1
JEE Main 2017 (Online) 8th April Morning Slot
+4
-1
If y = $${\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}},$$

then (x2 $$-$$ 1) $${{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is equal to :
A
125 y
B
124 y2
C
225 y2
D
225 y
2
JEE Main 2017 (Online) 8th April Morning Slot
+4
-1
The curve satisfying the differential equation, ydx $$-$$(x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :
A
$$\left( {{1 \over 4}, - {1 \over 2}} \right)$$
B
$$\left( { - {1 \over 3},{1 \over 3}} \right)$$
C
$$\left( {{1 \over 3}, - {1 \over 3}} \right)$$
D
$$\left( {{1 \over 4}, {1 \over 2}} \right)$$
3
JEE Main 2017 (Offline)
+4
-1
The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point:
A
$$\left( {{1 \over 2},{1 \over 2}} \right)$$
B
$$\left( {{1 \over 2}, - {1 \over 3}} \right)$$
C
$$\left( {{1 \over 2},{1 \over 3}} \right)$$
D
$$\left( { - {1 \over 2}, - {1 \over 3}} \right)$$
4
JEE Main 2017 (Offline)
+4
-1
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:
A
10
B
25
C
30
D
12.5
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