Application Of Derivatives Ques 69
69. The number of points in $(-\infty, \infty)$ for which
$x^{2}-x \sin x-\cos x=0$, is
(2013 Adv.)
6
4
2
0
Show Answer
Answer:
Correct Answer: 69.(c)
Solution:
Formula:
- PLAN The given equation contains algebraic and trigonometric functions called transcendental equation. To solve transcendental equations we should always plot the graph for LHS and RHS.
Here, $x^{2}=x \sin x+\cos x$
Let $f(x)=x^{2}$ and $g(x)=x \sin x+\cos x$
We know that the graph for $f(x)=x^{2}$
To plot,
$ \begin{aligned} g(x) & =x \sin x+\cos x \\ g^{\prime}(x) & =x \cos x+\sin x-\sin x \\ g^{\prime}(x) & =x \cos x ……(i) \\ g^{\prime \prime}(x) & =-x \sin x+\cos x ……(ii) \end{aligned} $
$ \begin{array}{llrl} \text { Put } & g^{\prime}(x) & =0 \Rightarrow x \cos x=0 \\ \therefore & x & =0, \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \end{array} $
At $x=0, \frac{3 \pi}{2}, \frac{7 \pi}{2}, \ldots, f^{\prime \prime}(x)>0$, so maximum
At $x=\frac{\pi}{2}, \frac{5 \pi}{2}, \frac{9 \pi}{2}, \ldots, f^{\prime}(x)>0$, so maximum
So, graphs of $f(x)$ and $g(x)$ are shown as
So, the number of solutions is 2.
BETA
