Application of Derivatives 4 Question 17
18. The number of points in $(-\infty, \infty)$ for which
$x^{2}-x \sin x-\cos x=0$, is
(2013 Adv.)
(a) 6
(b) 4
(c) 2
(d) 0
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Answer:
Correct Answer: 18. (c)
Solution:
- 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,
$$ \begin{aligned} g(x) & =x \sin x+\cos x \\ g^{\prime}(x) & =x \cos x+\sin x-\sin x \\ g^{\prime}(x) & =x \cos x \\ g^{\prime \prime}(x) & =-x \sin x+\cos x \end{aligned} $$
$$ \begin{array}{llrl} \text { Put } & g^{\prime}(x) & =0 \Rightarrow x \cos x=0 \\ \therefore & x & =0, \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \end{array} $$
At $x=0, \frac{3 \pi}{2}, \frac{7 \pi}{2}, \ldots, f^{\prime \prime}(x)>0$, so minimum
At $x=\frac{\pi}{2}, \frac{5 \pi}{2}, \frac{9 \pi}{2}, \ldots, f^{\prime}(x)<0$, so maximum
So, graph of $f(x)$ and $g(x)$ are shown as
So, number of solutions are 2.
