Question

Let $p$ and $q$ are real numbers such that $(\tan p-1{)}^3+2019(\tan p-1)=-1$ and $(1-\cot q{)}^3+2019(1–\cot q)=-1,$ $\tan p\ne \cot q$. Then number of possible values of $r$ which satisfy the equation $\tan p+\cot q+\sin r+\cos r=3,$ $r\in [-2\pi ,2\pi ]$ is

• A. 5.00
• B. 05
• C. 5
• D. 5.0

Answer 1

Answer: (A), (B), (C), (D)

Given,
$(\tan p-1{)}^3+2019(\tan p-1)=-1\dots (1)$
$(1-\cot q{)}^3+2019(1-\cot q)=-1\dots (2)$
Substract equation $\left(2\right)$ from $\left(1\right)$
$\left[\right(\tan p-1{)}^3-(1-\cot q{)}^3]+2019(\tan p-1)-2019(1-\cot q)=0$
Apply $[a^3-b^3=(a-b\left)\right(a^2+b^2+ab\left)\right]$
$\Rightarrow \left(\right(\tan p-1)-(1-\cot q\left)\right)\left[\right(\tan p-1{)}^2+(\tan p-1)(1-\cot q)+(1-\cot q{)}^2]+2019(\tan p+\cot q-2)=0$
$\Rightarrow [\tan p+\cot q-2]\left[\right(\tan p-1{)}^2+(1-\cot q{)}^2+(\tan p-1\left)\right(1-\cot q)+2019]=0$
This is possible if $\tan p+\cot q=2$
$\therefore \sin r+\cos r=1$
$\Rightarrow r=-2\pi ,-\frac{3\pi }{2},0,\frac{\pi }{2},2\pi$
No. of possible values $=5$

Alternate Method
Let $\tan p-1=t,1-\cot =k$. we get
$f\left(t\right)=t^3+2019t+1$ and
$g\left(k\right)=k^3+2019k+1$
$f^{\prime }\left(t\right)=3t^2+2019>0$, It means $f\left(t\right)=0$ has only one solution.
Similarly $g\left(k\right)=0$ has only one solution.
That is possible only when $t=k$
$\tan p-1=1-\cot q\Rightarrow \tan p+\cot q=2$
Now, we have $\sin r+\cos r=1$
$\Rightarrow \cos (r-\frac{\pi }{4})=\cos \frac{\pi }{4}$
$\Rightarrow r-\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4},n\in I$
$\Rightarrow r=2n\pi$ or $r=2n\pi +\frac{\pi }{2}$
$\because r\in [-2\pi ,2\pi ]$
$\therefore r=-2\pi ,0,2\pi$ or $r=\frac{\pi }{2},-\frac{3\pi }{2}$
Hence, $5$ values are there.