If f(5 + x) = f(5 − x) for every real x and f(x) = 0 has four distinct real roots, then the sum of the roots is
CAT 2026Papers & G StrategyQuantX maro (Modern Maths)
CAT 2020 Quant x maro G Strategy
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Solve the following CAT 2020 questions using "x" maro G strategy.
Question 1 of 23
If log4 5 = (log4 y) (log6 √5), then y equals
The number of real-valued solutions of the equation 2x + 2−x = 2 − (x − 2)2 is
If x = (4096)7+4√3, then which of the following equals 64?
The number of distinct real roots of the equation (x + 1/x)2 − 3(x + 1/x) + 2 = 0 equals
Among 100 students, x1 have birthdays in January, x2 have birthdays in February, and so on. If x0 = max(x1, x2, ..., x12), then the smallest possible value of x0 is
How many distinct positive integer-valued solutions exist to the equation (x2 − 7x + 11)(x2 − 13x + 42) = 1?
The area of the region satisfying the inequalities |x| − y ≤ 1, y ≥ 0, and y ≤ 1 is.
If y is a negative number such that 2y2log35 = 5log23, then y equals
The number of pairs of integers(x,y) satisfying x ≥ y ≥ −20 and 2x + 5y = 99 is
The value of logaa/b + logbb/a, for 1 < a ≤ b cannot be equal to
If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals
Let f(x) = x2 + ax + b and g(x) = f(x + 1) − f(x − 1). If f(x) ≥ 0 for all real x, and g(20) = 72, then the smallest possible value of b is
The number of integers that satisfy the equality (x2 − 5x + 7)x + 1 = 1 is
In how many ways can a pair of integers (x , a) be chosen such that x2 − 2 | x | + | a − 2 | = 0 ?
If x1 = −1 and xm = xm + 1 + (m + 1) for every positive integer m, then x100 equals
Let N, x and y be positive integers such that N = x + y, 2 25, then how many distinct values are possible for N?
Let loga30 = A, loga5/3= −B and log2a = 1/3, then log3a equals
The area, in sq. units, enclosed by the lines x = 2, y = |x − 2| + 4, the X-axis and the Y-axis is equal to
If f(x+y) = f(x)f(y) and f(5) = 4, then f(10) − f(−10) is equal to
Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if
Let m and n be positive integers, If x2 + mx + 2n = 0 and x2 + 2nx + m = 0 have real roots, then the smallest possible value of m + n is
The points (2 , 1) and (−3 , −4) are opposite vertices of a parellelogram. If the other two vertices lie on the line x + 9y + c = 0, then c is