ISI2004-MIII: 23

Question

If $\textit{f}(x)=x^{2}$ and $g(x)=x \sin x +\cos x$ then

• A. $f$ and $g$ agree at no point
• B. $f$ and $g$ agree at exactly one point
• C. $f$ and $g$ agree at exactly two point
• D. $f$ and $g$ agree at more then two point

Comments

https://gateoverflow.in/212449/isi-2004-23

Answer 1

$\begin{align*}&f(x) = x^2 \\
&g(x) = x\sin x + \cos x \end{align*}$

We are asked to make this two function equal and see at how many points they meet.

$\Rightarrow f(x) = g(x)$
$\Rightarrow x^2 = x \sin x + \cos x$
$\Rightarrow {\color{blue}{x^2 -x\sin x - \cos x}} = 0$ 
$\Rightarrow F(x) = 0$

This $F(x)$ is even and we just need to check at how many points $F(x)$ crosses the $X$ - axis for $x > 0$.

$F(0)$ = $-1$

$F^{'}(x) = 2x - x \cos x$

$F^{'}(x) = x\left ( {\color{red}{2 - \cos x}} \right )$

Here  ${\color{red}{2 - \cos x}}$ is greater than $0$ and $x\left ( {\color{red}{2 - \cos x}} \right )$ will be also greater than $0$ for $x > 0$.

$\Rightarrow F(x){\color{red}{ \text{ is strictly increasing function.}}}$

Therefore $F(x)$ will cut at only one point for $x > 0$.

As we have already seen that $F(x)$ is even, therefore $F(x)$ has only two real roots.

Correct Answer: $C$

Comments

As we have already seen that F(x) is even, therefore F(x) has only two real roots.

If F(X) would have been ODD instead of EVEN and still strictly increasing then how many roots possible??

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@jatin khachane 1 Then there would be only two solutions if F(0) does not exist. But if F(0) exists, then one solution, given the function is strictly increasing.

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"A polynomial of degree n has atmost n distinct real roots". As F(x) is of degree 2 so it will have atmost 2 roots. But how can we confirm that it will surely have 2 real roots if we don't solve this problem graphically.

Not getting this line.

As we have already seen that F(x) is even, therefore F(x) has only two real roots.