OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Chapter Test

Solve the following equations :

 

Question 1. \( 5^{x+1} + 5^{2-x} = 5^3 + 1 \)
Answer: The given equation is \( 5^{x+1} + 5^{2-x} = 5^3 + 1 \). We can rewrite this as \( 5^x \cdot 5^1 + 5^2 \cdot 5^{-x} = 125 + 1 \). So, \( 5 \cdot 5^x + \frac{25}{5^x} = 126 \). Now, let \( t = 5^x \). Substitute \( t \) into the equation: \( 5t + \frac{25}{t} = 126 \) Multiply the whole equation by \( t \) to clear the fraction: \( \implies 5t^2 + 25 = 126t \) Rearrange it into a standard quadratic equation form: \( \implies 5t^2 - 126t + 25 = 0 \) We can solve this quadratic equation by factoring. Break down the middle term \( -126t \): \( \implies 5t^2 - 125t - t + 25 = 0 \) Factor by grouping: \( \implies 5t(t - 25) - 1(t - 25) = 0 \) \( \implies (5t - 1)(t - 25) = 0 \) This gives us two possible values for \( t \): Either \( 5t - 1 = 0 \implies 5t = 1 \implies t = \frac{1}{5} \) Or \( t - 25 = 0 \implies t = 25 \) Now, substitute back \( 5^x \) for \( t \): Case 1: \( 5^x = \frac{1}{5} \) \( \implies 5^x = 5^{-1} \) \( \implies x = -1 \) Case 2: \( 5^x = 25 \) \( \implies 5^x = 5^2 \) \( \implies x = 2 \) So, the solutions for \( x \) are \( -1 \) and \( 2 \).In simple words: We changed the original equation by replacing \( 5^x \) with \( t \), which made it a simpler equation. We solved this simpler equation to find \( t \), then put \( 5^x \) back in place of \( t \) to find the values of \( x \).

🎯 Exam Tip: When solving exponential equations by substitution, remember to check both possible values of the substituted variable to ensure you find all solutions for the original variable.

 

Question 2. \( \sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6} \)
Answer: The given equation is \( \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = \frac{13}{6} \). Notice that the second term is the reciprocal of the first term. Let \( t = \sqrt{\frac{x}{1-x}} \). Then the equation becomes: \( t + \frac{1}{t} = \frac{13}{6} \) To solve for \( t \), multiply the entire equation by \( 6t \) to clear the denominators: \( \implies 6t^2 + 6 = 13t \) Rearrange this into a standard quadratic equation: \( \implies 6t^2 - 13t + 6 = 0 \) We can factor this quadratic equation. Find two numbers that multiply to \( 6 \times 6 = 36 \) and add up to \( -13 \) (these are \( -9 \) and \( -4 \) ): \( \implies 6t^2 - 9t - 4t + 6 = 0 \) Factor by grouping: \( \implies 3t(2t - 3) - 2(2t - 3) = 0 \) \( \implies (3t - 2)(2t - 3) = 0 \) This gives two possible values for \( t \): Either \( 3t - 2 = 0 \implies 3t = 2 \implies t = \frac{2}{3} \) Or \( 2t - 3 = 0 \implies 2t = 3 \implies t = \frac{3}{2} \) Now we substitute back \( \sqrt{\frac{x}{1-x}} \) for \( t \) for each case: Case 1: When \( t = \frac{2}{3} \) \( \sqrt{\frac{x}{1-x}} = \frac{2}{3} \) Square both sides to remove the square root: \( \implies \frac{x}{1-x} = \left(\frac{2}{3}\right)^2 \) \( \implies \frac{x}{1-x} = \frac{4}{9} \) Cross-multiply to solve for \( x \): \( \implies 9x = 4(1-x) \) \( \implies 9x = 4 - 4x \) Add \( 4x \) to both sides: \( \implies 13x = 4 \) \( \implies x = \frac{4}{13} \) Case 2: When \( t = \frac{3}{2} \) \( \sqrt{\frac{x}{1-x}} = \frac{3}{2} \) Square both sides: \( \implies \frac{x}{1-x} = \left(\frac{3}{2}\right)^2 \) \( \implies \frac{x}{1-x} = \frac{9}{4} \) Cross-multiply: \( \implies 4x = 9(1-x) \) \( \implies 4x = 9 - 9x \) Add \( 9x \) to both sides: \( \implies 13x = 9 \) \( \implies x = \frac{9}{13} \) Thus, the solutions for \( x \) are \( \frac{4}{13} \) and \( \frac{9}{13} \).In simple words: We saw that part of the equation was the inverse of another part, so we used a temporary variable \( t \) to make the equation simpler. We solved for \( t \), then put the original expression back in and squared both sides to find \( x \).

🎯 Exam Tip: When you encounter an equation where one part is the reciprocal of another, a substitution (like \( t \) for the expression) can simplify it into a solvable quadratic equation.

 

Question 3. \( (x + 1) (x + 2) (x + 3) (x + 4) = 120 \)
Answer: The given equation is \( (x+1)(x+2)(x+3)(x+4) = 120 \). To solve this, we group the terms cleverly. Multiply the first term by the last term, and the two middle terms together: \( \{(x+1)(x+4)\} \{(x+2)(x+3)\} = 120 \) Now, multiply the terms within each curly bracket: \( (x^2 + 4x + x + 4)(x^2 + 3x + 2x + 6) = 120 \) \( \implies (x^2 + 5x + 4)(x^2 + 5x + 6) = 120 \) Notice that \( x^2 + 5x \) appears in both factors. Let's make a substitution to simplify the equation. Let \( t = x^2 + 5x \). Substitute \( t \) into the equation: \( (t + 4)(t + 6) = 120 \) Expand the left side: \( \implies t^2 + 6t + 4t + 24 = 120 \) \( \implies t^2 + 10t + 24 = 120 \) Subtract \( 120 \) from both sides to form a quadratic equation: \( \implies t^2 + 10t - 96 = 0 \) Now, we factor this quadratic equation. We need two numbers that multiply to \( -96 \) and add up to \( 10 \) (these are \( 16 \) and \( -6 \) ): \( \implies t^2 + 16t - 6t - 96 = 0 \) Factor by grouping: \( \implies t(t + 16) - 6(t + 16) = 0 \) \( \implies (t - 6)(t + 16) = 0 \) This gives two possible values for \( t \): Either \( t - 6 = 0 \implies t = 6 \) Or \( t + 16 = 0 \implies t = -16 \) Now, we substitute back \( x^2 + 5x \) for \( t \) for each case: Case 1: When \( t = 6 \) \( x^2 + 5x = 6 \) Rearrange into a quadratic equation: \( \implies x^2 + 5x - 6 = 0 \) Factor this quadratic: \( \implies x^2 + 6x - x - 6 = 0 \) \( \implies x(x + 6) - 1(x + 6) = 0 \) \( \implies (x - 1)(x + 6) = 0 \) This gives two real solutions for \( x \): Either \( x - 1 = 0 \implies x = 1 \) Or \( x + 6 = 0 \implies x = -6 \) Case 2: When \( t = -16 \) \( x^2 + 5x = -16 \) Rearrange into a quadratic equation: \( \implies x^2 + 5x + 16 = 0 \) To find the roots of this quadratic, we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Here, \( a = 1, b = 5, c = 16 \). The discriminant \( D = b^2 - 4ac = (5)^2 - 4(1)(16) = 25 - 64 = -39 \). Since the discriminant is negative, the roots are complex (imaginary). \( x = \frac{-5 \pm \sqrt{-39}}{2} \) \( \implies x = \frac{-5 \pm i\sqrt{39}}{2} \) So, the solutions for \( x \) are \( 1, -6, \frac{-5 + i\sqrt{39}}{2}, \text{ and } \frac{-5 - i\sqrt{39}}{2} \).In simple words: We grouped terms in the original equation to find a common part, which we replaced with \( t \). After solving for \( t \), we put the original expression back in and solved two separate quadratic equations to find all the possible values for \( x \).

🎯 Exam Tip: For equations with many multiplied terms, look for ways to group them to create common expressions. This often allows for a substitution that simplifies the problem into a standard quadratic form.

 

Question 4. Prove that both the roots of the equation \( x^2 - x - 3 = 0 \) are irrational.
Answer: The given quadratic equation is \( x^2 - x - 3 = 0 \). To determine the nature of the roots, we need to calculate the discriminant \( D \). For a quadratic equation in the form \( ax^2 + bx + c = 0 \), the discriminant is given by the formula \( D = b^2 - 4ac \). In our equation, \( x^2 - x - 3 = 0 \), we have: \( a = 1 \) \( b = -1 \) \( c = -3 \) Now, substitute these values into the discriminant formula: \( D = (-1)^2 - 4(1)(-3) \) \( D = 1 - (-12) \) \( D = 1 + 12 \) \( D = 13 \) Since the discriminant \( D = 13 \), it is greater than zero ( \( D > 0 \) ). This means the roots are real and distinct. Furthermore, for the roots to be rational, the discriminant must be a perfect square (like 1, 4, 9, 16, etc.). Since \( 13 \) is not a perfect square, the roots must be irrational. Irrational roots always appear in pairs. Thus, both roots of the equation \( x^2 - x - 3 = 0 \) are irrational.In simple words: We checked the 'discriminant' of the equation. Because this number turned out to be positive but not a perfect square, it means the answers for \( x \) are real but cannot be written as simple fractions; they are irrational.

🎯 Exam Tip: Remember that for a quadratic equation \( ax^2 + bx + c = 0 \): if \( D > 0 \) and is a perfect square, roots are real and rational; if \( D > 0 \) and is not a perfect square, roots are real and irrational; if \( D = 0 \), roots are real and equal; if \( D < 0 \), roots are complex (imaginary).

 

Question 5. For what values of m will the equation \( x^2 - 2mx + 7m - 12 = 0 \) have (i) equal roots, (ii) reciprocal roots ?
Answer: The given quadratic equation is \( x^2 - 2mx + 7m - 12 = 0 \). We compare this with the standard quadratic form \( ax^2 + bx + c = 0 \). From the given equation, we have: \( a = 1 \) \( b = -2m \) \( c = 7m - 12 \)
(i) For equal roots: A quadratic equation has equal roots if its discriminant \( D \) is equal to zero ( \( D = 0 \) ). The discriminant formula is \( D = b^2 - 4ac \). Substitute the values of \( a, b, c \) into the formula: \( (-2m)^2 - 4(1)(7m - 12) = 0 \) \( \implies 4m^2 - 4(7m - 12) = 0 \) \( \implies 4m^2 - 28m + 48 = 0 \) Divide the entire equation by 4 to simplify: \( \implies m^2 - 7m + 12 = 0 \) Factor this quadratic equation. We need two numbers that multiply to \( 12 \) and add up to \( -7 \) (these are \( -3 \) and \( -4 \) ): \( \implies (m - 3)(m - 4) = 0 \) This gives two possible values for \( m \): Either \( m - 3 = 0 \implies m = 3 \) Or \( m - 4 = 0 \implies m = 4 \) So, for the equation to have equal roots, \( m \) must be \( 3 \) or \( 4 \).
(ii) For reciprocal roots: A quadratic equation has reciprocal roots if the product of its roots is equal to 1. For a quadratic equation \( ax^2 + bx + c = 0 \), the product of roots is \( \frac{c}{a} \). So, we set the product of roots to 1: \( \frac{c}{a} = 1 \) Substitute the values of \( a \) and \( c \): \( \frac{7m - 12}{1} = 1 \) \( \implies 7m - 12 = 1 \) Add 12 to both sides: \( \implies 7m = 13 \) Divide by 7: \( \implies m = \frac{13}{7} \) So, for the equation to have reciprocal roots, \( m \) must be \( \frac{13}{7} \).In simple words: To have equal answers for \( x \), the 'discriminant' of the equation must be zero, which gave us values of \( m \) as 3 or 4. To have answers for \( x \) that are inverses of each other (like 2 and 1/2), the product of the answers must be 1, which gave us \( m \) as 13/7.

🎯 Exam Tip: Remember the conditions for different types of roots: for equal roots, \( D=0 \); for real and distinct roots, \( D>0 \); for imaginary roots, \( D<0 \). For reciprocal roots, the product of roots (\( c/a \)) must be 1. For roots that are opposite in sign, the sum of roots (\( -b/a \)) must be 0.

 

Question 6. If one root of \( 2x^2 - 5x + k = 0 \) be double the other, find the value of k.
Answer: Let the roots of the quadratic equation \( 2x^2 - 5x + k = 0 \) be \( \alpha \) and \( 2\alpha \), since one root is double the other. For a standard quadratic equation \( Ax^2 + Bx + C = 0 \): The sum of the roots is \( -\frac{B}{A} \). The product of the roots is \( \frac{C}{A} \). From the given equation, we have \( A=2, B=-5, C=k \). Using the sum of roots: \( \alpha + 2\alpha = -\left(\frac{-5}{2}\right) \) \( \implies 3\alpha = \frac{5}{2} \) Solve for \( \alpha \): \( \implies \alpha = \frac{5}{6} \) Using the product of roots: \( \alpha \cdot (2\alpha) = \frac{k}{2} \) \( \implies 2\alpha^2 = \frac{k}{2} \) Now, substitute the value of \( \alpha = \frac{5}{6} \) into this equation: \( 2 \left(\frac{5}{6}\right)^2 = \frac{k}{2} \) \( \implies 2 \left(\frac{25}{36}\right) = \frac{k}{2} \) \( \implies \frac{25}{18} = \frac{k}{2} \) To find \( k \), multiply both sides by 2: \( \implies k = 2 \times \frac{25}{18} \) \( \implies k = \frac{25}{9} \) Thus, the value of \( k \) is \( \frac{25}{9} \).In simple words: We called the roots \( \alpha \) and \( 2\alpha \). Using the rules for the sum and product of roots in a quadratic equation, we first found the value of \( \alpha \) and then used it to calculate the value of \( k \).

🎯 Exam Tip: When given a relationship between roots (like one being double the other), express the roots in terms of a single variable (e.g., \( \alpha \) and \( 2\alpha \)) and use Vieta's formulas for sum and product of roots to set up equations and solve for the unknown constant.

 

Question 7. If \( \alpha, \beta \) be the roots of the equation \( x^2 - x - 1 = 0 \), determine the value of (i) \( \alpha^2 + \beta^2 \) and (ii) \( \alpha^3 + \beta^3 \).
Answer: The given quadratic equation is \( x^2 - x - 1 = 0 \). Let \( \alpha \) and \( \beta \) be its roots. For a quadratic equation \( ax^2 + bx + c = 0 \), we know from Vieta's formulas that: Sum of roots \( (\alpha + \beta) = -\frac{b}{a} \) Product of roots \( (\alpha\beta) = \frac{c}{a} \) In our equation, \( a=1, b=-1, c=-1 \). So, the sum of roots is: \( \alpha + \beta = -\left(\frac{-1}{1}\right) = 1 \) And the product of roots is: \( \alpha\beta = \frac{-1}{1} = -1 \)
(i) To determine the value of \( \alpha^2 + \beta^2 \): We use the algebraic identity \( (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta \). Rearranging this to find \( \alpha^2 + \beta^2 \): \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \) Now, substitute the values of \( (\alpha + \beta) \) and \( (\alpha\beta) \) we found: \( \alpha^2 + \beta^2 = (1)^2 - 2(-1) \) \( \implies \alpha^2 + \beta^2 = 1 + 2 \) \( \implies \alpha^2 + \beta^2 = 3 \)
(ii) To determine the value of \( \alpha^3 + \beta^3 \): We use the algebraic identity \( (\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta) \). Rearranging this to find \( \alpha^3 + \beta^3 \): \( \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \) Now, substitute the values of \( (\alpha + \beta) \) and \( (\alpha\beta) \): \( \alpha^3 + \beta^3 = (1)^3 - 3(-1)(1) \) \( \implies \alpha^3 + \beta^3 = 1 - (-3) \) \( \implies \alpha^3 + \beta^3 = 1 + 3 \) \( \implies \alpha^3 + \beta^3 = 4 \)In simple words: First, we found the sum and product of the roots using simple formulas. Then, we used algebraic rules to express \( \alpha^2 + \beta^2 \) and \( \alpha^3 + \beta^3 \) in terms of this sum and product, and finally calculated their values.

🎯 Exam Tip: Master the fundamental algebraic identities for powers of binomials. They are crucial for expressing symmetric polynomial expressions of roots (like sums of powers) in terms of the elementary symmetric polynomials (sum and product of roots), which can be directly found from the coefficients of the quadratic equation.

 

Question 8. If the roots of the equation \( ax^2 + bx + c = 0 \) be in the ratio 3 : 4, show that \( 12b^2 = 49ac \).
Answer: Let the roots of the quadratic equation \( ax^2 + bx + c = 0 \) be in the ratio \( 3:4 \). We can represent the roots as \( 3\alpha \) and \( 4\alpha \) for some non-zero value \( \alpha \). From Vieta's formulas for a quadratic equation \( ax^2 + bx + c = 0 \): Sum of roots \( = -\frac{b}{a} \) Product of roots \( = \frac{c}{a} \) Using the sum of roots with our chosen representation: \( 3\alpha + 4\alpha = -\frac{b}{a} \) \( \implies 7\alpha = -\frac{b}{a} \) Solve for \( \alpha \): \( \implies \alpha = -\frac{b}{7a} \) ...(1) Using the product of roots: \( (3\alpha)(4\alpha) = \frac{c}{a} \) \( \implies 12\alpha^2 = \frac{c}{a} \) ...(2) Now, substitute the value of \( \alpha \) from equation (1) into equation (2): \( 12 \left(-\frac{b}{7a}\right)^2 = \frac{c}{a} \) Square the term in the bracket: \( \implies 12 \left(\frac{b^2}{49a^2}\right) = \frac{c}{a} \) \( \implies \frac{12b^2}{49a^2} = \frac{c}{a} \) To clear the denominator on the left side, multiply both sides by \( 49a^2 \): \( \implies 12b^2 = \frac{c}{a} \cdot 49a^2 \) \( \implies 12b^2 = 49ac \) This proves the required relationship between the coefficients.In simple words: We used a simple way to write the roots based on their given ratio. Then, by using the rules for how roots relate to the coefficients in a quadratic equation (sum and product), we combined these rules to prove the given mathematical statement.

🎯 Exam Tip: When roots are given in a ratio, represent them as \( k\alpha \) and \( m\alpha \) (e.g., \( 3\alpha \) and \( 4\alpha \)). Then, form two equations using the sum and product of roots, eliminate \( \alpha \), and simplify to find the required relation between the coefficients.

 

Question 9. If x is real, prove that the quadratic expression (i) \( (x - 2) (x + 3) + 7 \) is always positive. (ii) \( 4x - 3x^2 - 2 \) is always negative.
Answer: To determine if a quadratic expression is always positive or always negative for real \( x \), we examine its leading coefficient \( a \) and its discriminant \( D \). For a quadratic expression \( ax^2 + bx + c \): - If \( a > 0 \) and \( D < 0 \), the expression is always positive. - If \( a < 0 \) and \( D < 0 \), the expression is always negative.
(i) For the expression \( (x - 2) (x + 3) + 7 \): First, expand and simplify the expression: \( (x - 2)(x + 3) + 7 = (x^2 + 3x - 2x - 6) + 7 \) \( = x^2 + x - 6 + 7 \) \( = x^2 + x + 1 \) Now, compare this with \( ax^2 + bx + c \): \( a = 1 \) \( b = 1 \) \( c = 1 \) Since \( a = 1 \), which is \( > 0 \), the parabola opens upwards. Next, calculate the discriminant \( D = b^2 - 4ac \): \( D = (1)^2 - 4(1)(1) \) \( D = 1 - 4 \) \( D = -3 \) Since \( D = -3 \), which is \( < 0 \), the quadratic equation \( x^2 + x + 1 = 0 \) has no real roots, meaning the parabola does not intersect the x-axis. Because \( a > 0 \) and \( D < 0 \), the expression \( x^2 + x + 1 \) is always positive for all real values of \( x \).
(ii) For the expression \( 4x - 3x^2 - 2 \): First, rearrange the expression in standard quadratic form: \( -3x^2 + 4x - 2 \) Now, compare this with \( ax^2 + bx + c \): \( a = -3 \) \( b = 4 \) \( c = -2 \) Since \( a = -3 \), which is \( < 0 \), the parabola opens downwards. Next, calculate the discriminant \( D = b^2 - 4ac \): \( D = (4)^2 - 4(-3)(-2) \) \( D = 16 - (24) \) \( D = 16 - 24 \) \( D = -8 \) Since \( D = -8 \), which is \( < 0 \), the quadratic equation \( -3x^2 + 4x - 2 = 0 \) has no real roots, meaning the parabola does not intersect the x-axis. Because \( a < 0 \) and \( D < 0 \), the expression \( -3x^2 + 4x - 2 \) is always negative for all real values of \( x \).In simple words: For the first expression, after simplifying it, we found that its leading number was positive and its 'discriminant' was negative. This means its graph is always above the x-axis, so it's always positive. For the second expression, its leading number was negative and its 'discriminant' was also negative. This means its graph is always below the x-axis, so it's always negative.

🎯 Exam Tip: To prove that a quadratic expression is always positive or always negative, calculate its discriminant \( D \) and observe the sign of the leading coefficient \( a \). If \( D < 0 \), the expression never crosses the x-axis. Its sign is then determined solely by \( a \).

 

Question 10. Draw the graph of the quadratic function \( x^2 - 4x + 3 \) and hence find the roots of the equation \( x^2 - 4x + 4 = 0 \). What is the minimum value of the function ?
Answer: Let the quadratic function be \( y = x^2 - 4x + 3 \). To draw the graph, we find some points by choosing values for \( x \) and calculating the corresponding \( y \) values:

These points include \( (0,3), (1,0), (2,-1), (3,0), (4,3), (5,8) \). Plotting these points and drawing a smooth curve through them gives the parabola for \( y = x^2 - 4x + 3 \).
From the graph, we can see: 1. The roots of \( x^2 - 4x + 3 = 0 \) are the x-intercepts of the graph, which are \( x = 1 \) and \( x = 3 \). These are the points \( (1,0) \) and \( (3,0) \). 2. The minimum value of the function \( y = x^2 - 4x + 3 \) is the y-coordinate of the vertex. From the table and graph, the vertex is \( (2, -1) \). So, the minimum value of \( y \) is \( -1 \).
Now, to find the roots of the equation \( x^2 - 4x + 4 = 0 \) using the graph: We know that \( y = x^2 - 4x + 3 \). The equation \( x^2 - 4x + 4 = 0 \) can be written as \( (x^2 - 4x + 3) + 1 = 0 \), which means \( y + 1 = 0 \), or \( y = -1 \). On the graph of \( y = x^2 - 4x + 3 \), the line \( y = -1 \) is a horizontal line that passes through the vertex \( (2, -1) \). This line touches the parabola at exactly one point, the vertex itself. Therefore, the equation \( x^2 - 4x + 4 = 0 \) has one repeated real root, which is \( x = 2 \).In simple words: We made a table of points for the first equation and drew its graph, which is a U-shaped curve. From this curve, we found the places where it touched the x-axis (the roots) and its lowest point (the minimum value). Then, by looking at the graph, we found the answers for the second equation by seeing where the curve touched the line \( y=-1 \).

🎯 Exam Tip: When using a graph to find roots of a modified equation (e.g., \( f(x)+k=0 \)), remember that this is equivalent to finding the x-values where the original graph \( y=f(x) \) intersects the horizontal line \( y=-k \). The minimum or maximum of a parabola \( y=ax^2+bx+c \) occurs at \( x = -b/(2a) \).

 

Question 11. For what values of 'a' will the expression \( x^2 - ax + 1 - 2a^2 \), for real \( x \), be always positive?
Answer: Let the given quadratic expression be \( E(x) = x^2 - ax + 1 - 2a^2 \). For a quadratic expression \( Ax^2 + Bx + C \) to be always positive for all real values of \( x \), two conditions must be met: 1. The leading coefficient \( A \) must be positive ( \( A > 0 \) ). This means the parabola opens upwards. 2. The discriminant \( D \) must be negative ( \( D < 0 \) ). This means the quadratic equation \( Ax^2 + Bx + C = 0 \) has no real roots, so the parabola never crosses the x-axis. Comparing \( E(x) \) with \( Ax^2 + Bx + C \): We have \( A = 1 \), \( B = -a \), and \( C = 1 - 2a^2 \). The first condition is already satisfied: \( A = 1 > 0 \). Now, we apply the second condition, \( D < 0 \). The discriminant is \( D = B^2 - 4AC \). Substitute the values of \( A, B, C \): \( D = (-a)^2 - 4(1)(1 - 2a^2) \) \( \implies D = a^2 - 4(1 - 2a^2) \) \( \implies D = a^2 - 4 + 8a^2 \) \( \implies D = 9a^2 - 4 \) For \( E(x) \) to be always positive, we need \( D < 0 \): \( 9a^2 - 4 < 0 \) \( \implies 9a^2 < 4 \) \( \implies a^2 < \frac{4}{9} \) Taking the square root of both sides, remember to consider both positive and negative roots: \( \implies \sqrt{a^2} < \sqrt{\frac{4}{9}} \) \( \implies |a| < \frac{2}{3} \) This inequality means that \( a \) must be between \( -\frac{2}{3} \) and \( \frac{2}{3} \). So, \( -\frac{2}{3} < a < \frac{2}{3} \). The values of 'a' for which the expression is always positive are in the interval \( \left(-\frac{2}{3}, \frac{2}{3}\right) \).In simple words: For the expression to always be positive, its graph must be a U-shape that never touches the x-axis. This happens when the number in front of \( x^2 \) is positive (which it is here, \( 1 \)) and when its 'discriminant' (a special calculation) is negative. We solved for 'a' to make the discriminant negative, finding that 'a' must be between \( -2/3 \) and \( 2/3 \).

🎯 Exam Tip: To ensure a quadratic expression is always positive or negative, always check two conditions: the sign of the leading coefficient (for the curve's direction) and the sign of the discriminant (for whether it crosses the x-axis). Both conditions must be satisfied simultaneously.

 

Question 12. If x be real, prove that the value of \( \frac{2x^2-2x+4}{x^2-4x+3} \) cannot lie between – 7 and 1.
Answer: Let \( y = \frac{2x^2-2x+4}{x^2-4x+3} \). We need to find the possible values of \( y \). Multiply both sides by \( (x^2-4x+3) \): \( \implies y(x^2-4x+3) = 2x^2-2x+4 \) Expand the left side: \( \implies yx^2 - 4yx + 3y = 2x^2 - 2x + 4 \) Rearrange the terms to form a quadratic equation in \( x \): \( \implies yx^2 - 2x^2 - 4yx + 2x + 3y - 4 = 0 \) Group terms by powers of \( x \): \( \implies (y-2)x^2 + (-4y+2)x + (3y-4) = 0 \) \( \implies (y-2)x^2 + (2-4y)x + (3y-4) = 0 \) Since \( x \) is a real number, the discriminant of this quadratic equation must be greater than or equal to zero ( \( D \ge 0 \) ). Here, \( A = (y-2) \), \( B = (2-4y) \), and \( C = (3y-4) \). The discriminant formula is \( D = B^2 - 4AC \). \( \implies (2-4y)^2 - 4(y-2)(3y-4) \ge 0 \) Factor out 4 from the first term: \( (2-4y)^2 = (2(1-2y))^2 = 4(1-2y)^2 \). \( \implies 4(1-2y)^2 - 4(y-2)(3y-4) \ge 0 \) Divide the entire inequality by 4: \( \implies (1-2y)^2 - (y-2)(3y-4) \ge 0 \) Expand both squared and multiplied terms: \( \implies (1 - 4y + 4y^2) - (3y^2 - 4y - 6y + 8) \ge 0 \) \( \implies 1 - 4y + 4y^2 - (3y^2 - 10y + 8) \ge 0 \) Remove the parenthesis, changing signs of terms inside: \( \implies 1 - 4y + 4y^2 - 3y^2 + 10y - 8 \ge 0 \) Combine like terms: \( \implies y^2 + 6y - 7 \ge 0 \) Factor this quadratic inequality: \( \implies (y+7)(y-1) \ge 0 \) To find the values of \( y \) that satisfy this inequality, we find the critical points where the expression equals zero: \( y = -7 \) and \( y = 1 \). We test intervals on a number line: - If \( y < -7 \), both \( (y+7) \) and \( (y-1) \) are negative, so their product is positive. - If \( -7 < y < 1 \), \( (y+7) \) is positive and \( (y-1) \) is negative, so their product is negative. - If \( y > 1 \), both \( (y+7) \) and \( (y-1) \) are positive, so their product is positive. So, for \( (y+7)(y-1) \ge 0 \) to be true, \( y \) must satisfy \( y \le -7 \) or \( y \ge 1 \). This means that \( y \) cannot take any value between \( -7 \) and \( 1 \). Therefore, the value of the given expression cannot lie between \( -7 \) and \( 1 \).In simple words: We changed the fraction into an equation, then rearranged it to look like a quadratic equation with \( x \) as the variable. Because \( x \) has to be a real number, the 'discriminant' of this equation must be positive or zero. Solving this condition gave us an inequality for \( y \), which showed that \( y \) must be less than or equal to \( -7 \) or greater than or equal to \( 1 \). This proves that \( y \) cannot be in between \( -7 \) and \( 1 \).

🎯 Exam Tip: When proving a range of values for a rational function like \( y = \frac{f(x)}{g(x)} \), convert it into a quadratic equation in \( x \). The condition that \( x \) is real implies the discriminant \( D \ge 0 \). Solving this inequality for \( y \) will give the valid range for \( y \).

 

Question 13. If the roots of the equation \( qx^2 + 2px + 2q = 0 \) are real and unequal, prove that the roots of the equation \( (p + q) x^2 + 2qx +(p - q) = 0 \) are imaginary.
Answer: We need to analyze two quadratic equations based on their discriminants.
Equation 1: \( qx^2 + 2px + 2q = 0 \) For this equation, let \( A_1=q, B_1=2p, C_1=2q \). It is given that its roots are real and unequal. This means its discriminant \( D_1 \) must be greater than zero. \( D_1 = B_1^2 - 4A_1C_1 \) \( D_1 = (2p)^2 - 4(q)(2q) \) \( D_1 = 4p^2 - 8q^2 \) Since \( D_1 > 0 \), we have: \( 4p^2 - 8q^2 > 0 \) Divide by 4: \( \implies p^2 - 2q^2 > 0 \) ...(A) This inequality tells us a relationship between \( p^2 \) and \( q^2 \).
Equation 2: \( (p + q) x^2 + 2qx + (p - q) = 0 \) For this equation, let \( A_2=(p+q), B_2=2q, C_2=(p-q) \). We need to prove that its roots are imaginary. This requires showing that its discriminant \( D_2 \) is less than zero. \( D_2 = B_2^2 - 4A_2C_2 \) \( D_2 = (2q)^2 - 4(p+q)(p-q) \) \( D_2 = 4q^2 - 4(p^2 - q^2) \) (Using the identity \( (a+b)(a-b) = a^2-b^2 \)) Expand and simplify: \( D_2 = 4q^2 - 4p^2 + 4q^2 \) \( D_2 = 8q^2 - 4p^2 \) Factor out 4: \( D_2 = 4(2q^2 - p^2) \) Now, let's use the inequality from (A): \( p^2 - 2q^2 > 0 \). Multiply this inequality by \( -1 \): \( -(p^2 - 2q^2) < 0 \) (Remember to reverse the inequality sign) \( \implies -p^2 + 2q^2 < 0 \) \( \implies 2q^2 - p^2 < 0 \) Since \( 2q^2 - p^2 \) is negative, and \( D_2 = 4(2q^2 - p^2) \), it means: \( D_2 = 4 \times (\text{a negative number}) \) Therefore, \( D_2 < 0 \). Since the discriminant of the second equation is negative, its roots are imaginary. This completes the proof.In simple words: We used the condition that the first equation has real roots to get an inequality relating \( p \) and \( q \). Then, for the second equation, we calculated its 'discriminant'. By using the inequality from the first part, we showed that the second discriminant must be a negative number. A negative discriminant means the roots are imaginary.

🎯 Exam Tip: This type of problem often involves using the discriminant condition (\( D>0 \), \( D=0 \), or \( D<0 \)) from one quadratic equation to prove a property about another. Always clearly define the discriminants for each equation and establish the relationship between their coefficients.

 

Question 14. If \( \alpha, \beta \) be the roots of \( x^2 - px + q = 0 \), find the value of \( \alpha^5\beta^7 + \alpha^7\beta^5 \) in terms of p and q.
Answer: Given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - px + q = 0 \). Using Vieta's formulas, we can find the sum and product of the roots in terms of the coefficients \( p \) and \( q \): Sum of roots: \( \alpha + \beta = -(-p)/1 = p \) Product of roots: \( \alpha\beta = q/1 = q \) We need to find the value of the expression \( \alpha^5\beta^7 + \alpha^7\beta^5 \). First, factor out the common terms from the expression: \( \alpha^5\beta^7 + \alpha^7\beta^5 = \alpha^5\beta^5 (\beta^2 + \alpha^2) \) This can be rewritten as: \( = (\alpha\beta)^5 (\alpha^2 + \beta^2) \) Now, we need to express \( (\alpha^2 + \beta^2) \) in terms of \( (\alpha+\beta) \) and \( (\alpha\beta) \). We know the identity: \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \) Substitute the values \( (\alpha+\beta) = p \) and \( (\alpha\beta) = q \) into this identity: \( \alpha^2 + \beta^2 = (p)^2 - 2(q) \) \( \implies \alpha^2 + \beta^2 = p^2 - 2q \) Now, substitute this result back into our factored expression: \( (\alpha\beta)^5 (\alpha^2 + \beta^2) = (q)^5 (p^2 - 2q) \) So, the value of \( \alpha^5\beta^7 + \alpha^7\beta^5 \) in terms of \( p \) and \( q \) is \( q^5(p^2 - 2q) \). This helps connect root properties to equation coefficients.In simple words: We started by finding the sum and product of the roots using the numbers \( p \) and \( q \) from the given equation. Then, we simplified the complex expression by taking out common factors. Finally, we used a known algebraic rule to replace \( \alpha^2 + \beta^2 \) with a combination of the sum and product, giving us the answer in terms of \( p \) and \( q \).

🎯 Exam Tip: When dealing with expressions involving high powers of roots, always try to factor out common terms and then use fundamental symmetric polynomial identities (like \( \alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta \) or \( \alpha^3+\beta^3 = (\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) \)) to reduce them to expressions involving only \( (\alpha+\beta) \) and \( (\alpha\beta) \), which can be directly replaced by the coefficients of the quadratic equation.

 

Question 21. If \(\alpha\) and \(\beta\) are the roots of \(ax^2 + bx + c = 0\) and if \(px^2 + qx + r = 0\) has roots \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\) then \(r\) =
(a) a + 2b
(b) a + b + c
(c) ab + bc + ca
(d) abc
Answer: (b) a + b + c
In simple words: When you set up the relationships between the roots and coefficients for both equations, you can solve for the unknown 'r' by comparing the product of the roots in the second equation. The product of roots for the new equation is found using the sum and product of roots from the first equation, leading to the simple expression for r.

🎯 Exam Tip: Remember the formulas for the sum and product of roots for a quadratic equation. For `Ax² + Bx + C = 0`, the sum of roots is `-B/A` and the product is `C/A`. This is key for solving such questions.

 

Question 22. The quadratic equations \(x^2 - 6x + a = 0\) and \(x^2 - cx + 6 = 0\) have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is
(a) 1
(b) 4
(c) 3
(d) 2
Answer: (d) 2
In simple words: We find the common root by using the given ratio of the other roots. By setting up equations for the sum and product of roots for both quadratic equations, we can solve for 'a' and then find the possible roots, checking which one satisfies all conditions.

🎯 Exam Tip: When dealing with common roots and ratios, express all roots in terms of common variables (like \(\alpha\) and \(\beta\)). Use the sum and product of roots relations to form a system of equations, which can then be solved. Always verify your solution with all conditions given in the problem.

 

Question 23. If \(\alpha, \beta\) are the roots of the equation \(\lambda(x^2 - x) + x + 5 = 0\) and if \(\lambda_1\) and \(\lambda_2\) are two values of \(\lambda\) obtained from \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{5}\), then \(\frac{\lambda_1}{\lambda_2^2}+\frac{\lambda_2}{\lambda_1^2}\) equals
(a) 4192
(b) 4144
(c) 4096
(d) 4048
Answer: (d) 4048
In simple words: First, we change the given equation into a standard quadratic form and find the sum and product of its roots, \(\alpha\) and \(\beta\), in terms of \(\lambda\). Then, we use the condition \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{5}\) to create a new quadratic equation for \(\lambda\). Once we find the two values of \(\lambda\) from this equation, we can use the sum and product of \(\lambda_1\) and \(\lambda_2\) to calculate the final expression.

🎯 Exam Tip: Questions involving symmetric expressions of roots often simplify using sum and product relationships. Remember the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) and \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\) to simplify such calculations.

 

Question 24. If \(\alpha, \beta\) be the roots of \(x^2 - a (x - 1) + b = 0\), then the value of \(\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}\) is
(b) \(\frac{1}{a+b}\)
(c) 0
(d) -1
Answer: (c) 0
In simple words: Since \(\alpha\) and \(\beta\) are roots of the given equation, they must satisfy it. This means we can replace parts of the expression \(\alpha^2 - a\alpha\) and \(\beta^2 - a\beta\) with a simpler term involving 'a' and 'b'. After making this substitution, the entire expression simplifies to zero.

🎯 Exam Tip: Always use the fact that roots satisfy their quadratic equation to simplify complex expressions. Rearranging the original equation in terms of \(\alpha^2 - a\alpha\) (or \(\beta^2 - a\beta\)) is a common trick to solve such problems efficiently.