Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
82 views
in Differential Equations by (94.8k points)
closed by
Let a solution `y=y(x)` of the differential equation
`xsqrt(x^(2)-1) dy-ysqrt(y^(2)-1)dx=0` satisfy `y(2)=(2)/(sqrt3)`
Statement-1, `y(x)=sec(sec^(-1)x-(pi)/(6))`
Statement-2 : y(x) is given by `(1)/(y)=(2sqrt3)/(x)-sqrt(1-(1)/(x^(2)))`
A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

1 Answer

0 votes
by (95.5k points)
selected by
 
Best answer
Correct Answer - C
We have,
`xsqrt(x^(2)-1)dy=ysqrt(y^(2)-1)dx`
`rArr" "(1)/(ysqrt(y^(2)-1))dy=(1)/(x sqrt(x^(2)-1))dx`
`rArr" "int(1)/(ysqrt(y^(2)-1))dy=int(1)/(xsqrt(x^(2)-1))dx`
`rArr" "sec^(-1)y=sec^(-1)x+C`
It is given that `y=(2)/(sqrt3)` when x = 2
`therefore" "sec^(-1).(2)/(sqrt3)=sec^(-1)2+CrArr(pi)/(6)=(pi)/(3)+CrArrC=-(pi)/(6)`
Putting `C=-(pi)/(6)` in (i), we get
`sec^(-1)y=sec^(-1)x-(pi)/(6)" ...(ii)"`
`rArr" "y=sec(sec^(-1)x-(pi)/(6))`
so, statement-1 is true.
From (ii), we have
`cos^(-1)((1)/(y))=cos^(-1)((1)/(x))-(pi)/(6)`
`rArr" "(1)/(y)=cos{cos^(-1)((1)/(x))-(pi)/(6)}`
`rArr" "(1)/(y)=cos{cos^(-1)((1)/(x))}cos((pi)/(6))+sin(cos^(-1).(1)/(x))sin(pi)/(6)`
`rArr" "(1)/(y)=(sqrt3)/(2x)+(1)/(2)sqrt(1-(1)/(x^(2)))`
So, statement-2 is false.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...