6 Most SAQ’s of Quadratic Expressions Chapter in Inter 2nd Year Maths-2A (TS/AP)

SAQ-1 : Find the range of (x+2)/2x²+3x+6) for x ∈ R

Let $$y = x + \frac{2}{2x^2 + 3x + 6}$$

$$\Rightarrow y(2x^2 + 3x + 6) = x + 2$$

$$\Rightarrow 2x^2y + 3xy + 6y – x – 2 = 0$$

$$\Rightarrow 2x^2y + x(3y – 1) + 6y – 2 = 0$$

$$\Rightarrow (2y)x^2 + (3y – 1)x + (6y – 2) = 0$$

(1) is a quadratic equation in x, and its roots are reals, which implies the discriminant ΔΔ must be greater than or equal to 0:

$$\Delta = b^2 – 4ac \geq 0$$

$$\Rightarrow (3y – 1)^2 – 4(2y)(6y – 2) \geq 0$$

$$\Rightarrow (9y^2 – 6y + 1) – 48y^2 + 16y \geq 0$$

$$\Rightarrow -39y^2 + 10y + 1 \geq 0$$

$$\Rightarrow 39y^2 – 10y – 1 \leq 0$$

$$\Rightarrow 39y^2 – 13y + 3y – 1 \leq 0$$

$$\Rightarrow 13y(3y – 1) + 1(3y – 1) \leq 0$$

$$\Rightarrow (13y + 1)(3y – 1) \leq 0$$

$$\Rightarrow y \in [-1/13, 1/3]$$

Therefore, the range of y is $$[-1/13, 1/3]$$

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SAQ-2 : Find the range of (x²+x+1)/(x²-x+1) for x ∈ R

Let $$y = x^2 + x + \frac{1}{x^2} – x + 1$$

$$\Rightarrow y(x^2 – x + 1) = x^2 + x + 1$$

$$\Rightarrow yx^2 – yx + y = x^2 + x + 1$$

$$\Rightarrow yx^2 – x^2 – yx – x + y – 1 = 0$$

$$\Rightarrow x^2(y – 1) – x(y + 1) + (y – 1) = 0$$

$$\Rightarrow (y – 1)x^2 – (y + 1)x + (y – 1) = 0$$

(1) is a quadratic equation in x, and its roots are reals, implying the discriminant ΔΔ must be greater than or equal to 0:

$$\Delta = b^2 – 4ac \geq 0$$

$$\Rightarrow (y + 1)^2 – 4(y – 1)^2 \geq 0$$

$$\Rightarrow (y + 1)^2 – (2y – 2)^2 \geq 0$$

Using the identity $$a^2 – b^2 = (a+b)(a-b)$$ we can simplify this to:

$$\Rightarrow (y + 1 + 2y – 2)(y + 1) – (2y – 2) \geq 0$$

$$\Rightarrow (3y – 1)(y + 1 – 2y + 2) \geq 0$$

$$\Rightarrow (3y – 1)(3 – y) \geq 0$$

$$\Rightarrow (3y – 1)(-y + 3) \geq 0$$

$$\Rightarrow (3y – 1)(y – 3) \leq 0$$

$$\Rightarrow y \in [1/3, 3]$$

Therefore, the range of y is $$[1/3, 3]$$

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SAQ-3 : Find the maximum value of the function (x²+14x+9)/(x²+2x+3) over R

Let $$y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$$

$$\Rightarrow y(x^2 + 2x + 3) = x^2 + 14x + 9$$

$$\Rightarrow yx^2 + 2xy + 3y = x^2 + 14x + 9$$

$$\Rightarrow yx^2 + 2xy + 3y – x^2 – 14x – 9 = 0$$

$$\Rightarrow x^2(y – 1) + x(2y – 14) + (3y – 9) = 0$$

$$\Rightarrow (y – 1)x^2 + (2y – 14)x + (3y – 9) = 0$$

Given that (1) is a quadratic equation in x and its roots are real, the discriminant (Δ) must be non-negative:

$$\Delta = b^2 – 4ac \geq 0$$

$$\Rightarrow (2y – 14)^2 – 4(y – 1)(3y – 9) \geq 0$$

$$\Rightarrow 4(y – 7)^2 – 4(y – 1)(3y – 9) \geq 0$$

$$y \in [-5, 4]$$

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SAQ-4 : If x is real, prove that x/(x²-5x+9) lies between 1 and (-1)/11

Let $$y = \frac{x}{x^2 – 5x + 9}$$

$$\Rightarrow y(x^2 – 5x + 9) = x$$

$$\Rightarrow yx^2 – 5yx + 9y – x = 0$$

$$\Rightarrow yx^2 – (5y + 1)x + 9y = 0$$

Given that (1) is a quadratic equation in x and its roots are real, the discriminant (Δ) must be non-negative:

$$\Delta = b^2 – 4ac \geq 0$$

$$\Rightarrow (5y + 1)^2 – 4(y)(9y) \geq 0$$

$$\Rightarrow (25y^2 + 10y + 1) – 36y^2 \geq 0$$

$$\Rightarrow -11y^2 + 10y + 1 \geq 0$$

$$\Rightarrow 11y^2 – 10y – 1 \leq 0$$

$$\Rightarrow 11y(y-1) + (y-1) \leq 0$$

$$\Rightarrow (11y + 1)(y – 1) \leq 0$$

$$\Rightarrow y \in [-1/11,1]$$

Therefore, the range of y is $$[-1/11, 1]$$

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SAQ-5 : Show that 1/3x+1+1/x+1-1/(3x+1)(x+1) does not lie between 1 and 4, if x is real

$$G.E = \frac{1}{3x} + 1 + \frac{1}{x} + 1 – \frac{1}{(3x + 1)(x + 1)} = x + 1 + 3x + 1 – \frac{1}{(3x + 1)(x + 1)} = 4x + \frac{1}{3x^2} + 4x + 1$$

Let: $$y = 4x + \frac{1}{3x^2} + 4x + 1$$

$$\Rightarrow y(3x^2 + 4x + 1) = 4x + 1$$

$$\Rightarrow 3yx^2 + 4yx + y = 4x + 1$$

$$\Rightarrow 3yx^2 + (4y – 4)x + (y – 1) = 0 \quad \text{(1)}$$

Since (1) is a quadratic equation in $�$ and its roots are real, we have:

$$\Delta = b^2 – 4ac \geq 0$$

$$(4y – 4)^2 – 4(3y)(y – 1) \geq 0$$

$$\Rightarrow 16y^2 + 16 – 32y – 12y^2 + 12y \geq 0$$

$$\Rightarrow 4y^2 – 20y + 16 \geq 0$$

$$\Rightarrow 4(y^2 – 5y + 4) \geq 0$$

$$\Rightarrow y^2 – 5y + 4 \geq 0$$

$$\Rightarrow (y – 1)(y – 4) \geq 0$$

$$\Rightarrow y \leq 1 \text{ or } y \geq 4$$

$$\Rightarrow y \text{ does not lie between 1 and 4}$$

Hence, the given expression does not lie between 1 and 4.

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SAQ-6 : If the expression (x-p)/(x²-3x+2) takes all real values for x ∈ R then find the bounds for p

$$\text{Let } y = x – \frac{p}{x^2 – 3x + 2}$$

$$\Rightarrow y(x^2 – 3x + 2) = x – p$$

$$\Rightarrow yx^2 – 3yx + 2y = x – p$$

$$\Rightarrow yx^2 + (-3y – 1)x + (2y + p) = 0 \quad \text{(1)}$$

Since (1) is a quadratic equation in x and its roots are reals, we have:

$$\Delta = b^2 – 4ac \geq 0$$

$$\Rightarrow (-3y – 1)^2 – 4y(2y + p) \geq 0$$

$$\Rightarrow 9y^2 + 6y + 1 – 8y^2 – 4py \geq 0$$

$$\Rightarrow y^2 + (6 – 4p)y + 1 \geq 0 \quad \text{(2)}$$

Given that y is real and the coefficient of y² is positive, (2) holds true only when the roots of $$y^2 + (6-4p)y + 1 = 0$$ are imaginary or real & equal, implying:

$$\Delta = b^2 – 4ac \leq 0$$

$$\Rightarrow (6 – 4p)^2 – 4 \leq 0$$

$$\Rightarrow 36 + 16p^2 – 48p – 4 \leq 0$$

$$\Rightarrow 16p^2 – 48p + 32 \leq 0$$

$$\Rightarrow 16(p^2 – 3p + 2) \leq 0$$

$$\Rightarrow p^2 – 3p + 2 \leq 0$$

$$\Rightarrow (p – 1)(p – 2) \leq 0$$

$$\Rightarrow 1 \leq p \leq 2$$

However, if $$x = p = 1$$ or 2, then $$x – \frac{p}{x^2 – 3x + 2} = x – \frac{p}{(x – 1)(x – 2)}$$

takes the indeterminate form 0/0.

$$1 < p < 2$$