SATHEE: Theory of Equations 4 Question 6

Theory of Equations 4 Question 6

7. If the roots of the equation $x^{2} - 2 a x + a^{2} + a - 3 = 0$ are real and less than 3 , then

(1999, 2M)

(a) $a < 2$

(b) $2 \leq a \leq 3$

(d) $a > 4$

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Answer:

Correct Answer: 7. (a)

Solution:

Let $f (x) = x^{2} - 2 a x + a^{2} + a - 3$

Since, both root are less than 3 .

$\begin{array}{lrl} \Rightarrow & α < 3, β & < 3 \\ \Rightarrow & Sum, S = α + β & < 6 \\ \Rightarrow & \frac{α + β}{2} < 3 \\ \Rightarrow & \frac{2 a}{2} < 3 \\ \Rightarrow & a < 3 \end{array}$

Again, product, $P = α β$

alt text

$\begin{array}{lc} \Rightarrow & P < 9 \Rightarrow α β < 9 \\ \Rightarrow & a^{2} + a - 3 < 9 \\ \Rightarrow & a^{2} + a - 12 < 0 \\ \Rightarrow & (a - 3) (a + 4) < 0 \\ \Rightarrow & - 4 < a < 3 \dots (ii) α \\ Again, D = B^{2} - 4 A C \geq 0 \\ \Rightarrow & (- 2 a)^{2} - 4 \cdot 1 (a^{2} + a - 3) \geq 0 \\ \Rightarrow & 4 a^{2} - 4 a^{2} - 4 a + 12 \geq 0 \\ \Rightarrow & - 4 a + 12 \geq 0 \Rightarrow a \leq 3 \\ Again, & a f (3) > 0 \\ \Rightarrow & 1 [(3)^{2} - 2 a (3) + a^{2} + a - 3] > 0 \\ \Rightarrow & 9 - 6 a + a^{2} + a - 3 > 0 \\ \Rightarrow & a^{2} - 5 a + 6 > 0 \\ \Rightarrow & (a - 2) (a - 3) > 0 \\ ∴ & a \in (- \infty, 2) \cup (3, \infty) \end{array}$

From Eqs. (i), (ii), (iii) and (iv), we get

$a \in (- 4, 2) .$

NOTE There is correction in answer $a < 2$ should be $- 4 < a < 2$ .

Theory of Equations 4 Question 6

7. If the roots of the equation $x^{2} - 2 a x + a^{2} + a - 3 = 0$ are real and less than 3 , then

Answer:

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Theory of Equations 4 Question 6

7. If the roots of the equation x2−2ax+a2+a−3=0 are real and less than 3 , then

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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7. If the roots of the equation $x^{2} - 2 a x + a^{2} + a - 3 = 0$ are real and less than 3 , then