JEE Advanced Mathematics Test 4

Result

JEE Advanced Mathematics Test 4

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
4 / -1

If x = 2 + 2^1/3 + 2^2/3, then the values of x³ – 6x² + 6x is-

A
–2

B
3

C
4

D
2

Solution

x = 2 + 2^1/3 + 2^2/3

(x – 2) = 2^1/3 + 2^2/3 …(1)

(x – 2)³ = (2^1/3 + 2^2/3)³

⇒ x³ – 8 – 6x (x – 2) = 2 + 4 + 3.2 (x – 2) from (1)

⇒x³ – 6x² + 12 x – 8 = 6 + 6x – 12

⇒x³ – 6x² + 6x = 8 – 6 = 2 Ans
Question 2
4 / -1

If |x – 1| + |x – 2| + |x – 3| ≥ 6 then.

A
0 ≤ x ≤ 4

B
x ≤ – 2 or x ≥ 4

C
x ≤ 0 or x ≥ 4

D
None of these

Solution

| x – 1 | + | x – 2 | + | x – 3 | ≥ 6

∴if x < 1, the inequation becomes

– (x – 1) – (x – 2) – (x – 3) ≥ 6

or – 3 x + 6 ≥ 6

∴x ≤ 0

∴x is such that x < 1 and x ≤ 0 ⇒x ≤ 0

If 1 ≤ x < 2, the inequation becomes

(x – 1) + (x – 2) – (x – 3) ≥ 6

or – x + 4 ≥ 6, ∴ x ≤ – 2

No such values satisfy 1 ≤ x ≤ 2

If 2 ≤ x ≤ 3, the inequation becomes

(x – 1) + (x – 2) – (x – 3) ≥ 6

or x ≥ 6, no such values satisfy 2 ≤ x < 3

If x ≥ 3, the inequation becomes

(x – 1) + (x – 2) + (x – 3) ≥ 6

or 3x – 6 ≥ 6 or 3 x ≥12

∴x ≥ 4 which satisfy x ≥ 3

∴The values of x satisfy x ≤ 0 or x ≥ 4

∴required solution is x ≤ 0 or x ≥ 4