Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
MATCH THE FOLLOWING:
(A) The value of $\int_{-2}^{2}(ax^3+bx+c)dx$ depends on
(B) $\int_{-1}^{1}(ax^3+bx)dx=0$ is true for all real values of $a, b$
(C) $\left(\sum_{n=1}^{10}\int_{2n-1}^{2n}\sin^2(x)dx + \sum_{n=1}^{10}\int_{2n}^{2n+1}\sin^2(x)dx\right)$
(D) $\int_{0}^{\pi/4}(\tan^n(x)+\tan^{n-2}(x))d(x-[x])$ where $[\cdot]$ is G.I.F.
Step-by-Step Solution
Key Concept: Recognize odd/even function properties and periodicity to simplify definite integrals and summations.
For part [A-u], integrate $∫(ax^3 + bx + c)dx = 4c$ to get $\frac{ax^4}{4} + \frac{bx^2}{2} + cx = 4c$. For part [B-s], evaluate $∫_{-1}(ax^3 + bx)dx = 0$ using the odd function property. Part [C-q] involves summing integrals of $\sin^{2n} x$ over intervals, using periodicity of sine. Part [D-p] evaluates $\int_0^{π/4} \frac{\tan^{n-1} x}{n-1}dx = \frac{1}{n-1}$ by direct integration of the tangent power.
Correct Answer: [A-u] [B-s] [C-q] [D-p]