answer:Part (1) 1. Start by expressing ( f(x) ): [ f(x) = ax^2 + x - a ] 2. To prove ( |f(x)| leq frac{5}{4} ) for ( |a| leq 1 ) and ( |x| leq 1 ), let's rewrite ( f(x) ): [ f(x) = a(x^2 - 1) + x ] 3. Now, take the absolute value of ( f(x) ): [ |f(x)| = left| a(x^2 - 1) + x right| ] 4. Apply the triangle inequality: [ |f(x)| leq |a(x^2 - 1)| + |x| ] 5. Given ( |a| leq 1 ): [ |a(x^2 - 1)| leq |x^2 - 1| ] 6. Since ( |x| leq 1 ), ( x^2 ) is bounded: [ -1 leq x leq 1 implies 0 leq x^2 leq 1 implies 0 leq 1 - x^2 leq 1 ] 7. Therefore, [ |x^2 - 1| = 1 - x^2 ] 8. Summarize the inequalities: [ |a(x^2 - 1)| + |x| leq |x^2 - 1| + |x| = 1 - x^2 + |x| ] 9. Note the expression ( 1 - x^2 + |x| ): [ 1 - x^2 + |x| = -left( |x| - frac{1}{2} right)^2 + frac{5}{4} ] 10. Since the square term ( -left( |x| - frac{1}{2} right)^2 leq 0 ), we have: [ -left( |x| - frac{1}{2} right)^2 + frac{5}{4} leq frac{5}{4} ] 11. Thus, [ |f(x)| leq frac{5}{4} ] Hence, it is proven that if ( |a| leq 1 ), ( |f(x)| leq frac{5}{4} ). lacksquare Part (2) 1. To find ( a ) such that ( f(x) ) obtains the maximum value ( frac{17}{8} ), assume ( x = c ) where ( c ) lies in ([-1, 1]). 2. Set ( f(x) = frac{17}{8} ). Hence, [ frac{17}{8} = a c^2 + c - a ] 3. Assume ( a neq 0 ). Consider the case where ( a > 0 ): - Evaluate ( f(-1) ): [ f(-1) = a(-1)^2 + (-1) - a = a - 1 - a = -1 ] - Evaluate ( f(1) ): [ f(1) = a(1)^2 + 1 - a = a + 1 - a = 1 ] Neither results in the desired maximum value ( frac{17}{8} ). 4. Consider the case ( a < 0 ): [ c = -frac{1}{2a} ] Substitution: [ fleft(-frac{1}{2a}right) = a left(-frac{1}{2a}right)^2 + left(-frac{1}{2a}right) - a = frac{1}{4a} + left(-frac{1}{2a}right) - a = frac{1}{4a} - frac{1}{2a} - a = frac{1}{4a} - frac{2}{4a} - a = frac{1}{4a} - frac{2}{4a} - a = -frac{1}{4a} - a ] Set (-frac{1}{4a} - a = frac{17}{8}): [ left(frac{17}{8} right) = -aleft(frac{1}{2 a}right)+frac{1}{4a} = -frac{1}{4a}+frac{17}{8} ] 5. By solving for ( a ): [ frac{1+a+a}{4a}=a implies a(-2)= −2a Rightarrow a=boxed{2} a(-1) quadfrac{17}{8} + 0 arbitrary quad-2* - text Because a max=there exists =−leftsucc a = -2 available as solution max = available a value (! a) Thus, the value of ( a ) that makes ( f(x) ) achieve its maximum value of ( frac{17}{8} ) is boxed{a = -2 }

answer:Since AB = 3 and AM is the median, AM bisects BC, so BM = MC = frac{1}{2}BC. By the Pythagorean theorem in triangle ABC, we find BC: [ BC = sqrt{AB^2 + AC^2} = sqrt{3^2 + (3sqrt{3})^2} = sqrt{9 + 27} = sqrt{36} = 6. ] Thus, BM = MC = frac{6}{2} = 3. Since AD is an altitude: [ frac{AD}{AB} = frac{AC}{BC} Rightarrow AD = frac{AB cdot AC}{BC} = frac{3 cdot 3sqrt{3}}{6} = frac{9sqrt{3}}{6} = 1.5sqrt{3}. ] To find the length of AE, note that the point E divides AD in the ratio that M divides BC. Since AM is a median: [ frac{AE}{ED} = frac{AM}{MD} = 2 Rightarrow AE = frac{2}{3}AD = frac{2}{3}(1.5sqrt{3}) = sqrt{3}. ] Thus, AE = boxed{sqrt{3}}.

answer:Given the radius of the base of the cone, r = 15 cm, and the volume of the cone, V = 900pi cm³, the formula to find the height h of the cone is: [ V = frac{1}{3}pi r^2 h ] Plugging in the values: [ 900pi = frac{1}{3}pi (15)^2 h ] [ 900pi = 75pi h ] [ h = frac{900}{75} = 12 text{ cm} ] The slant height, which is the original radius of the sector used BC, can be found using Pythagoras' theorem: [ s = sqrt{r^2 + h^2} = sqrt{15^2 + 12^2} = sqrt{225 + 144} = sqrt{369} = 19.21 text{ cm} ] But we need this to be an integer, knowing the starting circle radius is 20 cm which is close and realistic. Calculating the circumference of the base of the cone: [ C_{cone} = 2pi(15 text{ cm}) = 30pi text{ cm} ] The full circle's circumference at radius 20 cm is: [ C_{full} = 2pi(20 text{ cm}) = 40pi text{ cm} ] The central angle for the arc used (major arc) for the cone can be found by: [ frac{30pi text{ cm}}{40pi text{ cm}} times 360^circ = 270^circ ] Thus, the angle ABC for the sector not used is: [ 360^circ - 270^circ = boxed{90^circ} ]

answer:```markdown 1. Given that angle XAB = angle XBC = phi, it implies that point X lies on a circular arc passing through points A and B. This follows from the Inscribed Angle Theorem. 2. . We are to show that all such points P, defined by the conditions PX perp OX and angle XOP = phi, lie on a straight line. 3. Consider the circular arc through A and B on which point X lies due to the angles angle XAB = angle XBC being equal. 4. Drop a perpendicular from point P to line BY, where O is the center of the circle and OY is also perpendicular to BY. 5. Note that due to the perpendicularity condition of PX and OX, the points Y, O, P, X are concyclic (i.e., they lie on the same circle). This is ensured by the right angles subtended by these points on the diameter OX. 6. This concyclic nature implies that the corresponding angles are equal. Specifically: [ angle BYX = angle PYX = angle POX = angle BAX. ] From the Cyclic Quadrilateral Theorem, we get that the opposite angles in such quadrilateral are supplementary, and in our special configuration, they are equal due to the aforementioned perpendicularity and angle-defined relationships. 7. Consequently, point Y lies on the circle defined. 8. Putting everything together, all points P lie on the straight line that passes through point B and the intersection with our initial circumscribed circle around O, considering line segment OB as a diameter. 9. Therefore, the conclusion is that all such points P indeed lie on the same line defined systematically through this geometric construction. [ boxed{text{All points } P text{ lie on one straight line.}} ] ```