If the equation 2 operatorname{Cot}^{-1}(x^2+ | vedclass

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(D) Let $u = x^2 + 2x + k$. The equation becomes $2 \operatorname{Cot}^{-1}(u) = \pi - 3 \operatorname{Tan}^{-1}(u)$.Using the identity $\operatorname

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$(-\infty, 1)$

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(D) Let $u = x^2 + 2x + k$. The equation becomes $2 \operatorname{Cot}^{-1}(u) = \pi - 3 \operatorname{Tan}^{-1}(u)$.Using the identity $\operatorname{Cot}^{-1}(u) = \frac{\pi}{2} - \operatorname{Tan}^{-1}(u)$,we substitute:$2(\frac{\pi}{2} - \operatorname{Tan}^{-1}(u)) = \pi - 3 \operatorname{Tan}^{-1}(u)$$\pi - 2 \operatorname{Tan}^{-1}(u) = \pi - 3 \operatorname{Tan}^{-1}(u)$$\operatorname{Tan}^{-1}(u) = 0$,which implies $u = 0$.So,$x^2 + 2x + k = 0$.For this quadratic equation to have two distinct real solutions,the discriminant $D$ must be greater than $0$.$D = b^2 - 4ac = (2)^2 - 4(1)(k) = 4 - 4k$.Setting $D > 0$,we get $4 - 4k > 0$,which simplifies to $4 > 4k$,or $k Thus,$k \in (-\infty, 1)$.

Column $I$	Column $II$
$(A)$ If $a=1$ and $b=0$,then $(x, y)$	$(p)$ lies on the circle $x^2+y^2=1$
$(B)$ If $a=1$ and $b=1$,then $(x, y)$	$(q)$ lies on $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$	$(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$	$(s)$ lies on $(4x^2-1)(y^2-1)=0$

If the equation $2 \operatorname{Cot}^{-1}(x^2+2x+k) = \pi - 3 \operatorname{Tan}^{-1}(x^2+2x+k)$ has two distinct real solutions,then all the values of $k$ lie in the interval

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