NIMCET 2007 — Mathematics PYQ
NIMCET | Mathematics | 2007will be:
Preparing for Excellence...
(5−2)1/3−(5+2)1/3 will be:
Irrational
one
rational but not integer
none
(Correct Answer)none
Let the given expression be equal to a variable x:
x=(5−2)1/3−(5+2)1/3
To simplify this, let's assume:
a=(5−2)1/3
b=(5+2)1/3
So, our equation becomes:
x=a−b
Using the algebraic identity (a−b)3=a3−b3−3ab(a−b), let's cube both sides of the equation:
x3=(a−b)3
x3=a3−b3−3ab(a−b)
Now, replace a−b back with x:
x3=a3−b3−3ab(x)
Calculate a3 and b3:
a3=[(5−2)1/3]3=5−2
b3=[(5+2)1/3]3=5+2
Calculate the difference (a3−b3):
a3−b3=(5−2)−(5+2)
a3−b3=5−2−5−2=−4
Calculate the product ab:
ab=(5−2)1/3⋅(5+2)1/3
ab=[(5−2)(5+2)]1/3
Using the difference of squares identity (u−v)(u+v)=u2−v2:
ab=[(5)2−(2)2]1/3
ab=[5−4]1/3=[1]1/3=1
Substitute a3−b3=−4 and ab=1 into our cubed expression:
x3=−4−3(1)(x)
x3=−4−3x
Rearrange it into a standard cubic equation form:
x3+3x+4=0
Let's find an integer root by testing small factors of the constant term 4 (i.e., ±1,±2,±4):
If x=−1:
(−1)3+3(−1)+4=−1−3+4=0
Since x=−1 satisfies the equation perfectly, (x+1) is a factor. We can factorize the cubic polynomial as:
(x+1)(x2−x+4)=0
For the quadratic part x2−x+4=0, let's check its discriminant (D=b2−4ac):
D=(−1)2−4(1)(4)=1−16=−15
Since the discriminant is negative (D < 0), the quadratic part yields imaginary roots. Thus, the equation has only one real solution:
x=−1
The value of the expression is −1.
Since −1 is an integer, it falls under the category of rational numbers. Checking our options:
(a) Irrational (Incorrect)
(b) one (Incorrect, since it is negative one)
(c) rational but not integer (Incorrect, since it is an integer)
Therefore, none of the specific properties or values listed in (a), (b), or (c) are correct.
Correct Option: (d) none
Let the given expression be equal to a variable x:
x=(5−2)1/3−(5+2)1/3
To simplify this, let's assume:
a=(5−2)1/3
b=(5+2)1/3
So, our equation becomes:
x=a−b
Using the algebraic identity (a−b)3=a3−b3−3ab(a−b), let's cube both sides of the equation:
x3=(a−b)3
x3=a3−b3−3ab(a−b)
Now, replace a−b back with x:
x3=a3−b3−3ab(x)
Calculate a3 and b3:
a3=[(5−2)1/3]3=5−2
b3=[(5+2)1/3]3=5+2
Calculate the difference (a3−b3):
a3−b3=(5−2)−(5+2)
a3−b3=5−2−5−2=−4
Calculate the product ab:
ab=(5−2)1/3⋅(5+2)1/3
ab=[(5−2)(5+2)]1/3
Using the difference of squares identity (u−v)(u+v)=u2−v2:
ab=[(5)2−(2)2]1/3
ab=[5−4]1/3=[1]1/3=1
Substitute a3−b3=−4 and ab=1 into our cubed expression:
x3=−4−3(1)(x)
x3=−4−3x
Rearrange it into a standard cubic equation form:
x3+3x+4=0
Let's find an integer root by testing small factors of the constant term 4 (i.e., ±1,±2,±4):
If x=−1:
(−1)3+3(−1)+4=−1−3+4=0
Since x=−1 satisfies the equation perfectly, (x+1) is a factor. We can factorize the cubic polynomial as:
(x+1)(x2−x+4)=0
For the quadratic part x2−x+4=0, let's check its discriminant (D=b2−4ac):
D=(−1)2−4(1)(4)=1−16=−15
Since the discriminant is negative (D < 0), the quadratic part yields imaginary roots. Thus, the equation has only one real solution:
x=−1
The value of the expression is −1.
Since −1 is an integer, it falls under the category of rational numbers. Checking our options:
(a) Irrational (Incorrect)
(b) one (Incorrect, since it is negative one)
(c) rational but not integer (Incorrect, since it is an integer)
Therefore, none of the specific properties or values listed in (a), (b), or (c) are correct.
Correct Option: (d) none