JEE Main 18 March 2021 Shift 1 Solved Paper

Given, ‌‌y=mx+1 ...(i)
and 3x+4y=9...(ii)
From Eqs. (i) and (ii),
3x+4(mx+1)=9
⇒‌‌3x+4mx+4=9⇒x(3+4m)=5
⇒‌‌x=‌

5

3+4m

Given, that the abscissa of point of intersection of Eqs. (i) and (ii) i.e. x=‌

5

3+4m

is an integer.
∴ Possible values of x are
x=1,−1,5,−5
i.e. ‌

5

4m+3

=1‌‌ or ‌

5

4m+3

=−1
or ‌‌‌

5

4m+3

=5‌‌ or ‌

5

4m+3

=−5
⇒‌‌4m=2‌‌ or ‌‌−4m=8
Or 4m=−2‌‌‌ or ‌‌‌−4m=4
⇒‌‌m=‌

1

2

,−2,−‌

1

2

,−1
∵‌‌{−‌

1

2

,‌

1

2

}∉I
∴‌‌m={−1,−2}∈I
∴ Number of integral values of m are 2