💡Okay, so I have this function g(x) which is equal to (3x + 2) divided by (x - 3). And I need to find the value of g(-2). Hmm, let me think about how to approach this.First, I remember that to find the value of a function at a specific point, I just need to substitute that value into the function wherever I see the variable x. So, in this case, I should replace every x in the function with -2.Alright, let me write that out. So, g(-2) would be equal to (3 times -2 plus 2) divided by (-2 minus 3). Let me compute the numerator and the denominator separately to avoid mistakes.Starting with the numerator: 3 times -2 is -6, right? Then, adding 2 to that gives me -6 plus 2, which is -4. Okay, so the numerator simplifies to -4.Now, moving on to the denominator: -2 minus 3. Hmm, subtracting 3 from -2. That should be -5 because if I have a debt of 2 and then I go into more debt by 3, I end up with a total debt of 5. So, the denominator is -5.Putting it all together, g(-2) is equal to -4 divided by -5. Now, when I divide two negative numbers, the result is positive. So, -4 divided by -5 is the same as 4 divided by 5. That simplifies to 4/5.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting with the numerator: 3 times -2 is indeed -6, and adding 2 gives -4. The denominator: -2 minus 3 is -5. So, -4 divided by -5 is 4/5. Yep, that seems correct.I was a bit confused at first about the signs, especially when dealing with negative numbers, but breaking it down step by step helped me avoid mistakes. It's important to remember that a negative divided by a negative gives a positive result. So, I think I got it right.Just to recap, substituting x with -2 in the function g(x) gives me a numerator of -4 and a denominator of -5, which simplifies to 4/5. So, the value of g(-2) is 4/5.I don't think I made any errors here, but it's always good to verify. Maybe I can plug the value back into the original function to see if it makes sense. If I take 4/5 and multiply it by (x - 3), I should get 3x + 2 when x is -2.Let's test that: 4/5 times (-2 - 3) is 4/5 times -5, which is -4. And 3 times -2 plus 2 is indeed -6 + 2, which is -4. So, both sides match, which means my answer is correct.Alright, I feel confident that g(-2) is 4/5.

💡Okay, so I've got this problem about an ellipse, and I need to find its eccentricity. Let me try to break it down step by step. First, the ellipse is given by the equation ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where ( a > b > 0 ). The left focus is at point ( F ). There's a tangent line drawn at a point ( A ) on the ellipse, and this tangent intersects the y-axis at point ( Q ). The origin is ( O ). We're told that ( angle QFO = 45^circ ) and ( angle QFA = 30^circ ). We need to find the eccentricity of the ellipse.Alright, let's recall some properties of ellipses. The standard form is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where ( a ) is the semi-major axis and ( b ) is the semi-minor axis. The foci are located at ( (pm c, 0) ), where ( c = sqrt{a^2 - b^2} ). The eccentricity ( e ) is given by ( e = frac{c}{a} ). So, ultimately, we need to find ( e = frac{sqrt{a^2 - b^2}}{a} ).Now, let's think about the tangent line at point ( A ). The equation of the tangent to the ellipse at point ( A(x_0, y_0) ) is ( frac{x x_0}{a^2} + frac{y y_0}{b^2} = 1 ). This tangent intersects the y-axis at point ( Q ). To find the coordinates of ( Q ), we set ( x = 0 ) in the tangent equation, which gives ( frac{y y_0}{b^2} = 1 ), so ( y = frac{b^2}{y_0} ). Therefore, ( Q ) is at ( (0, frac{b^2}{y_0}) ).Similarly, if we set ( y = 0 ) in the tangent equation, we can find the x-intercept ( P ), but since the problem is about the y-intercept, we can focus on ( Q ).Now, let's consider the points ( Q ), ( F ), and ( O ). The origin ( O ) is at ( (0, 0) ), and the left focus ( F ) is at ( (-c, 0) ). Point ( Q ) is at ( (0, frac{b^2}{y_0}) ).We are given two angles: ( angle QFO = 45^circ ) and ( angle QFA = 30^circ ). Let's try to visualize this. First, ( angle QFO ) is the angle at point ( F ) between points ( Q ), ( F ), and ( O ). So, it's the angle between the lines ( FQ ) and ( FO ). Similarly, ( angle QFA ) is the angle at point ( F ) between points ( Q ), ( F ), and ( A ).To find these angles, we can use vector analysis. Let's define vectors for ( overrightarrow{FQ} ) and ( overrightarrow{FO} ), and then use the dot product to find the angle between them.First, ( overrightarrow{FO} ) is from ( F ) to ( O ), which is ( (0 - (-c), 0 - 0) = (c, 0) ).Next, ( overrightarrow{FQ} ) is from ( F ) to ( Q ), which is ( (0 - (-c), frac{b^2}{y_0} - 0) = (c, frac{b^2}{y_0}) ).Similarly, ( overrightarrow{FA} ) is from ( F ) to ( A ), which is ( (x_0 - (-c), y_0 - 0) = (x_0 + c, y_0) ).Now, let's compute the dot product for ( angle QFO ). The dot product of ( overrightarrow{FQ} ) and ( overrightarrow{FO} ) is:( overrightarrow{FQ} cdot overrightarrow{FO} = (c)(c) + left( frac{b^2}{y_0} right)(0) = c^2 ).The magnitudes of these vectors are:( | overrightarrow{FQ} | = sqrt{c^2 + left( frac{b^2}{y_0} right)^2 } ),( | overrightarrow{FO} | = sqrt{c^2 + 0} = c ).The cosine of the angle ( theta = 45^circ ) is given by:( cos theta = frac{ overrightarrow{FQ} cdot overrightarrow{FO} }{ | overrightarrow{FQ} | | overrightarrow{FO} | } = frac{c^2}{c sqrt{c^2 + left( frac{b^2}{y_0} right)^2 }} = frac{c}{sqrt{c^2 + left( frac{b^2}{y_0} right)^2 }} ).Since ( theta = 45^circ ), ( cos 45^circ = frac{sqrt{2}}{2} ). Therefore,( frac{c}{sqrt{c^2 + left( frac{b^2}{y_0} right)^2 }} = frac{sqrt{2}}{2} ).Let's square both sides to eliminate the square root:( frac{c^2}{c^2 + left( frac{b^2}{y_0} right)^2 } = frac{1}{2} ).Cross-multiplying:( 2c^2 = c^2 + left( frac{b^2}{y_0} right)^2 ).Subtracting ( c^2 ) from both sides:( c^2 = left( frac{b^2}{y_0} right)^2 ).Taking square roots:( c = frac{b^2}{y_0} ).So, ( y_0 = frac{b^2}{c} ).Alright, that's one equation relating ( y_0 ) and ( c ). Let's keep that in mind.Now, let's consider the second angle ( angle QFA = 30^circ ). This is the angle at ( F ) between points ( Q ), ( F ), and ( A ). So, we need to find the angle between vectors ( overrightarrow{FQ} ) and ( overrightarrow{FA} ).We already have ( overrightarrow{FQ} = (c, frac{b^2}{y_0}) ) and ( overrightarrow{FA} = (x_0 + c, y_0) ).First, let's compute the dot product ( overrightarrow{FQ} cdot overrightarrow{FA} ):( (c)(x_0 + c) + left( frac{b^2}{y_0} right)(y_0) = c x_0 + c^2 + b^2 ).Simplify:( c x_0 + c^2 + b^2 ).Now, the magnitudes of these vectors are:( | overrightarrow{FQ} | = sqrt{c^2 + left( frac{b^2}{y_0} right)^2 } ).But from earlier, we found that ( c = frac{b^2}{y_0} ), so ( frac{b^2}{y_0} = c ). Therefore, ( | overrightarrow{FQ} | = sqrt{c^2 + c^2} = sqrt{2 c^2} = c sqrt{2} ).Similarly, ( | overrightarrow{FA} | = sqrt{(x_0 + c)^2 + y_0^2} ).So, the cosine of the angle ( phi = 30^circ ) is:( cos phi = frac{ overrightarrow{FQ} cdot overrightarrow{FA} }{ | overrightarrow{FQ} | | overrightarrow{FA} | } = frac{c x_0 + c^2 + b^2}{c sqrt{2} cdot sqrt{(x_0 + c)^2 + y_0^2}} ).Since ( phi = 30^circ ), ( cos 30^circ = frac{sqrt{3}}{2} ). Therefore,( frac{c x_0 + c^2 + b^2}{c sqrt{2} cdot sqrt{(x_0 + c)^2 + y_0^2}} = frac{sqrt{3}}{2} ).Let's simplify this equation.First, note that ( c x_0 + c^2 + b^2 = c(x_0 + c) + b^2 ).Also, ( (x_0 + c)^2 + y_0^2 = x_0^2 + 2 c x_0 + c^2 + y_0^2 ).But since ( A(x_0, y_0) ) lies on the ellipse, it satisfies ( frac{x_0^2}{a^2} + frac{y_0^2}{b^2} = 1 ). Therefore, ( x_0^2 = a^2 (1 - frac{y_0^2}{b^2}) ).Substituting ( y_0 = frac{b^2}{c} ) from earlier, we get:( x_0^2 = a^2 left( 1 - frac{(b^2/c)^2}{b^2} right) = a^2 left( 1 - frac{b^2}{c^2} right) ).So, ( x_0^2 = a^2 - frac{a^2 b^2}{c^2} ).Therefore, ( (x_0 + c)^2 + y_0^2 = x_0^2 + 2 c x_0 + c^2 + y_0^2 ).Substituting ( x_0^2 ) and ( y_0^2 ):( = (a^2 - frac{a^2 b^2}{c^2}) + 2 c x_0 + c^2 + frac{b^4}{c^2} ).Simplify:( = a^2 - frac{a^2 b^2}{c^2} + 2 c x_0 + c^2 + frac{b^4}{c^2} ).Combine like terms:( = a^2 + c^2 + 2 c x_0 + left( - frac{a^2 b^2}{c^2} + frac{b^4}{c^2} right) ).Factor out ( frac{b^2}{c^2} ):( = a^2 + c^2 + 2 c x_0 + frac{b^2}{c^2} ( - a^2 + b^2 ) ).But ( c^2 = a^2 - b^2 ), so ( - a^2 + b^2 = - c^2 ). Therefore,( = a^2 + c^2 + 2 c x_0 - frac{b^2 c^2}{c^2} ).Simplify:( = a^2 + c^2 + 2 c x_0 - b^2 ).But ( a^2 - b^2 = c^2 ), so:( = c^2 + c^2 + 2 c x_0 ).Simplify:( = 2 c^2 + 2 c x_0 = 2 c (c + x_0) ).Therefore, ( | overrightarrow{FA} | = sqrt{2 c (c + x_0)} ).Now, going back to the cosine equation:( frac{c x_0 + c^2 + b^2}{c sqrt{2} cdot sqrt{2 c (c + x_0)}} = frac{sqrt{3}}{2} ).Simplify the denominator:( c sqrt{2} cdot sqrt{2 c (c + x_0)} = c sqrt{2} cdot sqrt{2 c} cdot sqrt{c + x_0} = c cdot sqrt{2} cdot sqrt{2 c} cdot sqrt{c + x_0} ).Simplify ( sqrt{2} cdot sqrt{2 c} = sqrt{4 c} = 2 sqrt{c} ).Therefore, the denominator becomes:( c cdot 2 sqrt{c} cdot sqrt{c + x_0} = 2 c^{3/2} sqrt{c + x_0} ).So, the equation becomes:( frac{c x_0 + c^2 + b^2}{2 c^{3/2} sqrt{c + x_0}} = frac{sqrt{3}}{2} ).Multiply both sides by 2:( frac{c x_0 + c^2 + b^2}{c^{3/2} sqrt{c + x_0}} = sqrt{3} ).Let me note that ( c x_0 + c^2 + b^2 = c(x_0 + c) + b^2 ). Also, ( c + x_0 ) is under the square root in the denominator.Let me denote ( s = c + x_0 ). Then, ( x_0 = s - c ).Substituting into the numerator:( c (s - c + c) + b^2 = c s + b^2 ).So, the equation becomes:( frac{c s + b^2}{c^{3/2} sqrt{s}} = sqrt{3} ).Simplify the numerator:( c s + b^2 = c s + b^2 ).So, the equation is:( frac{c s + b^2}{c^{3/2} sqrt{s}} = sqrt{3} ).Let me factor out ( c ) from the numerator:( frac{c (s) + b^2}{c^{3/2} sqrt{s}} = sqrt{3} ).Hmm, not sure if that helps. Maybe let's write it as:( frac{c s + b^2}{c^{3/2} sqrt{s}} = sqrt{3} ).Let me separate the terms:( frac{c s}{c^{3/2} sqrt{s}} + frac{b^2}{c^{3/2} sqrt{s}} = sqrt{3} ).Simplify each term:First term: ( frac{c s}{c^{3/2} sqrt{s}} = frac{c}{c^{3/2}} cdot frac{s}{sqrt{s}} = frac{1}{sqrt{c}} cdot sqrt{s} ).Second term: ( frac{b^2}{c^{3/2} sqrt{s}} ).So, the equation becomes:( frac{sqrt{s}}{sqrt{c}} + frac{b^2}{c^{3/2} sqrt{s}} = sqrt{3} ).Let me factor out ( frac{1}{sqrt{c}} ):( frac{1}{sqrt{c}} left( sqrt{s} + frac{b^2}{c sqrt{s}} right) = sqrt{3} ).Let me denote ( t = sqrt{s} ). Then, ( s = t^2 ), and ( sqrt{s} = t ), ( frac{1}{sqrt{s}} = frac{1}{t} ).Substituting into the equation:( frac{1}{sqrt{c}} left( t + frac{b^2}{c t} right) = sqrt{3} ).Multiply both sides by ( sqrt{c} ):( t + frac{b^2}{c t} = sqrt{3 c} ).Multiply both sides by ( t ):( t^2 + frac{b^2}{c} = sqrt{3 c} cdot t ).Rearrange:( t^2 - sqrt{3 c} cdot t + frac{b^2}{c} = 0 ).This is a quadratic equation in terms of ( t ). Let's write it as:( t^2 - sqrt{3 c} cdot t + frac{b^2}{c} = 0 ).Let me denote ( t = sqrt{s} = sqrt{c + x_0} ). So, solving for ( t ), we can use the quadratic formula:( t = frac{ sqrt{3 c} pm sqrt{ 3 c - 4 cdot 1 cdot frac{b^2}{c} } }{2} ).Simplify the discriminant:( sqrt{3 c - frac{4 b^2}{c}} ).So,( t = frac{ sqrt{3 c} pm sqrt{ 3 c - frac{4 b^2}{c} } }{2} ).But since ( t = sqrt{c + x_0} ) must be positive, we can consider both solutions.However, let's also recall that ( s = c + x_0 ), and ( x_0 ) is a coordinate on the ellipse, so ( x_0 ) must satisfy ( |x_0| leq a ). Therefore, ( s = c + x_0 ) must be positive, so ( t ) is positive.Now, let's see if we can find a relationship between ( a ), ( b ), and ( c ) to find the eccentricity ( e = frac{c}{a} ).We have from earlier that ( y_0 = frac{b^2}{c} ). Since ( A(x_0, y_0) ) lies on the ellipse, it must satisfy ( frac{x_0^2}{a^2} + frac{y_0^2}{b^2} = 1 ).Substituting ( y_0 = frac{b^2}{c} ):( frac{x_0^2}{a^2} + frac{(b^4 / c^2)}{b^2} = 1 ).Simplify:( frac{x_0^2}{a^2} + frac{b^2}{c^2} = 1 ).So,( frac{x_0^2}{a^2} = 1 - frac{b^2}{c^2} ).But ( c^2 = a^2 - b^2 ), so:( frac{x_0^2}{a^2} = 1 - frac{b^2}{a^2 - b^2} ).Simplify the right-hand side:( 1 - frac{b^2}{a^2 - b^2} = frac{(a^2 - b^2) - b^2}{a^2 - b^2} = frac{a^2 - 2 b^2}{a^2 - b^2} ).Therefore,( x_0^2 = a^2 cdot frac{a^2 - 2 b^2}{a^2 - b^2} ).So,( x_0 = pm a sqrt{ frac{a^2 - 2 b^2}{a^2 - b^2} } ).But since ( A ) is a point on the ellipse, ( x_0 ) can be positive or negative. However, since the tangent is drawn at ( A ) and intersects the y-axis at ( Q ), the position of ( A ) relative to the y-axis will determine the sign of ( x_0 ). But without loss of generality, let's assume ( x_0 ) is positive (if not, we can adjust later).So, ( x_0 = a sqrt{ frac{a^2 - 2 b^2}{a^2 - b^2} } ).Now, let's recall that ( s = c + x_0 ), so:( s = c + a sqrt{ frac{a^2 - 2 b^2}{a^2 - b^2} } ).But ( c = sqrt{a^2 - b^2} ), so:( s = sqrt{a^2 - b^2} + a sqrt{ frac{a^2 - 2 b^2}{a^2 - b^2} } ).Let me factor out ( sqrt{a^2 - b^2} ):( s = sqrt{a^2 - b^2} left( 1 + frac{a}{sqrt{a^2 - b^2}} sqrt{ frac{a^2 - 2 b^2}{a^2 - b^2} } right ) ).Simplify inside the parentheses:( 1 + frac{a}{sqrt{a^2 - b^2}} cdot sqrt{ frac{a^2 - 2 b^2}{a^2 - b^2} } = 1 + frac{a}{a^2 - b^2} sqrt{ (a^2 - 2 b^2)(a^2 - b^2) } ).Wait, this seems complicated. Maybe there's a better approach.Let me recall that ( e = frac{c}{a} ), so ( c = a e ). Also, ( b^2 = a^2 (1 - e^2) ).Let me substitute ( c = a e ) and ( b^2 = a^2 (1 - e^2) ) into our equations.From earlier, we have ( y_0 = frac{b^2}{c} = frac{a^2 (1 - e^2)}{a e} = frac{a (1 - e^2)}{e} ).Also, ( x_0^2 = a^2 cdot frac{a^2 - 2 b^2}{a^2 - b^2} ).Substituting ( b^2 = a^2 (1 - e^2) ):( x_0^2 = a^2 cdot frac{a^2 - 2 a^2 (1 - e^2)}{a^2 - a^2 (1 - e^2)} ).Simplify numerator and denominator:Numerator: ( a^2 - 2 a^2 (1 - e^2) = a^2 - 2 a^2 + 2 a^2 e^2 = - a^2 + 2 a^2 e^2 = a^2 (2 e^2 - 1) ).Denominator: ( a^2 - a^2 (1 - e^2) = a^2 - a^2 + a^2 e^2 = a^2 e^2 ).Therefore,( x_0^2 = a^2 cdot frac{a^2 (2 e^2 - 1)}{a^2 e^2} = a^2 cdot frac{2 e^2 - 1}{e^2} ).So,( x_0 = a sqrt{ frac{2 e^2 - 1}{e^2} } = a cdot frac{ sqrt{2 e^2 - 1} }{ e } ).Since ( x_0 ) is positive, we take the positive root.Now, let's go back to the quadratic equation we had earlier for ( t ):( t^2 - sqrt{3 c} cdot t + frac{b^2}{c} = 0 ).Substituting ( c = a e ) and ( b^2 = a^2 (1 - e^2) ):( t^2 - sqrt{3 a e} cdot t + frac{a^2 (1 - e^2)}{a e} = 0 ).Simplify:( t^2 - sqrt{3 a e} cdot t + frac{a (1 - e^2)}{e} = 0 ).But ( t = sqrt{c + x_0} = sqrt{a e + x_0} ).Substituting ( x_0 = frac{a sqrt{2 e^2 - 1}}{e} ):( t = sqrt{a e + frac{a sqrt{2 e^2 - 1}}{e}} = a sqrt{ e + frac{ sqrt{2 e^2 - 1} }{ e } } ).This seems quite complicated. Maybe instead of substituting ( x_0 ) in terms of ( e ), we can find another relationship.Wait, perhaps we can use the fact that ( s = c + x_0 ) and express ( t ) in terms of ( e ). Let me try that.We have ( s = c + x_0 = a e + frac{a sqrt{2 e^2 - 1}}{e} ).Factor out ( a ):( s = a left( e + frac{ sqrt{2 e^2 - 1} }{ e } right ) ).So,( t = sqrt{ s } = sqrt{ a left( e + frac{ sqrt{2 e^2 - 1} }{ e } right ) } ).But this seems messy. Maybe instead of going through this route, let's consider the ratio of the two cosine equations.We have:From ( angle QFO = 45^circ ):( frac{c}{sqrt{c^2 + left( frac{b^2}{y_0} right)^2 }} = frac{sqrt{2}}{2} ).From ( angle QFA = 30^circ ):( frac{c x_0 + c^2 + b^2}{c sqrt{2} cdot sqrt{(x_0 + c)^2 + y_0^2}} = frac{sqrt{3}}{2} ).Let me denote ( e = frac{c}{a} ), so ( c = a e ), ( b^2 = a^2 (1 - e^2) ), and ( y_0 = frac{b^2}{c} = frac{a^2 (1 - e^2)}{a e} = frac{a (1 - e^2)}{e} ).Also, ( x_0 = frac{a sqrt{2 e^2 - 1}}{e} ).Let me substitute all these into the second equation.First, compute the numerator ( c x_0 + c^2 + b^2 ):( c x_0 = a e cdot frac{a sqrt{2 e^2 - 1}}{e} = a^2 sqrt{2 e^2 - 1} ).( c^2 = a^2 e^2 ).( b^2 = a^2 (1 - e^2) ).So, numerator:( a^2 sqrt{2 e^2 - 1} + a^2 e^2 + a^2 (1 - e^2) = a^2 sqrt{2 e^2 - 1} + a^2 e^2 + a^2 - a^2 e^2 = a^2 sqrt{2 e^2 - 1} + a^2 ).Denominator:( c sqrt{2} cdot sqrt{(x_0 + c)^2 + y_0^2} ).Compute ( x_0 + c = frac{a sqrt{2 e^2 - 1}}{e} + a e = a left( frac{ sqrt{2 e^2 - 1} }{ e } + e right ) ).Compute ( y_0^2 = left( frac{a (1 - e^2)}{e} right )^2 = frac{a^2 (1 - e^2)^2}{e^2} ).So,( (x_0 + c)^2 + y_0^2 = a^2 left( frac{ sqrt{2 e^2 - 1} }{ e } + e right )^2 + frac{a^2 (1 - e^2)^2}{e^2} ).Factor out ( a^2 ):( a^2 left[ left( frac{ sqrt{2 e^2 - 1} }{ e } + e right )^2 + frac{(1 - e^2)^2}{e^2} right ] ).Let me compute the expression inside the brackets:First term: ( left( frac{ sqrt{2 e^2 - 1} }{ e } + e right )^2 ).Let me denote ( u = frac{ sqrt{2 e^2 - 1} }{ e } ), so the first term is ( (u + e)^2 = u^2 + 2 u e + e^2 ).Compute ( u^2 = frac{2 e^2 - 1}{e^2} ).So,First term: ( frac{2 e^2 - 1}{e^2} + 2 cdot frac{ sqrt{2 e^2 - 1} }{ e } cdot e + e^2 = frac{2 e^2 - 1}{e^2} + 2 sqrt{2 e^2 - 1} + e^2 ).Second term: ( frac{(1 - e^2)^2}{e^2} = frac{1 - 2 e^2 + e^4}{e^2} ).So, the entire expression inside the brackets is:( frac{2 e^2 - 1}{e^2} + 2 sqrt{2 e^2 - 1} + e^2 + frac{1 - 2 e^2 + e^4}{e^2} ).Combine the fractions:( frac{2 e^2 - 1 + 1 - 2 e^2 + e^4}{e^2} + 2 sqrt{2 e^2 - 1} + e^2 = frac{e^4}{e^2} + 2 sqrt{2 e^2 - 1} + e^2 = e^2 + 2 sqrt{2 e^2 - 1} + e^2 = 2 e^2 + 2 sqrt{2 e^2 - 1} ).Therefore,( (x_0 + c)^2 + y_0^2 = a^2 (2 e^2 + 2 sqrt{2 e^2 - 1}) = 2 a^2 (e^2 + sqrt{2 e^2 - 1}) ).So, the denominator becomes:( c sqrt{2} cdot sqrt{2 a^2 (e^2 + sqrt{2 e^2 - 1})} = a e sqrt{2} cdot sqrt{2 a^2 (e^2 + sqrt{2 e^2 - 1})} ).Simplify inside the square root:( sqrt{2 a^2 (e^2 + sqrt{2 e^2 - 1})} = a sqrt{2 (e^2 + sqrt{2 e^2 - 1})} ).Therefore, the denominator is:( a e sqrt{2} cdot a sqrt{2 (e^2 + sqrt{2 e^2 - 1})} = a^2 e sqrt{2} cdot sqrt{2 (e^2 + sqrt{2 e^2 - 1})} ).Simplify ( sqrt{2} cdot sqrt{2} = 2 ):( a^2 e cdot 2 sqrt{e^2 + sqrt{2 e^2 - 1}} = 2 a^2 e sqrt{e^2 + sqrt{2 e^2 - 1}} ).So, putting it all together, the second cosine equation becomes:( frac{a^2 sqrt{2 e^2 - 1} + a^2}{2 a^2 e sqrt{e^2 + sqrt{2 e^2 - 1}}} = frac{sqrt{3}}{2} ).Simplify numerator and denominator:Numerator: ( a^2 ( sqrt{2 e^2 - 1} + 1 ) ).Denominator: ( 2 a^2 e sqrt{e^2 + sqrt{2 e^2 - 1}} ).Cancel ( a^2 ):( frac{ sqrt{2 e^2 - 1} + 1 }{ 2 e sqrt{e^2 + sqrt{2 e^2 - 1}} } = frac{sqrt{3}}{2} ).Multiply both sides by 2:( frac{ sqrt{2 e^2 - 1} + 1 }{ e sqrt{e^2 + sqrt{2 e^2 - 1}} } = sqrt{3} ).Let me denote ( u = sqrt{2 e^2 - 1} ). Then, ( u^2 = 2 e^2 - 1 ), so ( e^2 = frac{u^2 + 1}{2} ).Substituting into the equation:( frac{ u + 1 }{ e sqrt{ frac{u^2 + 1}{2} + u } } = sqrt{3} ).Simplify the denominator inside the square root:( frac{u^2 + 1}{2} + u = frac{u^2 + 1 + 2 u}{2} = frac{(u + 1)^2}{2} ).Therefore, the denominator becomes:( e sqrt{ frac{(u + 1)^2}{2} } = e cdot frac{u + 1}{sqrt{2}} ).So, the equation becomes:( frac{ u + 1 }{ e cdot frac{u + 1}{sqrt{2}} } = sqrt{3} ).Simplify:( frac{ u + 1 }{ e } cdot frac{ sqrt{2} }{ u + 1 } = sqrt{3} ).Cancel ( u + 1 ):( frac{ sqrt{2} }{ e } = sqrt{3} ).Therefore,( e = frac{ sqrt{2} }{ sqrt{3} } = frac{ sqrt{6} }{ 3 } ).So, the eccentricity ( e ) is ( frac{ sqrt{6} }{ 3 } ).**Final Answer**The eccentricity of the ellipse is boxed{dfrac{sqrt{6}}{3}}.

💡Okay, so I have this problem about an equilateral triangle XYZ where each side is 13 units long. The medians XT and YZ intersect at right angles. I know that in an equilateral triangle, all medians are equal and they also act as altitudes and angle bisectors. The problem gives me the lengths of the medians XT and YZ as (13√3)/2 each. I need to find the area of triangle XYZ.First, let me recall some properties of medians in an equilateral triangle. In any triangle, the medians intersect at the centroid, which divides each median into a ratio of 2:1, with the longer part being closer to the vertex. Since the triangle is equilateral, all medians are equal, and they should intersect at 60 degrees, but in this problem, they intersect at right angles. That seems a bit unusual because in a regular equilateral triangle, the medians should intersect at 60 degrees, not 90 degrees. Hmm, maybe this is a special case or perhaps the triangle isn't regular? Wait, no, the problem says it's an equilateral triangle, so all sides are equal, and all angles are 60 degrees. So why are the medians intersecting at right angles? That must mean something specific about the triangle's dimensions or the way the medians are positioned.Wait, maybe I'm misunderstanding the problem. It says medians XT and YZ intersect at right angles. Let me visualize the triangle. Let's label the triangle with vertices X, Y, and Z. So, median XT would be from vertex X to the midpoint T of side YZ. Similarly, median YZ would be from vertex Y to the midpoint of side XZ. But wait, YZ is a side of the triangle, not a median. Maybe that's a typo? Or perhaps it's supposed to be another median. Let me double-check the problem statement."Medians XT and YZ intersect at right angles." Hmm, YZ is a side, not a median. That doesn't make sense because a median should connect a vertex to the midpoint of the opposite side. So, maybe it's supposed to be another median, like from Y to the midpoint of XZ? Let's assume that. So, median XT is from X to midpoint T of YZ, and median YW is from Y to midpoint W of XZ. These two medians intersect at right angles.Given that, I can proceed. So, in an equilateral triangle, all medians are equal, and they should intersect at the centroid, which divides each median into a 2:1 ratio. But in this case, they intersect at right angles, which is not typical for an equilateral triangle. So, perhaps the triangle isn't equilateral? Wait, no, the problem says it's equilateral. So, maybe the given lengths of the medians are different? Wait, no, the problem says XT and YZ are both (13√3)/2. Wait, but YZ is a side, not a median. So, perhaps the problem is misstated? Or maybe I'm misinterpreting.Wait, maybe YZ is a median. Let me think. If YZ is a median, then it must connect vertex Y to the midpoint of side XZ. But in an equilateral triangle, all sides are equal, so the median from Y to XZ would also be (13√3)/2. So, maybe the problem is correct, and both medians XT and YZ (the median from Y) are equal and intersect at right angles.So, let's proceed with that understanding. So, we have two medians, XT and YZ, each of length (13√3)/2, intersecting at right angles at the centroid G. The centroid divides each median into a 2:1 ratio, so from the vertex to the centroid is 2/3 of the median, and from the centroid to the midpoint is 1/3.So, the length from X to G would be (2/3)*(13√3)/2 = (13√3)/3, and similarly, the length from Y to G would be (2/3)*(13√3)/2 = (13√3)/3. Since the medians intersect at right angles, the triangle formed by points X, G, and Y is a right-angled triangle with legs of length (13√3)/3 each.Wait, but in reality, in an equilateral triangle, the medians should intersect at 60 degrees, not 90 degrees. So, if the medians intersect at right angles, that would imply that the triangle isn't equilateral? But the problem says it is. This is confusing.Wait, maybe I'm making a mistake here. Let me think again. In an equilateral triangle, the medians are also the altitudes and the angle bisectors, and they intersect at the centroid, which is also the orthocenter, circumcenter, and incenter. The angle between any two medians in an equilateral triangle is 60 degrees, not 90 degrees. So, if the problem states that the medians intersect at right angles, that would mean the triangle isn't equilateral, which contradicts the problem statement.Hmm, this is a problem. Maybe the problem is misstated, or perhaps I'm misinterpreting the medians. Let me check again. The problem says: "Medians XT and YZ intersect at right angles." So, XT is a median from X to the midpoint T of YZ, and YZ is a side. Wait, that can't be right because YZ is a side, not a median. So, perhaps it's a typo, and they meant another median, like from Y to the midpoint of XZ, which would be a median.Assuming that, let's say median XT and median YW (where W is the midpoint of XZ) intersect at right angles. Then, both medians are of length (13√3)/2, and they intersect at the centroid G, which divides each median into a 2:1 ratio. So, from X to G is (2/3)*(13√3)/2 = (13√3)/3, and similarly, from Y to G is (13√3)/3.Since the medians intersect at right angles, the triangle XGY is a right-angled triangle with legs of length (13√3)/3 each. Therefore, the area of triangle XGY is (1/2)*[(13√3)/3]^2.Calculating that: (1/2)*[(169*3)/9] = (1/2)*(507/9) = (1/2)*(169/3) = 169/6.But wait, the area of the whole triangle XYZ is 6 times the area of triangle XGY because the centroid divides the triangle into 6 smaller triangles of equal area. So, the area of XYZ would be 6*(169/6) = 169.But wait, the side length is 13, so the area of an equilateral triangle is (√3/4)*a^2, which would be (√3/4)*169 ≈ 42.25. Hmm, that's one of the answer choices, B) 42.25.But wait, in my calculation above, I got 169, which is much larger. That can't be right. So, where did I go wrong?Ah, I see. I think I made a mistake in assuming that the area of triangle XGY is 169/6. Let me recalculate that.The area of triangle XGY is (1/2)*base*height. The base and height are both (13√3)/3. So, area = (1/2)*[(13√3)/3]*[(13√3)/3] = (1/2)*(169*3)/9 = (1/2)*(507/9) = (1/2)*(169/3) = 169/6 ≈ 28.1667.Then, the area of the whole triangle XYZ would be 6*(169/6) = 169. But that contradicts the formula for the area of an equilateral triangle, which should be (√3/4)*13^2 ≈ 42.25.So, clearly, something is wrong here. Maybe my assumption that both medians are (13√3)/2 is incorrect? Wait, no, the problem states that both medians XT and YZ are (13√3)/2. But in an equilateral triangle, the length of a median is also the altitude, which is (a√3)/2, so for a=13, the median should be (13√3)/2, which matches the given. So, that part is correct.But then, why is the area coming out to 169 when it should be 42.25? Because 169 is the square of 13, which is the side length. That doesn't make sense.Wait, maybe I'm misunderstanding the role of the centroid. If the medians intersect at right angles, then the triangle formed by the centroid and the two vertices is a right-angled triangle, but the area calculation might be different.Alternatively, perhaps I should use coordinate geometry to solve this problem. Let me try that.Let me place the triangle XYZ in a coordinate system. Let me assume point X is at (0, 0), point Y is at (13, 0), and point Z is at (6.5, (13√3)/2), since it's an equilateral triangle. The midpoint T of YZ would be at ((13 + 6.5)/2, (0 + (13√3)/2)/2) = (9.75, (13√3)/4). Similarly, the midpoint W of XZ would be at ((0 + 6.5)/2, (0 + (13√3)/2)/2) = (3.25, (13√3)/4).Now, the median XT is the line from X(0,0) to T(9.75, (13√3)/4). The median YW is the line from Y(13,0) to W(3.25, (13√3)/4). Let me find the equations of these two medians and check if they intersect at right angles.First, the slope of XT: ( (13√3)/4 - 0 ) / (9.75 - 0) = (13√3)/4 / 9.75. Since 9.75 is 39/4, so slope = (13√3)/4 / (39/4) = (13√3)/39 = √3/3.Similarly, the slope of YW: ( (13√3)/4 - 0 ) / (3.25 - 13) = (13√3)/4 / (-9.75). Again, 9.75 is 39/4, so slope = (13√3)/4 / (-39/4) = -13√3/39 = -√3/3.Now, the product of the slopes of XT and YW is (√3/3)*(-√3/3) = (-3)/9 = -1/3. For two lines to be perpendicular, the product of their slopes should be -1. Here, it's -1/3, which is not -1. So, the medians XT and YW are not perpendicular in this coordinate system, which is consistent with the fact that in an equilateral triangle, medians intersect at 60 degrees, not 90 degrees.But the problem states that medians XT and YZ intersect at right angles. Wait, YZ is a side, not a median. So, perhaps the problem is misstated, or I'm misinterpreting which medians are intersecting at right angles.Alternatively, maybe the triangle isn't equilateral? But the problem says it is. This is confusing.Wait, perhaps the triangle is not in a standard position, and the medians are arranged differently. Maybe it's a different kind of equilateral triangle where the medians intersect at right angles. But in reality, in an equilateral triangle, the medians cannot intersect at right angles because the angles between them are 60 degrees.So, perhaps the problem is incorrect, or maybe I'm misunderstanding the configuration. Alternatively, maybe the triangle isn't equilateral, but the problem says it is. Hmm.Wait, maybe the problem is referring to a different kind of median. Or perhaps it's a typo, and one of the medians is actually an altitude or something else. Alternatively, maybe the triangle is not in a plane, but that's unlikely.Alternatively, perhaps the triangle is equilateral, but the medians are not the standard ones. Wait, no, in an equilateral triangle, all medians are the same.Wait, maybe the problem is referring to the medians from X and Y, which are XT and YZ, but YZ is a side, not a median. So, perhaps it's a typo, and they meant YW, the median from Y to the midpoint of XZ. If that's the case, then as I calculated earlier, the slopes of XT and YW are √3/3 and -√3/3, respectively, and their product is -1/3, not -1, so they are not perpendicular.Therefore, perhaps the problem is incorrect, or maybe I'm missing something.Alternatively, maybe the triangle isn't equilateral, but the problem says it is. So, perhaps I need to approach this differently.Wait, let's think about the properties of medians intersecting at right angles. In any triangle, if two medians are perpendicular, there is a relationship between the sides. Maybe I can use that.In general, for a triangle with sides a, b, c, and medians m_a, m_b, m_c, if two medians are perpendicular, say m_a and m_b, then the following relation holds: a² + b² = 5c². Wait, is that correct? Let me recall.Actually, the formula for the length of a median is m_a = (1/2)√(2b² + 2c² - a²). If two medians are perpendicular, say m_a and m_b, then the relation is a² + b² = 5c². Wait, no, I think it's a different relation.Wait, let me derive it. Let me consider vectors. Let me place the triangle in a coordinate system with the centroid at the origin. Then, the position vectors of the vertices can be represented as vectors a, b, and c, such that a + b + c = 0.The medians can be represented as vectors from the centroid to the midpoints. Wait, maybe this is getting too complicated.Alternatively, let me recall that in a triangle, if two medians are perpendicular, then the sum of the squares of two sides is equal to five times the square of the third side. So, if medians from A and B are perpendicular, then c² = (2a² + 2b²)/5.Wait, let me check that. Suppose in triangle ABC, medians from A and B are perpendicular. Then, the formula is c² = (2a² + 2b²)/5. Is that correct? Let me verify.Yes, I think that's correct. So, if two medians are perpendicular, then the side opposite to the third vertex squared is equal to (2 times the sum of the squares of the other two sides) divided by 5.So, in our case, if medians XT and YZ are perpendicular, then perhaps the side opposite to the third vertex squared is equal to (2 times the sum of the squares of the other two sides) divided by 5.But in our problem, the triangle is equilateral, so all sides are equal, a = b = c = 13. So, plugging into the formula, we get c² = (2a² + 2b²)/5 => 13² = (2*13² + 2*13²)/5 => 169 = (2*169 + 2*169)/5 => 169 = (4*169)/5 => 169 = 676/5 => 169 = 135.2, which is not true. So, this is a contradiction.Therefore, in an equilateral triangle, it's impossible for two medians to be perpendicular because it would require the sides to satisfy a condition that they don't. Therefore, the problem as stated is impossible, unless the triangle isn't equilateral, which contradicts the problem statement.Wait, but the problem says it's an equilateral triangle, so perhaps I'm misunderstanding the configuration. Maybe the medians are not from the vertices to the midpoints of the opposite sides, but something else? No, that's the definition of a median.Alternatively, maybe the problem is referring to the medians as lines, not segments, so they extend beyond the centroid, and perhaps the segments beyond the centroid are considered. But even so, the angle between the medians would still be 60 degrees, not 90 degrees.Alternatively, perhaps the problem is referring to the medians from X and Z, not Y. Wait, but the problem says medians XT and YZ. Hmm.Wait, maybe the problem is referring to the median from X to YZ, which is XT, and the median from Y to XZ, which is YW, and these two medians intersect at right angles. But as I calculated earlier, their slopes are √3/3 and -√3/3, so the angle between them is 60 degrees, not 90 degrees.Therefore, perhaps the problem is incorrect, or maybe I'm misinterpreting it. Alternatively, maybe the triangle isn't equilateral, but the problem says it is. Hmm.Wait, let me try to calculate the area using the given information, regardless of the contradiction. The problem gives the lengths of the medians as (13√3)/2 each, and they intersect at right angles. So, perhaps I can use the formula for the area of a triangle in terms of two medians and the angle between them.The formula is: Area = (4/3) * (1/2) * m1 * m2 * sin(theta), where m1 and m2 are the lengths of the medians, and theta is the angle between them. The factor of 4/3 comes from the fact that the area of the triangle is 4/3 times the area of the triangle formed by the two medians and the side connecting their endpoints.Wait, let me recall the exact formula. The area of the triangle in terms of two medians m1 and m2 and the angle theta between them is (4/3) times the area of the parallelogram formed by the medians, which is (4/3)*(m1*m2*sin(theta))/2 = (2/3)*m1*m2*sin(theta).So, in this case, m1 = m2 = (13√3)/2, and theta = 90 degrees, so sin(theta) = 1.Therefore, Area = (2/3)*[(13√3)/2]*[(13√3)/2]*1 = (2/3)*(169*3)/4 = (2/3)*(507/4) = (1014)/12 = 84.5.But wait, the area of an equilateral triangle with side 13 is (√3/4)*13² ≈ 42.25, which is exactly half of 84.5. So, that suggests that perhaps the formula I used is incorrect, or I misapplied it.Wait, maybe the formula is different. Let me check.I think the correct formula for the area of a triangle in terms of two medians m1 and m2 and the angle theta between them is (4/3) times the area of the triangle formed by the two medians. The area formed by the two medians is (1/2)*m1*m2*sin(theta). Therefore, the area of the original triangle is (4/3)*(1/2)*m1*m2*sin(theta) = (2/3)*m1*m2*sin(theta).So, plugging in the values: (2/3)*[(13√3)/2]*[(13√3)/2]*sin(90°) = (2/3)*(169*3)/4*1 = (2/3)*(507/4) = (1014)/12 = 84.5.But again, this is double the expected area of the equilateral triangle, which is 42.25. So, perhaps the formula is not applicable here, or I'm misapplying it.Alternatively, maybe the formula is for the area of the triangle formed by the medians, not the original triangle. Wait, no, the formula is supposed to give the area of the original triangle.Wait, perhaps I need to consider that in an equilateral triangle, the medians are also the altitudes, and their lengths are (13√3)/2, which is correct. So, the area of the triangle should be (base * height)/2 = (13 * (13√3)/2)/2 = (169√3)/4 ≈ 42.25.But according to the formula using the medians and the angle between them, I get 84.5, which is double that. So, perhaps the formula is incorrect, or I'm misapplying it.Alternatively, maybe the formula is for the area of the triangle formed by the medians, not the original triangle. If that's the case, then the area of the original triangle would be (3/4) times that, but I'm not sure.Wait, let me think differently. If the medians intersect at right angles, then the triangle formed by the medians is a right-angled triangle. But in reality, the medians of an equilateral triangle form a smaller equilateral triangle themselves, not a right-angled one. So, perhaps the formula is not applicable here.Alternatively, maybe I should use vector methods. Let me assign coordinates to the triangle and try to calculate the area.Let me place the centroid at the origin (0,0). Let me denote the position vectors of the vertices as vectors a, b, and c. Since it's an equilateral triangle, the vectors a, b, and c have equal magnitudes and are separated by 120 degrees.The medians from X and Y would be vectors from X to the midpoint of YZ and from Y to the midpoint of XZ. Let me denote the midpoints as T and W, respectively.The midpoint T of YZ has position vector (b + c)/2, and the midpoint W of XZ has position vector (a + c)/2.The vectors representing the medians XT and YW would be T - X = (b + c)/2 - a and W - Y = (a + c)/2 - b.Given that these medians are perpendicular, their dot product is zero.So, [(b + c)/2 - a] · [(a + c)/2 - b] = 0.Let me expand this:[(b + c)/2 - a] · [(a + c)/2 - b] = 0Let me compute each term:First vector: (b + c)/2 - a = (-a + b + c)/2Second vector: (a + c)/2 - b = (a - 2b + c)/2Now, their dot product:[(-a + b + c)/2] · [(a - 2b + c)/2] = 0Multiply both sides by 4 to eliminate denominators:(-a + b + c) · (a - 2b + c) = 0Now, expand the dot product:(-a)·a + (-a)·(-2b) + (-a)·c + b·a + b·(-2b) + b·c + c·a + c·(-2b) + c·c = 0Simplify each term:-|a|² + 2a·b - a·c + a·b - 2|b|² + b·c + a·c - 2b·c + |c|² = 0Combine like terms:-|a|² + (2a·b + a·b) + (-a·c + a·c) + (-2|b|²) + (b·c - 2b·c) + |c|² = 0Simplify:-|a|² + 3a·b - 2|b|² - b·c + |c|² = 0But since the triangle is equilateral, |a| = |b| = |c| = 13, and the angle between any two vectors is 120 degrees, so a·b = |a||b|cos(120°) = 13*13*(-1/2) = -169/2.Similarly, b·c = a·c = -169/2.So, plugging these into the equation:-13² + 3*(-169/2) - 2*13² - (-169/2) + 13² = 0Calculate each term:-169 + 3*(-169/2) - 2*169 - (-169/2) + 169 = 0Simplify:-169 - (507/2) - 338 + (169/2) + 169 = 0Combine like terms:(-169 + 169) + (-507/2 + 169/2) - 338 = 0Simplify:0 + (-338/2) - 338 = 0Which is:-169 - 338 = -507 ≠ 0This is a contradiction, which suggests that our assumption that the medians are perpendicular in an equilateral triangle is impossible. Therefore, the problem as stated is impossible because in an equilateral triangle, the medians cannot intersect at right angles.But the problem says it's an equilateral triangle with medians intersecting at right angles, so perhaps the problem is incorrect, or maybe I'm misinterpreting it.Alternatively, maybe the triangle isn't equilateral, but the problem says it is. So, perhaps the answer is that the area is 42.25, which is the area of an equilateral triangle with side 13, and the condition about the medians intersecting at right angles is a red herring or a mistake.Alternatively, perhaps the problem is referring to a different kind of median or a different configuration. But given the information, I think the area should be 42.25, which is option B.

💡Alright, so I have this problem where I need to find the sum of several inverse function values for a piecewise function. The function f(x) is defined as:[ f(x) = left{ begin{aligned} 2x + 1 & quad text{if } x < 1 x^2 & quad text{if } x ge 1 end{aligned} right. ]And I need to find ( f^{-1}(-3) + f^{-1}(-1) + f^{-1}(1) + f^{-1}(3) + f^{-1}(9) ).Okay, so first, I remember that to find the inverse function, I need to essentially reverse the mapping of f(x). That means for each y-value, I need to find the corresponding x-value such that f(x) = y. Since f(x) is piecewise, I'll have to handle each piece separately.Let me start by understanding the function better. For x less than 1, f(x) is a linear function: 2x + 1. For x greater than or equal to 1, f(x) is a quadratic function: x squared. So, the function has two different behaviors depending on whether x is less than 1 or not.To find the inverse, I need to consider each piece separately and find their inverses. Then, for each y-value I'm given (-3, -1, 1, 3, 9), I'll determine which piece of the function maps to that y-value and use the corresponding inverse.Starting with the first piece: f(x) = 2x + 1 for x < 1. Let's find its inverse.Let me set y = 2x + 1. To find the inverse, I solve for x in terms of y:y = 2x + 1Subtract 1 from both sides:y - 1 = 2xDivide both sides by 2:x = (y - 1)/2So, the inverse function for this piece is g^{-1}(y) = (y - 1)/2.Now, for the second piece: f(x) = x^2 for x >= 1. To find the inverse, I set y = x^2 and solve for x:y = x^2Take the square root of both sides:x = sqrt(y)But since x >= 1, we only consider the positive square root. So, the inverse function for this piece is h^{-1}(y) = sqrt(y).Alright, so now I have the inverse functions for both pieces:- For y-values that come from x < 1, use g^{-1}(y) = (y - 1)/2.- For y-values that come from x >= 1, use h^{-1}(y) = sqrt(y).Now, I need to determine which inverse function to use for each y-value: -3, -1, 1, 3, 9.Let me start with f^{-1}(-3):First, I need to figure out if -3 comes from the linear piece or the quadratic piece.Looking at the linear piece: f(x) = 2x + 1 for x < 1. What is the range of this piece? Since x < 1, let's see what values f(x) can take.If x approaches negative infinity, 2x + 1 approaches negative infinity. When x approaches 1 from the left, 2x + 1 approaches 2*1 + 1 = 3. So, the range of the linear piece is (-∞, 3).The quadratic piece: f(x) = x^2 for x >= 1. The range here is [1, ∞) because x^2 is always positive and increases as x increases.So, for y = -3: it's less than 3 and also negative, so it must come from the linear piece. Therefore, we use g^{-1}(-3) = (-3 - 1)/2 = (-4)/2 = -2.Next, f^{-1}(-1):Similarly, y = -1 is less than 3 and negative, so it comes from the linear piece. So, g^{-1}(-1) = (-1 - 1)/2 = (-2)/2 = -1.Now, f^{-1}(1):Here, y = 1. Let's see which piece this comes from. The linear piece has a range up to 3, so y =1 is within (-∞, 3). But wait, the quadratic piece also has y >=1. So, y=1 is in the overlap of both ranges. Hmm, that might be tricky.Wait, actually, for y=1, we need to check if it's achieved by both pieces.From the linear piece: f(x) = 2x + 1 =1Solving for x: 2x +1 =1 => 2x=0 => x=0. Since x=0 <1, it's valid for the linear piece.From the quadratic piece: f(x) =x^2=1Solving for x: x=1 or x=-1. But since x >=1, only x=1 is valid.So, y=1 is achieved by both pieces at x=0 and x=1. But since the function is defined as f(x)=2x+1 for x<1 and f(x)=x^2 for x>=1, at x=1, it's defined by the quadratic piece. So, for y=1, the inverse would be x=1 because f(1)=1^2=1.Therefore, f^{-1}(1)=1.Wait, but earlier, I thought y=1 is achieved by both pieces, but since x=1 is included in the quadratic piece, we take that as the inverse. So, f^{-1}(1)=1.Moving on to f^{-1}(3):y=3. Let's see which piece this comes from.From the linear piece: f(x)=2x+1=3Solving for x: 2x+1=3 => 2x=2 => x=1. But x=1 is the boundary point. The linear piece is defined for x<1, so x=1 is not included in the linear piece. Therefore, y=3 must come from the quadratic piece.From the quadratic piece: f(x)=x^2=3Solving for x: x=sqrt(3) or x=-sqrt(3). But since x>=1, we take x=sqrt(3).Therefore, f^{-1}(3)=sqrt(3).Finally, f^{-1}(9):y=9. This is clearly greater than 1, so it comes from the quadratic piece.From the quadratic piece: f(x)=x^2=9Solving for x: x=3 or x=-3. Since x>=1, we take x=3.Therefore, f^{-1}(9)=3.Now, let's summarize all the inverse values:- f^{-1}(-3) = -2- f^{-1}(-1) = -1- f^{-1}(1) = 1- f^{-1}(3) = sqrt(3)- f^{-1}(9) = 3Now, adding them all together:-2 + (-1) + 1 + sqrt(3) + 3Let's compute step by step:-2 + (-1) = -3-3 + 1 = -2-2 + sqrt(3) = -2 + sqrt(3)-2 + sqrt(3) + 3 = ( -2 + 3 ) + sqrt(3) = 1 + sqrt(3)So, the sum is 1 + sqrt(3).Wait, let me double-check the addition:-2 (from f^{-1}(-3)) +-1 (from f^{-1}(-1)) +1 (from f^{-1}(1)) +sqrt(3) (from f^{-1}(3)) +3 (from f^{-1}(9)).Adding them up:-2 -1 +1 + sqrt(3) +3Combine the constants:(-2 -1) + (1 +3) = (-3) + 4 = 1Then add sqrt(3):1 + sqrt(3)Yes, that seems correct.So, the final answer is 1 + sqrt(3).

💡To determine the largest four-digit number that is divisible by 5, I start by identifying the largest four-digit number, which is 9999.Next, I recall that a number is divisible by 5 if its last digit is either 0 or 5. Since 9999 ends with a 9, it is not divisible by 5.To find the nearest number that meets the divisibility criteria, I subtract 4 from 9999, resulting in 9995. This number ends with a 5, making it divisible by 5.Therefore, the largest four-digit number that is divisible by 5 is 9995.

💡Okay, so I have this problem where I need to prove that the sum of f(1) + f(2) + ... + f(n-1) is equal to n times [f(n) - 1] for n greater than or equal to 2. Here, f(n) is the harmonic series up to 1/n. Hmm, harmonic series, okay, so f(n) = 1 + 1/2 + 1/3 + ... + 1/n.First, I need to understand what exactly is being asked. I have to show that when I add up all these f(k) terms from k=1 to k=n-1, it equals n multiplied by [f(n) - 1]. That seems a bit abstract, but maybe if I write out the terms, it will become clearer.Let me start by writing out f(1), f(2), f(3), etc., to see if I can spot a pattern or something.f(1) = 1f(2) = 1 + 1/2f(3) = 1 + 1/2 + 1/3f(4) = 1 + 1/2 + 1/3 + 1/4And so on, up to f(n-1) = 1 + 1/2 + 1/3 + ... + 1/(n-1)So, if I sum all these f(k) from k=1 to k=n-1, it's like adding up all these harmonic series up to each term. That sounds complicated, but maybe there's a way to rearrange or group the terms.Wait, maybe I can think of it as a double summation. Let's denote S = f(1) + f(2) + ... + f(n-1). Then,S = sum_{k=1}^{n-1} f(k) = sum_{k=1}^{n-1} [sum_{m=1}^k 1/m]So, S is a double sum where for each k from 1 to n-1, I'm summing 1/m from m=1 to k.I can switch the order of summation. Instead of summing over k first and then m, I can sum over m first and then k. Let me try that.So, S = sum_{m=1}^{n-1} [sum_{k=m}^{n-1} 1/m]Because for each m, k goes from m to n-1. So, for each m, how many times does 1/m appear in the sum? It appears (n-1 - m + 1) times, which is (n - m) times.Therefore, S = sum_{m=1}^{n-1} (n - m)/mSo, S = sum_{m=1}^{n-1} (n/m - 1)Which can be split into two sums:S = n * sum_{m=1}^{n-1} 1/m - sum_{m=1}^{n-1} 1The first sum is n times the harmonic series up to n-1, which is n * f(n-1). The second sum is just adding 1, (n-1) times, so it's (n-1).So, S = n * f(n-1) - (n - 1)But wait, the problem states that S should be equal to n * [f(n) - 1]. Let me see if these two expressions are equal.We have S = n * f(n-1) - (n - 1) and we need to show that this is equal to n * [f(n) - 1].Let me compute n * [f(n) - 1]:n * [f(n) - 1] = n * [ (1 + 1/2 + 1/3 + ... + 1/n) - 1 ] = n * (1/2 + 1/3 + ... + 1/n)So, n * [f(n) - 1] = n * (f(n) - 1) = n * (f(n) - 1)But f(n) = f(n-1) + 1/n, so f(n) - 1 = f(n-1) + 1/n - 1 = f(n-1) - 1 + 1/nTherefore, n * [f(n) - 1] = n * (f(n-1) - 1 + 1/n) = n * f(n-1) - n + 1Wait, so n * [f(n) - 1] = n * f(n-1) - n + 1But earlier, we had S = n * f(n-1) - (n - 1) = n * f(n-1) - n + 1So, yes, S = n * [f(n) - 1]Therefore, the two expressions are equal, which proves the statement.Hmm, that seems to work out. Let me check with a small n to make sure.Take n=2:Left side: f(1) = 1Right side: 2 * [f(2) - 1] = 2 * [ (1 + 1/2) - 1 ] = 2 * (1/2) = 1Okay, that matches.Take n=3:Left side: f(1) + f(2) = 1 + (1 + 1/2) = 2.5Right side: 3 * [f(3) - 1] = 3 * [ (1 + 1/2 + 1/3) - 1 ] = 3 * (1/2 + 1/3) = 3 * (5/6) = 2.5Good, that works too.Maybe n=4:Left side: f(1) + f(2) + f(3) = 1 + (1 + 1/2) + (1 + 1/2 + 1/3) = 1 + 1.5 + 1.833... ≈ 4.333...Right side: 4 * [f(4) - 1] = 4 * [ (1 + 1/2 + 1/3 + 1/4) - 1 ] = 4 * (1/2 + 1/3 + 1/4) = 4 * (6/12 + 4/12 + 3/12) = 4 * (13/12) ≈ 4.333...Perfect, it holds.So, my approach was to express the sum S as a double summation, switch the order of summation, and then simplify it to get the desired expression. Then, verifying with small n's to ensure it works.I think this makes sense. I didn't see any mistakes in the reasoning, and the examples check out. So, I'm confident that this proof is correct.