Limit and Continuity MCQ Quiz in मराठी - Objective Question with Answer for Limit and Continuity - मोफत PDF डाउनलोड करा

गणना:

दिलेले आहे, \(\lim_{x\to0^+}\left(\sin(kx)+\cos(kx)+x\right)^\frac{3}{x}\) = e 9

समजा, L = \(\lim_{x\to0^+}\frac{3}{x}\left(\sin(kx)+\cos(kx)+x-1\right)\)

= \(\lim_{x\to0^+}\frac{3}{x}\left(\sin(kx)+\cos(kx)+x-1\right)\)

= \(\lim_{x\to0^+}3\left(\frac{\sin(kx)}{x}+1-\frac{1-\cos(kx)}{x}\right)\)

= \(\lim_{x\to0^+}3\left(\frac{\sin(kx)}{(kx)}\cdot k+1-\frac{2\sin^2\frac{kx}{2}}{\frac{k^2x^2}{4}}\cdot \frac{k^2x}{4}\right)\)

= \(\lim_{x\to0^+}3\left(\frac{\sin(kx)}{(kx)}\cdot k+1-2\left(\frac{\sin\frac{kx}{2}}{\frac{kx}{2}}\right)\cdot \frac{k^2x}{4}\right)\)

= \(\lim_{x\to0^+}3\left(k+1-\frac{k^2x}{2}\right)\)

= ३(के + १)

∴ \(\lim_{x\to0^+}\left(\sin(kx)+\cos(kx)+x\right)^\frac{3}{x}\) = \(e^{\frac{3}{x}(\lim_{x\to0^+}\sin(kx)+\cos(kx)+x-1)}\)

= e3(k + 1) = e9 

⇒ 3(k + 1) = 9

⇒ k + 1 = 3

⇒ k = 2

∴ k चे मूल्य 2 आहे.

पर्याय 1 योग्य आहे.

गणना:

दिलेले आहे, \(\lim_{x\to0^+}\left(\sin(kx)+\cos(kx)+x\right)^\frac{3}{x}\) = e 9

समजा, L = \(\lim_{x\to0^+}\frac{3}{x}\left(\sin(kx)+\cos(kx)+x-1\right)\)

= \(\lim_{x\to0^+}\frac{3}{x}\left(\sin(kx)+\cos(kx)+x-1\right)\)

= \(\lim_{x\to0^+}3\left(\frac{\sin(kx)}{x}+1-\frac{1-\cos(kx)}{x}\right)\)

= \(\lim_{x\to0^+}3\left(\frac{\sin(kx)}{(kx)}\cdot k+1-\frac{2\sin^2\frac{kx}{2}}{\frac{k^2x^2}{4}}\cdot \frac{k^2x}{4}\right)\)

= \(\lim_{x\to0^+}3\left(\frac{\sin(kx)}{(kx)}\cdot k+1-2\left(\frac{\sin\frac{kx}{2}}{\frac{kx}{2}}\right)\cdot \frac{k^2x}{4}\right)\)

= \(\lim_{x\to0^+}3\left(k+1-\frac{k^2x}{2}\right)\)

= ३(के + १)

∴ \(\lim_{x\to0^+}\left(\sin(kx)+\cos(kx)+x\right)^\frac{3}{x}\) = \(e^{\frac{3}{x}(\lim_{x\to0^+}\sin(kx)+\cos(kx)+x-1)}\)

= e3(k + 1) = e9 

⇒ 3(k + 1) = 9

⇒ k + 1 = 3

⇒ k = 2

∴ k चे मूल्य 2 आहे.

पर्याय 1 योग्य आहे.