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Miscellaneous equation

Q.No: 1
Test Name : CAT 2017 Actual Paper Slot 1

If x + 1 = x2 and x > 0, then 2x4 is

A
6 + 4√5
B
3 + 5√5
C
5 + 3√5
D
7 + 3√5
Solution:
Q.No: 2
Test Name : CAT 2017 Actual Paper Slot 1

If 92x – 1 – 81x – 1 = 1944, then x is

A
3
B
9/4
C
4/9
D
1/3
Solution:
Q.No: 3
Test Name : CAT 2017 Actual Paper Slot 2


A
3/2
B
2/5
C
3/4
D
4/9
Solution:
Q.No: 4
Test Name : CAT 2017 Actual Paper Slot 2

The numbers 1, 2, ..... 9 are arranged in a 3 × 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.
If the top left and the top right entries of the grid are 6 and 2, respectively, then the bottom middle entry is

Solution:
Q.No: 5
Test Name : CAT 2018 Actual Paper Slot 1

Given that x2018y2017 = 1/2 and x2016y2019 = 8, the value of x2 + y3 is

A
33/4
B
35/4
C
37/4
D
31/4
Solution:
Q.No: 6
Test Name : CAT 2019 Actual Paper Slot 1

The number of the real roots of the equation 2cos (x (x + 1)) = 2x + 2–x is

A
1
B
0
C
infinite
D
2
Solution:
Q.No: 7
Test Name : CAT 2019 Actual Paper Slot 2

How many pairs (m, n) of positive integers satisfy the equation m2 + 105 = n2?

Solution:
Q.No: 8
Test Name : CAT Actual Paper 2020 Slot-1


Solution:
Q.No: 9
Test Name : CAT Actual Paper 2020 Slot-1

The number of real-valued solutions of the equation 2x + 2–x = 2 – (x – 2)2 is

A
infinite
B
0
C
2
D
1
Solution:
Q.No: 10
Test Name : CAT Actual Paper 2020 Slot-2

The number of integers that satisfy the equality (x2 – 5x + 7)x+1 = 1 is

A
2
B
3
C
4
D
5
Solution:
Q.No: 11
Test Name : CAT Actual Paper 2020 Slot-3


Solution:
Q.No: 12
Test Name : CAT Actual Paper 2020 Slot-3


A
4
B
3
C
1
D
2
Solution:
Q.No: 13
Test Name : CAT Actual Paper 2022 Slot-2

Let r and c be real numbers. If r and –r are roots of 5x3 + cx2 – 10x + 9 = 0, then c equals

A
4
B
–9/2
C
–4
D
9/2
Solution:
Q.No: 14
Test Name : CAT Actual Paper 2022 Slot-3


A
2
B
–1/2
C
1/2
D
–2
Solution:
Q.No: 15
Test Name : CAT Actual Paper 2023 Slot 1

The equation x3 + (2r + 1)x2 + (4r – 1)x + 2 = 0 has –2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is

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