 2
Question 5.
Find the range of values of x for which the expression 12x² + 7x – 10 is negative.
Solution:
For range of x for which 12x² + 7x – 10 < 0
⇒ 12\(\left[x^2+\frac{7}{12} x-\frac{5}{6}\right]\) < 0
⇒ x² + \(\frac{7}{12} x+\frac{49}{576}-\frac{49}{576}-\frac{5}{6}\) < 0
⇒ \(\left(x+\frac{7}{24}\right)^2-\left(\frac{49+480}{576}\right)\) < 0
⇒ \(\left(x+\frac{7}{24}\right)^2<\left(\frac{23}{24}\right)^2\) ⇒ \(\left|x+\frac{7}{24}\right|<\frac{23}{24}\)
⇒ \(\frac{-23}{24}<x+\frac{7}{24}<\frac{23}{24}\)
⇒ \(\frac{-23}{24}-\frac{7}{24}<x<\frac{23}{24}-\frac{7}{24}\)
⇒ \(\frac{-30}{24}<x<\frac{16}{24} \text { i.e } \frac{-5}{4}<x<\frac{2}{3}\)
Question 6.
(i) Find the values of ‘a’ for which the expression x² – (3a – 1) x + 2a² + 2a – 11 is always positive.
(ii) If x² + 4ox + 2 > 0 for all values of x, then a lies in the interval
(a) ( – 2,4)
(b) (1, 2)
(c) (- \(\sqrt{2}\), \(\sqrt{2}\))
(d) \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
(e) (- 4, 2)
Solution:
(i) On comparing x² – (3a – 1) x + 2a² + 2a – 11 with Ax² + Bx + C, we have
A = 1 ; B = – (3a – 1) and C = 2a² + 2a – 11
Here D = B² – 4AC = (3a – 1)² – 4 (2a²+ 2a-ll) = (9a² – 6a + 1) – (8a² + 8a – 44)
= a²- 14a + 45
Here A = 1 > 0 and if D < 0 then given expression will be always positive
if a² – 14a + 45 < 0 if (a – 9) (a – 5) < 0
The critical points are a = 5, 9
Then by method of intervals, 5 < a < 9.
(ii) On comparing x² + 4ax + 2 with Ax² + Bx + C.
Here A = 1 ; B = 4a and C = 2 and discriminant
D = b² – 4AC = (4a)² – 4 x 1 x 2 = 16a² – 8
Here A = 1 > 0.
Then x² + 4ax + 2 > 0
if D < 0 if b² – 4AC < 0 if 16a² – 8 < 0
if a² < \(\frac { 1 }{ 2 }\) if | a | < \(\frac { 1 }{ 2 }\) if – \(\frac{1}{\sqrt{2}}<a<\frac{1}{\sqrt{2}}\) if a ∈ \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
Question 7.
Find the greatest value of 3 + 5x – 2x² for all real values of x.
Solution:
Let y = 3 + 5x – 2x² ⇒ – 2x² + 5x + 3 – y = 0
Since x is real
∴ D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ 5² – 4 x (- 2) (3 – y) ≥ 0
⇒ 25 + 24 – 8y ≥ 0
⇒ 49 – 8y ≥ 0
⇒ 8y ≤ 49
⇒ y ≤ \(\frac { 49 }{ 8 }\) = 6\(\frac { 1 }{ 8 }\)
∴ greatest value of y = 6\(\frac { 1 }{ 8 }\)
Question 8.
Find the least value of \(\frac{6 x^2-22 x+21}{5 x^2-18 x+17}\) for real values of x.
Solution:
Let y = \(\frac{6 x^2-22 x+21}{5 x^2-18 x+17}\)
⇒ y (5x² – 18x + 17) = 6x² – 22x + 21
⇒ x² (6 – 5y) + x (18y – 22) + 21 – 17y = 0
Since x is given to be real
∴ D ≥ 0
⇒ (18y – 22)² – 4 x (6 – 5y) (21 – 17y) ≥ 0
⇒ 4 (9y- 11 )² – 4 (6 – 5y) (21 – 17y) ≥ 0
⇒ (81y² + 121 – 198y) – (126 – 102y – 105y + 85y²) ≥ 0
⇒ – 4y² + 9y – 5 ≥ 0
⇒ 4y² – 9y + 5 ≤ 0
⇒ (y – 1) (4y – 5) ≤ 0
The critical points are y – 1 = 0 and 4y – 5 = 0 i.e. y = 1, \(\frac { 5 }{ 4 }\)
Then by method of intervals, we have 1 ≤ y ≤ \(\frac { 5 }{ 4 }\) ⇒ y ∈ [1, \(\frac { 5 }{ 4 }\) ]
∴ least value of y = 1
Question 9.
If x be real, prove that the value of \(\frac{11 x^2+12 x+6}{x^2+4 x+2}\) cannot lie between – 5 and 3.
Solution:
Let y = \(\frac{11 x^2+12 x+6}{x^2+4 x+2}\)
⇒ y (x² + 4x + 2) = 11x² + 12x + 6
⇒ x²(y – 11) + x (4y – 12) + 2y – 6 = 0
For x is to be real ∴ Disc ≥ 0 ⇒ (4y – 12)² – 4 (y – 11) (2y – 6) ≥ 0
⇒ 16 (y – 3)² – 8(y – 11) (y – 3) ≥ 0
⇒ 8[2(y² – 6y + 9) – (y² – 14y + 33)] ≥ 0
⇒ y² + 2y – 15 ≥ 0
⇒ (y – 3) (y + 5) ≥ 0
The critical points are given by y = 3, – 5.
Then by method of intervals, we have y ≤ – 5 or y ≥ 3
Hence y cannot lie between – 5 and 3.