QJEEM21Maa8e9: JEE Main 2021 Mathematics - Class 11 JEEM21Maa8e9 Solution: The set of all values of K > 1, for which the equation has real …

Question

Verified Solution • 2025–26 Session moderate

The set of all values of K >

-

1, for which the equation

{(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0

has real roots, is :

Accepted Answer

{(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0

Let

3{x^2} + 4x + 3 = a

and

3{x^2} + 4x + 2 = b \Rightarrow b = a - 1

Given equation becomes

\Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0

\Rightarrow a(a - kb) - b(a - kb) = 0

\Rightarrow (a - kb)(a - b) = 0 \Rightarrow a = kb

or a = b (reject)

\because

a = kb

\Rightarrow 3{x^2} + 4x + 3 = k(3{x^2} + 4x + 2)

\Rightarrow 3(k - 1){x^2} + 4(k - 1)x + (2k - 3) = 0

for real roots

D

\ge

0

\Rightarrow 16{(k - 1)^2} - 4(3(k - 1))(2k - 3) \ge 0

\Rightarrow 4(k - 1)\{ 4(k - 1) - 3(2k - 3)\} \ge 0

\Rightarrow 4(k - 1)\{ - 2k + 5\} \ge 0

\Rightarrow - 4(k - 1)\{ 2k - 5\} \ge 0

\Rightarrow (k - 1)(2k - 5) \le 0

\therefore

k \in \left[ {1,{5 \over 2}} \right]

\therefore

k \ne 1

\therefore

k \in \left( {1,{5 \over 2}} \right]

Step-by-Step Verified Solution