undefined First, <=t's recall a key relationship between inverse tan>=nt and inverse cotan>=nt: ^-1(z) = ^-1(1z) This relationship exists because cotan>=nt is the reciprocal of tan>=nt. Geometrically, if an ang<= has a tan>=nt value of z, then the same ang<= has a cotan>=nt value of 1/z.

Using this relationship, <=t's rewrite the right side of our equation: ^-1(33^x+1) = ^-1(3^x+13) Our equation becomes: ^-1(23^-x+1) = ^-1(3^x+13)

For two inverse tan>=nt expressions to be equal, their arguments must be equal (since inverse tan>=nt is a one-to-one function in its principal domain): 23^-x+1 = 3^x+13

Cross-multiplying to solve: 2 3 = (3^-x+1)(3^x+1) 6 = 3^-x 3^x + 3^-x + 3^x + 1 6 = 1 + 3^-x + 3^x + 1 4 = 3^-x + 3^x

Remember that 3^-x = 13^x, so we can rewrite: 4 = 13^x + 3^x 4 3^x = 1 + (3^x)^2 (3^x)^2 - 4(3^x) + 1 = 0

Let's substitute y = 3^x to make this easier to solve: y^2 - 4y + 1 = 0 Using the quadratic formula: y = 4 \± sqrt(16-4)2 = 4 \± sqrt(12)2 = 4 \± 2sqrt(3)2 = 2 \± sqrt(3) So we have two solutions: y = 3^x = 2 + sqrt(3) y = 3^x = 2 - sqrt(3)

Now <=t's find the x-values by taking logarithm base 3: For the first solution: 3^x = 2 + sqrt(3) x = _3(2 + sqrt(3)) Since 2 + sqrt(3) 3.732 > 1, this value of x is positive. For the second solution: 3^x = 2 - sqrt(3) x = _3(2 - sqrt(3)) Since 2 - sqrt(3) 0.268 < 1, this value of x is negative.

A quick conceptual check: When we have an equation with exponential terms like 3^x and 3^-x, it's common to >=t solutions in pairs - one positive and one negative. This happens because of the symmetrical nature of the equation. We can see this symmetry in the final form 4 = 3^-x + 3^x where replacing x with -x gives us the same equation.

Therefore, there is one positive real value x = _3(2 + sqrt(3)) and one negative real value x = _3(2 - sqrt(3)) satisfying the equation. The correct option is: There is one positive and one negative real value of x satisfying the above equation.

CUET Mathematics 2024 Slot 1 PYQs

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