(c) (i) Recall the definition of Riemann integration:
∫baf(x)dx=limn→∞n∑k=0b−anf(a+k(b−a)n)
As b−an>0∧f(a+k(b−a)n)>0, every term of the Riemann sum is greater than 0.
Hence, the integral is greater than 0.
x2+tx−10=(x+a)(x+b)x2+tx−10=x2+(a+b)x+ab∴ab=−10For a,b∈Z,(a,b)=(10,−1),(5,−2),(2,−5),(1,−10),(−1,10),(−2,5),(−5,2),(−10,1)∴t=9,3,−3,−9Product of all t=92⋅32=729
tan(cos−1(45)+sin−1(1))=tan(π2+cos−1(45))=1tan(cos−1(45))=1√52−424=43
cos−1(1)=0tan(sin−1(45)+cos−1(1))=tan(sin−1(45))=4√52−42=43
Presumably, both f(x) and g(x) has inverse functions.
g(1+g−1(a))=f(1+f−1(1+f(−1+f−1(b))))f−1(g(1+g−1(a)))=1+f−1(1+f(−1+f−1(b)))f(f−1(g(1+g−1(a)))−1)=1+f(−1+f−1(b))f(−1+f−1(b))=f(f−1(g(1+g−1(a)))−1)−1
But I can't solve for f(1 + f^{-1}(b))
x4+2x3−4x2−8x+ax2+4ax+4a=x3(x+2)−4x(x+2)+a(x+2)2=(x3−4x)(x+2)+a(x+2)2=x(x+2)2(x−2)+a(x+2)2=(x+2)2(x2−2x+a)
If the odd function is continuous, it must pass through (0,0) as the property of odd function is that the graph of the function is rotational symmetrical about the origin.
Substitute (2,-2) into g(x) = f(x+3) - 5.
−2=f(5)−5f(5)=3
From this, we get f(−5)=−3.
When x = -8,
g(−8)=f(−5)−5g(−8)=−3−5=−8.
So it must also pass through (-8,-8).
Answer: (0,0), (-8,-8)
I have been on this forum for 3 years. There are always these kind of low-quality posts. Like you did, I have also observed an increasing trend of low-quality posts. Whenever I see them, I just ignore them so it is not much of a problem.
