−tan−1x So, given equation reduces to ‌2(‌
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2
−tan−1(x2+2x+k)) ‌‌‌=π−3tan−1(x2+2x+k) ‌⇒‌‌tan−1(x2+2x+k)=0 ‌⇒‌‌x2+2x+k=0 Given Eq. (i) has two distinct real solution. So, its discriminant value is always greater than zero. So, (4) −4k>0 ⇒‌‌k<1⇒k∈(−∞,1)
