1
WB JEE 2019
MCQ (Single Correct Answer)
+1
-0.25
Change Language
General solution of $${(x + y)^2}{{dy} \over {dx}} = {a^2},a \ne 0$$ is (C is an arbitrary constant)
A
$${x \over a} = \tan {y \over a} + C$$
B
$$\tan xy = C$$
C
$$\tan (x + y) = C$$
D
$$\tan {{y + C} \over a} = {{x + y} \over a}$$
2
WB JEE 2019
MCQ (Single Correct Answer)
+1
-0.25
Change Language
Let P(4, 3) be a point on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. If the normal at P intersects the X-axis at (16, 0), then the eccentricity of the hyperbola is
A
$${{\sqrt 5 } \over 2}$$
B
2
C
$${\sqrt 2 }$$
D
$${\sqrt 3 }$$
3
WB JEE 2019
MCQ (Single Correct Answer)
+1
-0.25
Change Language
If the radius of a spherical balloon increases by 0.1%, then its volume increases approximately by
A
0.2%
B
0.3%
C
0.4%
D
0.05%
4
WB JEE 2019
MCQ (Single Correct Answer)
+1
-0.25
Change Language
The three sides of a right angled triangle are in GP (geometric progression). If the two acute angles be $$\alpha$$ and $$\beta$$, then tan$$\alpha$$ and tan$$\beta$$ are
A
$${{\sqrt 5 + 1} \over 2}$$ and $${{\sqrt 5 - 1} \over 2}$$
B
$$\sqrt {{{\sqrt 5 + 1} \over 2}} $$ and $$\sqrt {{{\sqrt 5 - 1} \over 2}} $$
C
$${\sqrt 5 }$$ and $${1 \over {\sqrt 5 }}$$
D
$${{\sqrt 5 } \over 2}$$ and $${2 \over {\sqrt 5 }}$$
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