The option(s) with the values of a and L that | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(t) = e^t(\sin^6 at + \cos^4 at)$. We observe that for $a=2$ and $a=4$,the function $f(t)$ satisfies $f(t+\pi) = e^{t+\pi}(\sin^6 a(t+\pi) +

Vedclass · Accepted Answer

$(A, C)$

Vedclass · Answer

(A) Let $f(t) = e^t(\sin^6 at + \cos^4 at)$. We observe that for $a=2$ and $a=4$,the function $f(t)$ satisfies $f(t+\pi) = e^{t+\pi}(\sin^6 a(t+\pi) + \cos^4 a(t+\pi)) = e^{\pi} e^t(\sin^6 at + \cos^4 at) = e^{\pi} f(t)$,since $\sin(a(t+\pi)) = \sin(at + a\pi) = \sin(at)$ and $\cos(a(t+\pi)) = \cos(at)$ for even integers $a$.Let $I_k = \int_{(k-1)\pi}^{k\pi} f(t) dt$. Then $I_{k+1} = \int_{k\pi}^{(k+1)\pi} f(t) dt$. Substituting $t = u + \pi$,we get $I_{k+1} = \int_{(k-1)\pi}^{k\pi} f(u+\pi) du = e^{\pi} \int_{(k-1)\pi}^{k\pi} f(u) du = e^{\pi} I_k$.Thus,$I_1 = A$,$I_2 = e^{\pi} A$,$I_3 = e^{2\pi} A$,and $I_4 = e^{3\pi} A$,where $A = \int_0^{\pi} f(t) dt$.The numerator is $\int_0^{4\pi} f(t) dt = I_1 + I_2 + I_3 + I_4 = A(1 + e^{\pi} + e^{2\pi} + e^{3\pi}) = A \frac{e^{4\pi}-1}{e^{\pi}-1}$.Therefore,$L = \frac{A \frac{e^{4\pi}-1}{e^{\pi}-1}}{A} = \frac{e^{4\pi}-1}{e^{\pi}-1}$.This holds for both $a=2$ and $a=4$. Thus,options $(A)$ and $(C)$ are correct.

The option$(s)$ with the values of $a$ and $L$ that satisfy the following equation is(are) $\frac{\int_0^{4 \pi} e^t(\sin^6 at + \cos^4 at) dt}{\int_0^{\pi} e^t(\sin^6 at + \cos^4 at) dt} = L$.

Similar Questions

$\int_{-1}^1 \sin ^7 x \cdot \cos ^6 x \, dx = $ . . . . . . .

By using the properties of definite integrals,evaluate the integral $\int_{2}^{8}|x-5| d x$.

$\int_{-\pi}^{\pi} (1-x^2) \sin x \cdot \cos^2 x \, dx$ is equal to:

If $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$,then $27 I^2$ equals . . . . . . . .

The value of $\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module