Question

Match the following by appropriately matching the lists based on the information given in Column $1$ and Column $2$
$\begin{array}{cc}Column1:& Column2:\\ \text{Equation}& \text{Number}\\ & \text{of roots}\\ \text{a.}{x}^2\tan x=1,x\in [0,2\pi ]& \text{p.}5\\ \text{b.}{2}^{\cos x}=|\sin x|,x\in [0,2\pi ]& \text{q.}2\\ \text{c.}\text{If}f\left(x\right)\text{is a polynomial of degree}& \\ 5\text{with real coefficients such that}& \\ f\left(|x|\right)=0\text{has}8\text{real roots, then the}& \\ \text{number of roots of}f\left(x\right)=0\text{is/are}& \text{r.}3\\ \text{d.}{7}^{|x|}(|5-|x||)=1& \text{s.}4\end{array}$

Then which of the following is correct ?

• A. $a\to s,b\to r,c\to p,d\to q$
• B. $a\to q,b\to s,c\to p,d\to s$
• C. $a\to q,b\to s,c\to p,d\to r$
• D. $a\to p,b\to r,c\to q,d\to s$

Answer 1

The correct option is B $a\to q,b\to s,c\to p,d\to s$

$a.$
$y=\tan x=\frac{1}{x^2}$


From the graph, it is clear that it will have two real roots.

$b.$
$y=2^{\cos x}\text{and}y=|\sin x|$


From the graph, two curves meet at four points for $x\in [0,2\pi ]$. So the equation $2^{\cos x}=|\sin x|$ has four solutions.

$c.$
Given that $f\left(|x|\right)=0$ has $8$ real roots or $f\left(x\right)=0$ has four positive roots.
Since, $f\left(x\right)$ is a polynomial of degree $5,f\left(x\right)$ can not have even number of real roots.
Hence, $f\left(x\right)$ has all the five roots real and one root is negative.

$d.$
$7^{|x|}(|5-|x||)=1$
or $|5-|x||=7^{-|x|}$
Draw the graph of $y=7^{-|x|}\text{and}y=|5-|x||$


Hence, the number of roots is $4$.