OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Exercise 10 (F)

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(f)

Question 1. Show that
(a) \( x^2 - 3x + 6 > 0 \) for all x
(b) \( 4x - x^2 - 6 < 0 \) for all x
(c) \( 2x^2 - 4x + 7 \) is always +ve
(d) \( - 2x^2 + 3x - 4 \) is always -ve
Answer:
(a) For the expression \( x^2 - 3x + 6 \):
First, we find the discriminant \( D \). Here, \( a = 1 \), \( b = -3 \), \( c = 6 \).
The discriminant is \( D = b^2 - 4ac = (-3)^2 - 4 \times 1 \times 6 = 9 - 24 = -15 \).
Since \( D = -15 \), which is less than 0, the roots are imaginary. This means the quadratic expression will always have the same sign as its leading coefficient, 'a'.
Here, the coefficient \( a = 1 \), which is greater than 0.
So, \( x^2 - 3x + 6 > 0 \) for all real values of x.

(b) For the expression \( 4x - x^2 - 6 \):
Rewrite it as \( -x^2 + 4x - 6 \). Here, \( a = -1 \), \( b = 4 \), \( c = -6 \).
The discriminant is \( D = b^2 - 4ac = (4)^2 - 4 \times (-1) \times (-6) = 16 - 24 = -8 \).
Since \( D = -8 \), which is less than 0, the roots are imaginary. The expression will always have the same sign as its leading coefficient, 'a'.
Here, the coefficient \( a = -1 \), which is less than 0.
So, \( 4x - x^2 - 6 < 0 \) for all real values of x.

(c) For the expression \( 2x^2 - 4x + 7 \):
Here, \( a = 2 \), \( b = -4 \), \( c = 7 \).
The discriminant is \( D = b^2 - 4ac = (-4)^2 - 4 \times 2 \times 7 = 16 - 56 = -40 \).
Since \( D = -40 \), which is less than 0, the roots are imaginary. The expression will always have the same sign as its leading coefficient, 'a'.
Here, the coefficient \( a = 2 \), which is greater than 0.
So, \( 2x^2 - 4x + 7 \) is always positive (always +ve) for all real values of x.

(d) For the expression \( -2x^2 + 3x - 4 \):
Here, \( a = -2 \), \( b = 3 \), \( c = -4 \).
The discriminant is \( D = b^2 - 4ac = (3)^2 - 4 \times (-2) \times (-4) = 9 - 32 = -23 \).
Since \( D = -23 \), which is less than 0, the roots are imaginary. The expression will always have the same sign as its leading coefficient, 'a'.
Here, the coefficient \( a = -2 \), which is less than 0.
So, \( -2x^2 + 3x - 4 \) is always negative (always -ve) for all real values of x.

(e) For the expression \( -x^2 + 3x - 3 \):
Here, \( a = -1 \), \( b = 3 \), \( c = -3 \).
The discriminant is \( D = b^2 - 4ac = (3)^2 - 4 \times (-1) \times (-3) = 9 - 12 = -3 \).
Since \( D = -3 \), which is less than 0, the roots are imaginary. The expression will always have the same sign as its leading coefficient, 'a'.
Here, the coefficient \( a = -1 \), which is less than 0.
So, \( -x^2 + 3x - 3 < 0 \) for all real values of x.
In simple words: For a quadratic equation, if the discriminant (the part under the square root in the quadratic formula) is less than zero, it means the graph never crosses the x-axis. So, if the 'a' value (coefficient of \( x^2 \)) is positive, the whole expression is always positive. If 'a' is negative, the whole expression is always negative.

🎯 Exam Tip: When proving a quadratic expression is always positive or negative, always calculate the discriminant first. If \( D < 0 \), the sign of the expression is determined by the sign of the coefficient of \( x^2 \).

Question 2. Explain why \( 3x^2 + kx - 1 \) is never always positive for any value of k.
Answer:
For the quadratic expression \( 3x^2 + kx - 1 \):
Here, \( a = 3 \), \( b = k \), \( c = -1 \).
Let's find the discriminant \( D = b^2 - 4ac = k^2 - 4 \times 3 \times (-1) = k^2 + 12 \).
Since \( k^2 \) is always greater than or equal to 0 for any real value of k, \( k^2 + 12 \) will always be greater than 0.
This means \( D > 0 \), so the roots of the quadratic equation are always real and distinct (unequal).
When the roots are real, the parabola (graph of the quadratic) will always cross the x-axis at two different points. Let these roots be \( \alpha \) and \( \beta \), with \( \alpha > \beta \).
The coefficient \( a = 3 \) is positive. For a positive 'a', the parabola opens upwards.
This means:
When \( x > \alpha \) or \( x < \beta \), the expression \( 3x^2 + kx - 1 \) will have the same sign as 'a', which is positive.
However, when \( x \) is between the roots (i.e., \( \beta < x < \alpha \)), the expression \( 3x^2 + kx - 1 \) will have the opposite sign to 'a', meaning it will be negative.
Since there is always a range of x-values where the expression is negative (between the roots), it can never be "always positive" for all values of x. It will be positive sometimes and negative at other times.
In simple words: No matter what value 'k' has, this math problem will always have real answers where it crosses the x-axis. Because the graph crosses the x-axis and then goes below it, it can't always stay positive. It has to dip into negative values sometimes.

🎯 Exam Tip: To show an expression is not always positive, it is sufficient to prove that its discriminant is greater than zero and the 'a' coefficient is positive, which implies it will have negative values between its real roots.

Question 3. Under what conditions is \( 2x^2 + kx + 2 \) always positive ?
Answer:
For the quadratic expression \( 2x^2 + kx + 2 \):
Compare this with the general quadratic form \( ax^2 + bx + c \). We have \( a = 2 \), \( b = k \), \( c = 2 \).
For a quadratic expression to be always positive for all real values of x, two conditions must be met:
1. The coefficient 'a' must be positive. (Here, \( a = 2 \), which is \( > 0 \), so this condition is already satisfied).
2. The discriminant 'D' must be less than zero. (Meaning, the roots are imaginary and the parabola does not cross the x-axis).

Let's calculate the discriminant \( D = b^2 - 4ac \):
\( D = k^2 - 4 \times 2 \times 2 = k^2 - 16 \).
For the expression to be always positive, we need \( D < 0 \):
\( k^2 - 16 < 0 \)
\( k^2 < 16 \)
Taking the square root of both sides, we get:
\( \sqrt{k^2} < \sqrt{16} \)
\( |k| < 4 \)
This inequality \( |k| < 4 \) means that k must be between -4 and 4.
So, the condition is \( -4 < k < 4 \).
Hence, \( 2x^2 + kx + 2 > 0 \) for all real values of x when \( -4 < k < 4 \).
In simple words: For a hill-shaped graph (parabola) to always stay above the ground (be positive), two things must happen: the 'a' number (the one with \( x^2 \)) must be positive, and the special number called 'discriminant' must be less than zero. If the 'a' is positive and the discriminant is negative, the graph is always above zero.

🎯 Exam Tip: Remember the two conditions for a quadratic \( ax^2 + bx + c \) to be always positive: \( a > 0 \) and \( D < 0 \). Both must be met simultaneously.

Question 4. Find the values of a so that the expression \( x^2 - (a + 2) x + 4 \) is always positive.
Answer:
For the expression \( x^2 - (a + 2) x + 4 \) to be always positive, we need to satisfy two conditions:
1. The coefficient of \( x^2 \) must be positive. Here, it is \( a = 1 \), which is \( > 0 \). This condition is satisfied.
2. The discriminant (D) must be less than zero.

Let's identify the coefficients for the expression: \( A = 1 \), \( B = -(a + 2) \), \( C = 4 \).
Now, calculate the discriminant \( D = B^2 - 4AC \):
\( D = (-(a + 2))^2 - 4 \times 1 \times 4 \)
\( D = (a + 2)^2 - 16 \)
For the expression to be always positive, we must have \( D < 0 \):
\( (a + 2)^2 - 16 < 0 \)
\( (a + 2)^2 < 16 \)
Take the square root of both sides:
\( \sqrt{(a + 2)^2} < \sqrt{16} \)
\( |a + 2| < 4 \)
This inequality means that \( (a + 2) \) must be between -4 and 4:
\( -4 < a + 2 < 4 \)
To find 'a', subtract 2 from all parts of the inequality:
\( -4 - 2 < a < 4 - 2 \)
\( -6 < a < 2 \)
Therefore, the expression \( x^2 - (a + 2) x + 4 \) is always positive when 'a' is in the interval \( (-6, 2) \).
In simple words: To make a math problem always positive, its 'a' number must be positive (which it is here, 1). The other rule is that its 'D' number (discriminant) must be less than zero. When we solve for 'a' using this rule, we find that 'a' has to be a number between -6 and 2.

🎯 Exam Tip: Pay close attention to the absolute value inequality \( |x| < k \implies -k < x < k \). This is a common step in solving for parameter ranges.

Question 5. Find the range of values of x for which the expression \( 12x^2 + 7x - 10 \) is negative.
Answer:
We need to find the values of x for which \( 12x^2 + 7x - 10 < 0 \).
First, find the roots of the quadratic equation \( 12x^2 + 7x - 10 = 0 \). We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) or factorize the expression.

Let's factorize or use completing the square for simplicity in this context:
\( 12x^2 + 7x - 10 < 0 \)
Divide by 12:
\( x^2 + \frac{7}{12}x - \frac{10}{12} < 0 \)
\( x^2 + \frac{7}{12}x - \frac{5}{6} < 0 \)
Complete the square for the x terms:
\( x^2 + \frac{7}{12}x + (\frac{7}{24})^2 - (\frac{7}{24})^2 - \frac{5}{6} < 0 \)
\( (x + \frac{7}{24})^2 - (\frac{49}{576}) - \frac{5 \times 96}{6 \times 96} < 0 \)
\( (x + \frac{7}{24})^2 - (\frac{49}{576} + \frac{480}{576}) < 0 \)
\( (x + \frac{7}{24})^2 - (\frac{529}{576}) < 0 \)
\( (x + \frac{7}{24})^2 < \frac{529}{576} \)
\( (x + \frac{7}{24})^2 < (\frac{23}{24})^2 \)
Taking the square root of both sides:
\( |x + \frac{7}{24}| < \frac{23}{24} \)
This means that \( (x + \frac{7}{24}) \) must be between \( -\frac{23}{24} \) and \( \frac{23}{24} \):
\( -\frac{23}{24} < x + \frac{7}{24} < \frac{23}{24} \)
To find 'x', subtract \( \frac{7}{24} \) from all parts:
\( -\frac{23}{24} - \frac{7}{24} < x < \frac{23}{24} - \frac{7}{24} \)
\( -\frac{30}{24} < x < \frac{16}{24} \)
Simplify the fractions:
\( -\frac{5}{4} < x < \frac{2}{3} \)
So, the expression \( 12x^2 + 7x - 10 \) is negative when x is in the interval \( (-\frac{5}{4}, \frac{2}{3}) \).
In simple words: To find when the given math problem gives a negative answer, we first find the special 'x' values where the answer is zero. These are \( -\frac{5}{4} \) and \( \frac{2}{3} \). Since the \( x^2 \) term is positive, the graph looks like a smile. So, it dips below zero (becomes negative) for all 'x' values that are in between these two special numbers.

🎯 Exam Tip: When solving quadratic inequalities, first find the roots of the corresponding equation. Then, consider the sign of the leading coefficient (coefficient of \( x^2 \)) to determine the intervals where the expression is positive or negative.

Question 6.
(i) Find the values of 'a' for which the expression \( x^2 - (3a - 1) x + 2a^2 + 2a - 11 \) is always positive.
(ii) If \( x^2 + 4ax + 2 > 0 \) for all values of x, then a lies in the interval
(a) (-2,4)
(b) (1, 2)
(c) \( (-\sqrt{2}, \sqrt{2}) \)
(d) \( (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \)
(e) (-4, 2)
Answer:
(i) For the expression \( x^2 - (3a - 1) x + 2a^2 + 2a - 11 \) to be always positive:
Compare with \( Ax^2 + Bx + C \). Here, \( A = 1 \), \( B = -(3a - 1) \), and \( C = 2a^2 + 2a - 11 \).
Condition 1: \( A > 0 \). Here, \( A = 1 \), which is \( > 0 \). This condition is satisfied.
Condition 2: The discriminant \( D \) must be less than zero. \( D = B^2 - 4AC \).
\( D = (-(3a - 1))^2 - 4 \times 1 \times (2a^2 + 2a - 11) \)
\( D = (3a - 1)^2 - 4(2a^2 + 2a - 11) \)
\( D = (9a^2 - 6a + 1) - (8a^2 + 8a - 44) \)
\( D = 9a^2 - 6a + 1 - 8a^2 - 8a + 44 \)
\( D = a^2 - 14a + 45 \)
For the expression to be always positive, we need \( D < 0 \):
\( a^2 - 14a + 45 < 0 \)
Factorize the quadratic:
\( (a - 5)(a - 9) < 0 \)
The critical points where the expression equals zero are \( a = 5 \) and \( a = 9 \).
Since the quadratic \( a^2 - 14a + 45 \) opens upwards (coefficient of \( a^2 \) is 1, which is positive), the expression is negative between its roots.
Therefore, \( 5 < a < 9 \).
The expression is always positive when 'a' is in the interval \( (5, 9) \).

(ii) For the expression \( x^2 + 4ax + 2 > 0 \) for all values of x:
Compare with \( Ax^2 + Bx + C \). Here, \( A = 1 \), \( B = 4a \), and \( C = 2 \).
Condition 1: \( A > 0 \). Here, \( A = 1 \), which is \( > 0 \). This condition is satisfied.
Condition 2: The discriminant \( D \) must be less than zero. \( D = B^2 - 4AC \).
\( D = (4a)^2 - 4 \times 1 \times 2 \)
\( D = 16a^2 - 8 \)
For the expression to be always positive, we need \( D < 0 \):
\( 16a^2 - 8 < 0 \)
\( 16a^2 < 8 \)
\( a^2 < \frac{8}{16} \)
\( a^2 < \frac{1}{2} \)
Taking the square root of both sides:
\( \sqrt{a^2} < \sqrt{\frac{1}{2}} \)
\( |a| < \frac{1}{\sqrt{2}} \)
This inequality means that 'a' must be between \( -\frac{1}{\sqrt{2}} \) and \( \frac{1}{\sqrt{2}} \):
\( -\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}} \)
So, a lies in the interval \( (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \).
The correct option is (d).
In simple words: For the first part, we need the 'a' number to make the whole math problem always positive. We found that the discriminant must be negative, which tells us 'a' must be between 5 and 9. For the second part, we use the same rules to find the range for 'a', and it turns out 'a' must be between \( -\frac{1}{\sqrt{2}} \) and \( \frac{1}{\sqrt{2}} \).

🎯 Exam Tip: Always remember to check both conditions for an expression to be always positive: \( a > 0 \) and \( D < 0 \). For MCQ questions, use these conditions to quickly narrow down the options.

Question 7. Find the greatest value of \( 3 + 5x - 2x^2 \) for all real values of x.
Answer:
Let the expression be \( y = 3 + 5x - 2x^2 \).
We can rearrange this as a quadratic equation in x:
\( 2x^2 - 5x + (y - 3) = 0 \)
For x to be a real number, the discriminant (D) of this quadratic equation must be greater than or equal to 0.
Here, \( a = 2 \), \( b = -5 \), \( c = (y - 3) \).
So, \( D = b^2 - 4ac \ge 0 \)
\( (-5)^2 - 4 \times 2 \times (y - 3) \ge 0 \)
\( 25 - 8(y - 3) \ge 0 \)
\( 25 - 8y + 24 \ge 0 \)
\( 49 - 8y \ge 0 \)
\( 49 \ge 8y \)
\( 8y \le 49 \)
\( y \le \frac{49}{8} \)
Converting the fraction to a mixed number:
\( y \le 6\frac{1}{8} \)
This means the value of y can be 6 and 1/8, or any number smaller than that. Therefore, the greatest possible value of y is \( 6\frac{1}{8} \).
In simple words: To find the highest value this math problem can reach, we treat it like a quadratic equation. Because 'x' must be a real number, the discriminant (D) must be zero or more. When we solve this, we find that the expression's value can be 6 and 1/8, or less, meaning its biggest possible value is exactly 6 and 1/8.

🎯 Exam Tip: To find the greatest or least value of a quadratic expression \( ax^2 + bx + c \) for real values of x, set the expression equal to y and form a quadratic in x. Then, apply the condition \( D \ge 0 \) for real roots to find the range of y.

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.
Answer:
Let \( y = \frac{6 x^2-22 x+21}{5 x^2-18 x+17} \).
Multiply both sides by the denominator:
\( y(5x^2 - 18x + 17) = 6x^2 - 22x + 21 \)
Expand and rearrange into a quadratic equation in x:
\( 5yx^2 - 18yx + 17y = 6x^2 - 22x + 21 \)
\( (5y - 6)x^2 + (-18y + 22)x + (17y - 21) = 0 \)
For x to be a real number, the discriminant (D) of this quadratic equation must be greater than or equal to 0.
Here, \( A = (5y - 6) \), \( B = (-18y + 22) \), \( C = (17y - 21) \).
So, \( D = B^2 - 4AC \ge 0 \)
\( (-18y + 22)^2 - 4(5y - 6)(17y - 21) \ge 0 \)
Factor out 2 from \( (-18y + 22) = 2(-9y + 11) \):
\( (2(-9y + 11))^2 - 4(5y - 6)(17y - 21) \ge 0 \)
\( 4(9y - 11)^2 - 4(5y - 6)(17y - 21) \ge 0 \)
Divide by 4:
\( (9y - 11)^2 - (5y - 6)(17y - 21) \ge 0 \)
\( (81y^2 - 198y + 121) - (85y^2 - 105y - 102y + 126) \ge 0 \)
\( (81y^2 - 198y + 121) - (85y^2 - 207y + 126) \ge 0 \)
\( 81y^2 - 198y + 121 - 85y^2 + 207y - 126 \ge 0 \)
\( -4y^2 + 9y - 5 \ge 0 \)
Multiply by -1 and reverse the inequality sign:
\( 4y^2 - 9y + 5 \le 0 \)
Factorize the quadratic:
\( (4y - 5)(y - 1) \le 0 \)
The critical points where the expression equals zero are \( 4y - 5 = 0 \implies y = \frac{5}{4} \) and \( y - 1 = 0 \implies y = 1 \).
Since the quadratic \( 4y^2 - 9y + 5 \) opens upwards (coefficient of \( y^2 \) is 4, which is positive), the expression is less than or equal to zero (negative) between its roots.
Therefore, \( 1 \le y \le \frac{5}{4} \).
The least value of y is 1.
In simple words: We want to find the smallest number this fraction can become. We set the fraction equal to 'y' and rearrange it to get a quadratic equation involving 'x'. Since 'x' must be a real number, the discriminant of this new equation must be greater than or equal to zero. Solving this inequality gives us the range for 'y', which is between 1 and \( 1\frac{1}{4} \). The smallest value in this range is 1.

🎯 Exam Tip: When finding the range of a rational expression, cross-multiply, rearrange to form a quadratic in x, and use the condition \( D \ge 0 \) to find the range of the variable y. Always be careful when multiplying or dividing an inequality by a negative number, as it reverses the inequality sign.

Question 9. If x be real, prove that the value of \( \frac{11 x^2+12 x+6}{x^2+4x+2} \) cannot lie between -5 and 3.
Answer:
Let \( y = \frac{11 x^2+12 x+6}{x^2+4x+2} \).
Multiply both sides by the denominator:
\( y(x^2 + 4x + 2) = 11x^2 + 12x + 6 \)
Expand and rearrange into a quadratic equation in x:
\( yx^2 + 4yx + 2y = 11x^2 + 12x + 6 \)
\( (y - 11)x^2 + (4y - 12)x + (2y - 6) = 0 \)
For x to be a real number, the discriminant (D) of this quadratic equation must be greater than or equal to 0.
Here, \( A = (y - 11) \), \( B = (4y - 12) \), \( C = (2y - 6) \).
So, \( D = B^2 - 4AC \ge 0 \)
\( (4y - 12)^2 - 4(y - 11)(2y - 6) \ge 0 \)
Factor out 4 from \( (4y - 12) = 4(y - 3) \). Factor out 2 from \( (2y - 6) = 2(y - 3) \).
\( (4(y - 3))^2 - 4(y - 11)2(y - 3) \ge 0 \)
\( 16(y - 3)^2 - 8(y - 11)(y - 3) \ge 0 \)
Factor out \( 8(y - 3) \):
\( 8(y - 3)[2(y - 3) - (y - 11)] \ge 0 \)
\( 8(y - 3)[2y - 6 - y + 11] \ge 0 \)
\( 8(y - 3)[y + 5] \ge 0 \)
Divide by 8:
\( (y - 3)(y + 5) \ge 0 \)
The critical points where the expression equals zero are \( y - 3 = 0 \implies y = 3 \) and \( y + 5 = 0 \implies y = -5 \).
Since the quadratic \( (y - 3)(y + 5) \) opens upwards (coefficient of \( y^2 \) would be 1, which is positive), the expression is greater than or equal to zero (positive) outside its roots.
Therefore, \( y \le -5 \) or \( y \ge 3 \).
This means that the value of y can be less than or equal to -5, or greater than or equal to 3.
Consequently, the value of y cannot lie between -5 and 3.
In simple words: To show that the fraction's value cannot be between -5 and 3, we set the fraction equal to 'y' and form a new quadratic equation for 'x'. For 'x' to be a real number, the discriminant must be greater than or equal to zero. When we solve this, we find that 'y' must either be -5 or less, or 3 or more. This proves that 'y' can never fall in the gap between -5 and 3.

🎯 Exam Tip: For "prove that value cannot lie between..." questions, set the expression equal to 'y', rearrange into a quadratic in x, and use the condition \( D \ge 0 \). The resulting inequality for 'y' will directly show the permissible range, thus proving what is excluded.