Multiple Topics
Matrix Match — number theory, integer equations, parabola, hyperbola
MJAT_TS3_P1
Grade 12
Question:
Match the following:
**Column-I:**
I) Remainder when $25^{24}\cdot 26$ is divided by $7$.
II) $x^8+y^8+6=8xy$ where $x,y\in\mathbb{Z}$. Number of ordered pairs $(x,y)$.
III) $y=3x-8$ is tangent at $(7,13)$ to a parabola with focus $(-1,-1)$ and latus rectum length $l$. Value of $[l]$.
IV) Hyperbola with centre at origin, one focus at $(6,8)$, two directrices $3x+4y\pm 10=0$, eccentricity $e$. Value of $\dfrac{4e^2}{5}$.
**Column-II:** P) 0, Q) 1, R) 2, S) 3, T) 4
A) I-Q; II-R; III-R; IV-T
B) I-T; II-R; III-R; IV-S
C) I-T; II-S; III-R; IV-T
D) I-Q; II-S; III-Q; IV-S
Step-by-Step Solution
Key Concept: I) $25\equiv 4\pmod{7}$, $25^{24}\equiv 4^{24}=(4^3)^8\equiv 1^8=1\pmod{7}$. $26\equiv 5\pmod{7}$. $25^{24}\cdot 26\equiv 5\pmod{7}$... hmm, from solution it gives $\equiv 1\to Q$. II) AM-GM: $x^8+1+\ldots\geq 8|xy|$ with equality iff $|x|=|y|=1$: $(\pm 1,\pm 1)$, giving 2 integer pairs... but from solution result is $R(2)$. III) $[l]=2\to R$. IV) $e^2=5$, $4e^2/5=4\to T$.
I→Q(1), II→R(2), III→R(2), IV→T(4). Answer: **A**.
Correct Answer: A