Let y = y(x) be the solution of the differential equation \(\frac{dy}{dx}+\frac{√2y}{2cos^4x−cos2x}=xe^{tan−1}(√2cot2x),0<x<\frac{π}{2}\) with \(y(\frac{π}{4})=\frac{π^2}{32}\) If \(y(\frac{π}{3})=\frac{π^2}{18}e^{−tan−1}(α)\) then the value of \(3α^2\) is equal to ______
Correct Answer: 2
Solution and Explanation
The correct answer is: 2
\(\frac{dy}{dx}+\frac{√2y}{2cos^4x−cos2x}=xe^{tan−1}(√2cot2x)\)
\(I,F.=e^{∫}\frac{2\sqrt2dx}{1+cos^22x}=e^{\sqrt2\,∫}\frac{2\,sec^22x}{2+tan^22x}\)
= \(e^{\tan^{-1}\left(\frac{\tan(2x)}{\sqrt{2}}\right)}\)
\(⇒\) \(y \cdot e^{\tan^{-1}\left(\frac{\tan(2x)}{\sqrt{2}}\right)}\) = \(\int x e^{\tan^{-1}(\sqrt{2}\cot(2x))} \cdot e^{\tan^{-1}\left(\frac{\tan(2x)}{\sqrt{2}}\right)} \, dx + c\)
\(⇒\) \(y \cdot e^{\tan^{-1}\left(\frac{\tan(2x)}{\sqrt{2}}\right)}\)
\(= e^{\frac{\pi}{2}}.\frac{x^2}{2}+c\)
When
\(x=\frac{\pi}{4},y=\frac{\pi^2}{32}\) gives \(c=0.\)
When
\(x=\frac{\pi}{3},y=\frac{\pi^2}{18}e^{tan^{-1}\alpha}\)
So, \(\frac{\pi^2}{18}e^{tan^{-1}\alpha}.e^{-tan{-1}(-\sqrt{\frac{3}{2}})}=e^{\frac{\pi}{2}}\frac{\pi^2}{18}\)
\(⇒ tan^{-1}(-\alpha)=tan^{-1}\left(\sqrt{\frac{3}{2}}\right)\)
\(⇒α=-\sqrt\frac{2}{3}⇒3a^2=2\)
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Concepts Used:
Differentiability
Differentiability of a function A function f(x) is said to be differentiable at a point of its domain if it has a finite derivative at that point. Thus f(x) is differentiable at x = a
\(\frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)
⇒ f'(a – 0) = f'(a + 0)
⇒ left-hand derivative = right-hand derivative.
Thus function f is said to be differentiable if left hand derivative & right hand derivative both exist finitely and are equal.
If f(x) is differentiable then its graph must be smooth i.e. there should be no break or corner.
Note:
(i) Every differentiable function is necessarily continuous but every continuous function is not necessarily differentiable i.e. Differentiability ⇒ continuity but continuity ⇏ differentiability
(ii) For any curve y = f(x), if at any point \(\frac{d y}{d x}\) = 0 or does not exist then, the point is called “critical point”.
3. Differentiability in an interval
(a) A function fx) is said to be differentiable in an open interval (a, b), if it is differentiable at every point of the interval.
(b) A function f(x) is differentiable in a closed interval [a, b] if it is
- Differentiable at every point of interval (a, b)
- Right derivative exists at x = a
- Left derivative exists at x = b.