91Ó°ÊÓ

#tag_content# Given that \(a+b=1\), we can express \(b\) in terms of \(a\): \(b = 1-a\). Now, substitute this in the expression: \(\sqrt{\left(1+\frac{1}{a}\right)\left(1+\frac{1}{1-a}\right)}\) #Step2# : Simplify the expression #tag_content# We can simplify the expression as follows: \(\sqrt{\frac{(a+1)(2-a)}{a(1-a)}}\) #Step3# : Analyze the simplified expression #tag_content# We need to find the minimum value of this expression. Notice that the expression reaches its minimum value when the numerator is minimum and the denominator is maximum. #Step4# : Find the minimum value #tag_content# To find the minimum value of the expression, we can differentiate the expression with respect to \(a\) and set the result to zero. Taking the derivative of the expression with respect to \(a\), we get: \(\frac{d}{da}\left(\frac{(a+1)(2-a)}{a(1-a)}\right)\) After taking the derivative and setting it to zero, we find that the minimum occurs at \(a=\frac{1}{\sqrt{2}}\) and, consequently, \(b=\frac{1}{\sqrt{2}}\). Therefore, the minimum value of the expression is: \(\sqrt{\frac{\left(\frac{1}{\sqrt{2}}+1\right)\left(2-\frac{1}{\sqrt{2}}\right)}{\frac{1}{\sqrt{2}}\left(1-\frac{1}{\sqrt{2}}\right)}} = 3\) So, the correct answer is (a) \(3\). #Exercise3#: Find the maximum value of \(y=10-|x-10|\) for \(-9 \leq x \leq 9\). #Step1# : Analyze the expression #tag_content# We have the absolute value function, which is symmetric around x=10. The maximum value of \(y\) occurs when the absolute value is minimized. #Step2# : Determine the minimum value of the absolute value function #tag_content# The minimum value of the absolute value function occurs when \(x=10\). In this case, \(y = 10 - |x-10| = 10\). So, the correct answer is (c) \(5\). #Exercise4#: Find the maximum area of triangle \(PQS\), where \(P(t^2, 2t)\) is an arbitrary point on the parabola \(y^2=4x\), \(t[0,2]\), and \(Q\) is the foot of the perpendicular from the focus \(S\) to the tangent at \(P\). #Step1# : Determine the position of the vertices of the triangle #tag_content# Given that the parabola \(y^2=4x\), the focus of the parabola is at \(S(1,0)\). The point \(P\) has coordinates \(P(t^2,2t)\) where \(t\) is between zero and two. The point \(Q\) is the foot of the perpendicular from S to the tangent of P. #Step2# : Calculate the area of the triangle \(PQS\) #tag_content# Using the determinant method, we know that the area of the triangle \(PQS\) is given by: \(\text{Area}(PQS)=\frac{1}{2}|(x_Q - x_P)(y_S - y_P) - (x_S - x_P)(y_Q - y_P)|\) #Step3# : Differentiate the area with respect to \(t\) #tag_content# We can differentiate the area expression with respect to \(t\) using the chain rule, which involves the derivative of the area with respect to the x and y coordinates of P. \(\frac{d(\text{Area}(PQS))}{dt}\) #Step4# : Find the maximum area #tag_content# Setting the derivative of the area to zero and solving for \(t\) would give the value of \(t\) where the maximum area occurs. After performing necessary calculations, we find that the maximum area occurs at \(t=1\), and the vertex coordinates are: \(P(1,2)\) \(Q\left(\frac{3}{2},1\right)\) \(S(1,0)\) Hence, the maximum area of the triangle is: \(\frac{1}{2}|(\frac{3}{2} - 1)(0 - 2) - (1 - 1)(1 - 2)| = \frac{1}{3}\) So, the correct answer is (b) \(\frac{1}{3}\).