Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

For the two circles x^{2} + y^{2} = 16 and x^{2} + y^{2} $$-$$ 2y = 0 there is/are

A

one pair of common tangents

B

only one common tangent

C

three common tangents

D

no common tangent

Centre C_{1} of x^{2} + y^{2} = 16 is (0, 0) and radius (r_{1}) = 4

Centre C_{2} of x^{2} + y^{2} $$-$$ 2y = 0 is (0, 1) and radius (r_{2}) = $$\sqrt {{1^2} + 0 - 0} = 1$$

$${C_1}{C_2} = \sqrt {{{(0 - 0)}^2} + {{(0 - 1)}^2}} = 1$$

$${r_1} - {r_2} = 4 - 1 = 3$$

Since $${C_1}{C_2} < {r_1} - {r_2}$$.

So second circle lies completely inside the first circle, so there is no common tangent possible.

2

MCQ (Single Correct Answer)

The circles x^{2} + y^{2} $$-$$ 10x + 16 = 0 and x^{2} + y^{2} = a^{2} intersect at two distinct points if

A

a < 2

B

2 < a < 8

C

a > 8

D

a = 2

Two circles having centres C_{1}, C_{2} and radii r_{1} and r_{2} are intersecting when

$$|{r_1} - {r_2}| < {C_1}{C_2} < {r_1} + {r_2}$$

Here, C_{1} $$\equiv$$ (5, 0), C_{2} $$\equiv$$ (0, 0),

$${r_1} = \sqrt {25 + 0 - 16} = 3$$, $${r_2} = a$$

$$\therefore$$ $$|3 - a| < \sqrt {{{(5 - 0)}^2} + {{(0 - 0)}^2}} < 3 + a$$

$$ \Rightarrow |3 - a| < 5 < 3 + a \Rightarrow |3 - a| < 5$$ and $$5 < 3 + a$$

$$ \Rightarrow - 5 < 3 - a < 5$$ and $$a > 2 \Rightarrow - 2 < a < 8$$ and $$a > 2$$

$$ \Rightarrow 2 < a < 8$$.

3

MCQ (Single Correct Answer)

The equation of the chord of he circle $${x^2} + {y^2} - 4x = 0$$ whose midpoint is (1, 0) is

A

y = 2

B

y = 1

C

x = 2

D

x = 1

Equation of chord of circle

x^{2} + y^{2} + 2gx + 2fy + c = 0, where midpoint is (x_{1}, y_{1}) is T = S_{1}

where T = xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c

and S_{1} = x$$_1^2$$ + y$$_1^2$$ + 2gx_{1} + 2fy_{1} + c

$$\therefore$$ Equation of chord of the circle x^{2} + y^{2} $$-$$ 4x = 0 whose midpoint is (1, 0) is

1x + 0y $$-$$ 2(x + 1) + 0(y + 0) = 1^{2} + 0^{2} $$-$$ 4 . 1

$$\Rightarrow$$ $$-$$ x $$-$$ 2 = $$-$$ 3 $$\Rightarrow$$ x = 1

4

MCQ (Single Correct Answer)

The locus of the centres of the circles which touch both the axes is given by

A

$${x^2} - {y^2} = 0$$

B

$${x^2} + {y^2} = 0$$

C

$${x^2} - {y^2} = 1$$

D

$${x^2} + {y^2} = 1$$

Let the coordinates of centre of circle are (h, k) and radius is a unit. Circle touches both the axes

h = a ..... (i)

k = a ..... (ii)

Eliminating a by squaring (i) and (ii) and then subtracting we get $${h^2} - {k^2} = 0$$

$$ \Rightarrow {x^2} - {y^2} = 0$$, which is required equation of locus.

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

WB JEE 2022 (5)

WB JEE 2021 (1)

WB JEE 2020 (2)

WB JEE 2019 (2)

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Logarithms

Sequence and Series

Quadratic Equations

Complex Numbers

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Matrices and Determinants

Vector Algebra

Three Dimensional Geometry

Probability

Statistics

Sets and Relations

Functions

Definite Integration

Application of Integration

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Differential Equations

Straight Lines and Pair of Straight Lines

Circle

Parabola

Ellipse and Hyperbola