The set of all values of $k>-1$, for which the equation $\left(3 x^{2}+4 x+3\right)^{2}-(k+1)\left(3 x^{2}+4 x+3\right)\left(3 x^{2}+4 x+2\right)+k\left(3 x^{2}+4 x+2\right)^{2}=0$ has real roots is:

(B) $\left(1, \frac{5}{2}\right]$

(A) $\left(\frac{1}{2}, \frac{3}{2}\right]-\{-1\}$

(C) $\left[-\frac{1}{2}, 1\right)$

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Quadratic Equations and Inequalities

The set of all values of

k > - 1

, for which the equation

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

has real roots is:

Answer: (B)

With

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

and

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

(so

u = v + 1

) the equation factors as

(u - v) (u - k v) = 0

. Since

u - v = 1 \neq = 0

, we need

u = k v

, i.e.

v + 1 = k v

, giving

v = \frac{1}{k - 1}

. The quadratic

3 x^{2} + 4 x + 2

has least value

\frac{2}{3}

, so it attains every value in

[\frac{2}{3}, \infty)

. A real

x

exists exactly when

\frac{1}{k - 1} \geq \frac{2}{3}

with

k > 1

, which gives

k - 1 \leq \frac{3}{2}

, i.e.

1 < k \leq \frac{5}{2}

. The required set is

(1, \frac{5}{2}]

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option