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Graduate Aptitude Test in Engineering

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1

Subjective

The equations to the pairs of opposite sides of a parallelogram are x^{2} $$-$$ 5x + 6 = 0 and y^{2} $$-$$ 6y + 5 = 0. Find the equations of its diagonals.

.

$${x^2} - 5x + 6 = 0 \Rightarrow (x - 3)(x - 2) = 0$$

$$\Rightarrow$$ x = 3, x = 2 are the equations of lines $${y^2} - 6y + 5 = 0$$

$$ \Rightarrow (y - 5)(y - 1) = 0$$

$$\Rightarrow$$ y = 5, y = 1 are the equations of lines ABCD is parallelogram formed by above lines

co-ordinates of A(2, 1)

co-ordinates of B(3, 1)

co-ordinates of C(3, 5)

co-ordinates of D(2, 5)

Equation of diagonal $$AC:{{{y_2} - {y_1}} \over {{x_2} - {x_1}}} = {{y - {y_1}} \over {x - {x_1}}}$$

$$ \Rightarrow {{5 - 1} \over {3 - 2}} = {{y - 1} \over {x - 2}} \Rightarrow {4 \over 1} = {{y - 1} \over {x - 2}}$$

$$ \Rightarrow 4x - 8 = y - 1 \Rightarrow 4x - y = 7$$

Equation of diagonal $$BD:{{5 - 1} \over {2 - 3}} = {{y - 1} \over {x - 3}}$$

$$ \Rightarrow {4 \over { - 1}} = {{y - 1} \over {x - 3}} \Rightarrow 4x - 12 = - y + 1$$

$$ \Rightarrow 4x + y = 13$$

$$\therefore$$ Required equations of diagonals are

4x $$-$$ y = 7 and 4x + y = 13

2

Subjective

If 2a $$-$$ 5b $$-$$ 3c = 0, show that the straight line ax + by + c = 0 always passes through a fixed point. Find the equation of one straight line passing through this fixed point and parallel to the line 9x + 12y = 20.

.

We have, $$2a - 5b - 3c = 0$$.

or, $$c = {{2a - 5b} \over 3}$$ ..... (i)

Hence, equation of the straight line can be written as

$$ax + by + {{2a - 5b} \over 3} = 0$$

or, $$3ax + 3by + 2a - 5b = 0$$

or, $$a(3x + 2) + b(3y - 5) = 0$$.

This will be true for all values of a and b, only if (3x + 2) and (3y $$-$$ 5) are simultaneously zero which means, for x = $$-$$2/3 and y = 5/3, the equation is always satisfied for any values of a and b.

$$\therefore$$ The straight line passes through the point ($$-$$2/3, 5/3)

Now, slope of the given line 9x + 12y = 20 is $$-$$3/4

Any straight line parallel to this line will also have slope of $$-$$3/4

$$\therefore$$ Equation of the line parallel to 9x + 12y = 20 and passing through ($$-$$2/3, 5/3) is

$$y - {5 \over 3} = - {3 \over 4}\left( {x + {2 \over 3}} \right)$$

$$y - {5 \over 3} = - {3 \over 4}\left( {x + {2 \over 3}} \right)$$ or, $$ - 4y + {{20} \over 3} = 3x + 2$$

or, $$3x + 4y - {{14} \over 3} = 0$$ or, $$9x + 12y - 14 = 0$$.

On those following papers in Subjective

Number in Brackets after Paper Indicates No. of Questions

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