
If the roots of the equation ${x^2} - 2ax + {a^2} + a - 3 = 0$ are real and less than 3, then
A. $a < 2$
B. $2 \leqslant a \leqslant 3$
C. $3 < a \leqslant 4$
D. $a > 4$
Answer
161.4k+ views
Hint: Check the concavity of the given quadratic equation and using this information check whether $f(3)$ is positive or negative. Also use the fact that the discriminant must be greater than or equal to 0 as real roots exist. With the two inequalities, find the range of a.
Formula used: Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$
Complete step-by-step solution:
Let $f(x) = {x^2} - 2ax + {a^2} + a - 3$
The coefficient of ${x^2}$ in the equation ${x^2} - 2ax + {a^2} + a - 3 = 0$ is greater than 0. Therefore, it is a concave upwards graph. Since it is a concave upwards graph, $f(x)$ will be negative only when $x \in \left[ {p,q} \right]$ where $p,q$ are the roots.

Therefore, $f(3)$ is positive. We also know that real roots exist. Therefore, the discriminant of the quadratic equation, ${x^2} - 2ax + {a^2} + a - 3 = 0$ must also be greater than or equal to 0.
Since $f(3) > 0$,
${3^2} - 2a(3) + {a^2} + a - 3 > 0$
$9 - 6a + {a^2} + a - 3 > 0$
${a^2} - 5a + 6 > 0$
Solving further we get,
${a^2} - 2a - 3a + 6 > 0$
$a\left( {a - 2} \right) - 3\left( {a - 2} \right) > 0$
$\left( {a - 2} \right)\left( {a - 3} \right) > 0$
$a \in \left( { - \infty ,2} \right) \cup \left( {3,\infty } \right)$
Since discriminant, $D \geqslant 0$,
${\left( { - 2a} \right)^2} - 4\left( 1 \right)\left( {{a^2} + a - 3} \right) \geqslant 0$
$4{a^2} - 4{a^2} - 4a + 12 \geqslant 0$
Simplifying the above inequality,
$4a \leqslant 12$
$a \leqslant 3$
$a \in ( - \infty ,3]$
Taking the intersection of $\left( { - \infty ,2} \right) \cup \left( {3,\infty } \right)$ and $( - \infty ,3]$, we get $a \in \left( { - \infty ,2} \right)$.
Therefore, the correct answer is option A. $a < 2$.
Note: Given a quadratic polynomial $a{x^2} + bx + c$, if $a > 0$ then the graph of the quadratic polynomial will be a concave upwards graph and if $a < 0$ then the graph of the quadratic polynomial will be a concave downwards graph.
Formula used: Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$
Complete step-by-step solution:
Let $f(x) = {x^2} - 2ax + {a^2} + a - 3$
The coefficient of ${x^2}$ in the equation ${x^2} - 2ax + {a^2} + a - 3 = 0$ is greater than 0. Therefore, it is a concave upwards graph. Since it is a concave upwards graph, $f(x)$ will be negative only when $x \in \left[ {p,q} \right]$ where $p,q$ are the roots.

Therefore, $f(3)$ is positive. We also know that real roots exist. Therefore, the discriminant of the quadratic equation, ${x^2} - 2ax + {a^2} + a - 3 = 0$ must also be greater than or equal to 0.
Since $f(3) > 0$,
${3^2} - 2a(3) + {a^2} + a - 3 > 0$
$9 - 6a + {a^2} + a - 3 > 0$
${a^2} - 5a + 6 > 0$
Solving further we get,
${a^2} - 2a - 3a + 6 > 0$
$a\left( {a - 2} \right) - 3\left( {a - 2} \right) > 0$
$\left( {a - 2} \right)\left( {a - 3} \right) > 0$
$a \in \left( { - \infty ,2} \right) \cup \left( {3,\infty } \right)$
Since discriminant, $D \geqslant 0$,
${\left( { - 2a} \right)^2} - 4\left( 1 \right)\left( {{a^2} + a - 3} \right) \geqslant 0$
$4{a^2} - 4{a^2} - 4a + 12 \geqslant 0$
Simplifying the above inequality,
$4a \leqslant 12$
$a \leqslant 3$
$a \in ( - \infty ,3]$
Taking the intersection of $\left( { - \infty ,2} \right) \cup \left( {3,\infty } \right)$ and $( - \infty ,3]$, we get $a \in \left( { - \infty ,2} \right)$.
Therefore, the correct answer is option A. $a < 2$.
Note: Given a quadratic polynomial $a{x^2} + bx + c$, if $a > 0$ then the graph of the quadratic polynomial will be a concave upwards graph and if $a < 0$ then the graph of the quadratic polynomial will be a concave downwards graph.
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

Magnetic Force on a Current-Carrying Wire - Important Concepts and Tips for JEE

A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Trending doubts
JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

JEE Main B.Arch Cut Off Percentile 2025

JoSAA Counselling 2025: Registration Dates OUT, Eligibility Criteria, Cutoffs

NIT Cutoff Percentile for 2025

JEE Mains 2025 Cutoff: Expected and Category-Wise Qualifying Marks for NITs, IIITs, and GFTIs

Other Pages
NCERT Solutions for Class 10 Maths Chapter 13 Statistics

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related To Circles

NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume

NCERT Solutions for Class 10 Maths Chapter 14 Probability

List of Fastest Century in IPL History

NEET 2025 – Every New Update You Need to Know
