JEE Main Test 42

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JEE Main Test 42

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Weekly Quiz Competition

Question 1
4 / -1

In the rectangular Cartesian plane, the equation of a curve is given by the equation ( x² + y² ) ² = 4x²y. The exhaustive set of y – coordinates of the points on this curve is given by

A
R

B
[ 0, 1 ]

C
[ – 1, 1 ]

D
[ 0,∞]

Solution

The equation is x + 2 ( y² - 2y ) x² + y⁴ = 0.

Put x² = t then t² + 2 ( y² - 2y ) t + y² = 0 .............. ( i )

Equation ( i ) will be non - negative roots if

4 ( y² - 2y ) ² - 4y² ≥ 0

⇒ y ≤ 1

Also, y4≥ 0 and, 0 ≤ - ( y2 - 2y ) .

⇒ 0 ≤ y ≤ 2. So y ∈ [ 0, 1 ]

The correct answer is: [ 0, 1 ]
Question 2
4 / -1

If b > a, then the equation (x– a) (x– b) – 1 = 0 has

A
both roots in [a, b]

B
both roots in (b, ∞)

C
one root in (- ∞, a) and other in (b, ∞).

D
both roots in [– ∞, a]

Solution

Let, f (x) = (x - a) (x - b) - 1

⇒f (a) = -1 and f (b) = -1.

Also, The coefficient x² = 1 > 0

Hence a and b both lie between the roots of the equation f (x) = 0.

∴ The equation ( x - a ) ( x - b) - 1 = 0 has one root

In (- ∞, a) and other in ( b, ∞ ) [ ∵ b > a ]

The correct answer is: one root in (- ∞, a) and other in (b, ∞).
Question 3
4 / -1

if a² + b² – c² – 2ab = 0 then the point of concurrency of family of straight lines ax + by + c = 0 lies on the line

A
y = –x

B
y = x

C
x + y =1.

D
y = x + 1

Solution

(a - b)² - c² = 0

(a + b + c)(a - b - c ) = 0.

Line ax + by + c = 0 passes through

either of two points (1, -1) and (-1, 1 ).

The correct answer is: y = –x