The number of values of a∈R for which the equation 5tan−1(x2+x+a)+3cot−1(x2+x+a)=2π has unique solution, is

The given equation is 5tan−1(x2+x+a)+3cot−1(x2+x+a)=2π2tan−1(x2+x+a)=2π-3π22tan−1(x2+x+a)=π2tan−1(x2+x+a)=π4It implies that x2+x+a=1Given that it has unique solution it means, the discriminant is zero1-4a-1=04a-1=1a-1=14a=54Therefore, the number of vlaues of a is 12π=5tan−1(x2+x+a)+3cot−1(x2+x+a)=3π2+2tan−1(x2+x+a)⇒x2+x+a=1. For required condition Δ= 1−4(a−1)=0⇒a=54

The number of values of a∈R for which the equation 5tan−1(x2+x+a)+3cot−1(x2+x+a)=2π has unique solution, is

The given equation is 5tan−1(x2+x+a)+3cot−1(x2+x+a)=2π2tan−1(x2+x+a)=2π-3π22tan−1(x2+x+a)=π2tan−1(x2+x+a)=π4It implies that x2+x+a=1Given that it has unique solution it means, the discriminant is zero1-4a-1=04a-1=1a-1=14a=54Therefore, the number of vlaues of a is 12π=5tan−1(x2+x+a)+3cot−1(x2+x+a)=3π2+2tan−1(x2+x+a)⇒x2+x+a=1. For required condition Δ= 1−4(a−1)=0⇒a=54

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The number of values of $a \in R$ for which the equation $5 \tan^{- 1} (x^{2} + x + a) + 3 \cot^{- 1} (x^{2} + x + a) = 2 π$ has unique solution, is

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Detailed Solution

The given equation is

$5 \tan^{- 1} (x^{2} + x + a) + 3 \cot^{- 1} (x^{2} + x + a) = 2 π$

$\begin{array}{rcl} 2 \tan^{- 1} (x^{2} + x + a) & = & 2 π - \frac{3 π}{2} \\ 2 \tan^{- 1} (x^{2} + x + a) & = & \frac{π}{2} \\ \tan^{- 1} (x^{2} + x + a) & = & \frac{π}{4} \end{array}$

It implies that

$x^{2} + x + a = 1$

Given that it has unique solution

it means, the discriminant is zero

$\begin{array}{rcl} 1 - 4 (a - 1) & = & 0 \\ 4 (a - 1) & = & 1 \\ a - 1 & = & \frac{1}{4} \\ a & = & \frac{5}{4} \end{array}$

Therefore, the number of vlaues of $a$ is $1$

$2 π = 5 \tan^{- 1} (x^{2} + x + a) + 3 \cot^{- 1} (x^{2} + x + a) = \frac{3 π}{2} + 2 \tan^{- 1} (x^{2} + x + a)$ $\Rightarrow x^{2} + x + a = 1. F o r r e q u i r e d c o n d i t i o n Δ = 1 - 4 (a - 1) = 0 \Rightarrow a = \frac{5}{4}$