The document contains solutions to various math problems including functions, quadratic equations, profit maximization, and more.
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Q 22. f(x) =x −15 andg(x) = √x −3 a) f g(x) =x −15 √x −3 [Put x=12] ∴f g(12) =12−15 √12−3 =−3 √9=−1 b) Domain:It is the interval over which the function is defined. The functionf ghas a square root in the denominator which is defined only non-negative numbers.Also the function is undefined when the denominator is zero, therefore x −3>0 =⇒ x >3 The domain of the function f gis {x ∈R| x >3} Q 23. Answer a) The end points of the diameter are: (x1, y1) = (−6,5) and (x2, y2) = (4,3) Diameter of the circle is given by d=p(∆y)2+ (∆x)2=p(y2−y1)2+ (x2−x1)2=p(3−5)2+ (4−(−6))2=√104 b) CenterCof the circle is the midpoint of diameter, which is C=x1+x2 2,y1+y2 2=−6 + 4 2,5 + 3 2= (−1,4) 1
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c) Symmetry abouty- axis is obtained by changing the sign ofx- coordinate. Therefore, the point (1,4) is symmetric toCabouty- axis. Q 24. Answer Line passes through points:(x1, y1) = (1,-7) and (x2, y2) =(3,-9). Slope of the line is given by m=∆y ∆x=y2−y1 x2−x1 =−9−(−7) 3−1=−1 Equation of line passing through point (x1, y1) =(1,-7),having slopem= −1, is y − y1=m(x − x1)=⇒ y+ 7 =−1(x −1) Re-arranging the equation in slope-intercept form: y=−x −6 Q 25. Answer Let the annual income bex$. Tax paid is a fixed amount of 90 $ and 1.4 % ofx.Therefore, Total tax paid = 90 + 1.4 100 x The total tax paid for the year was 1035 $.Implies, 90 +1.4 100 x= 1035 =⇒1.4 100 x= 945 =⇒ x= 67500 Therefore, the annual income is 67,500 $ Q 26. f(x) = 4x2−20 andg(x) =x −3 Answer a) 2
Therefore, the inverse off(x) is f−1(x) = 35−5x Q 29. P(x) =−0.002x2+ 4.3x −1400 Answer a) Profit whenx= 850, P(850) =−0.002(850)2+ 4.3×850−1400 = 810 Therefore, the profit made with 850 doughnuts is 810$ b) Maximum ofP(x) occurs at a stationary point which is obtained by solving d dxP(x) = 0 Derivative ofP(x) is d dx(−0.002x2+ 4.3x −1400) =−0.004x+ 4.3 Therefore, −0.004x+ 4.3 = 0 =⇒ x= 4.3 0.004 = 1075 For maximum to occur at a point the required condition is that the second derivative at that point should be negative, that is d2 dx2P(x)x<0 But, d2 dx2P(x) =−0.004 which is always negative. Therefore,the company should make 1075 doughnuts daily to maximize profit. Q 30. 5
Answer x −3 x+ 2=20 x2−4 [x2−4 = (x+ 2)(x −2)] ∴x −3 x+ 2=20 (x+ 2)(x −2) [Multiply throughout by (x+2)] x −3 x+ 2×(x+ 2) =20 (x+ 2)(x −2) ×(x+ 2) [If, x6= -2, cancel(x+ 2)] ∴x −3 =20 x −2 [Multiply throughout by (x −2)] (x −3)(x −2) =20 x −2×(x −2) [If, x6= 2, cancel(x −2)] x2−5x+ 6 = 20 [Subtract 20 from both sides] x2−5x −14 = 0 =⇒ x2−7x+ 2x −14 = 0 x(x −7) + 2(x −7) = 0 ∴(x −7)(x+ 2) = 0 =⇒ x= 7 or−2 But,x=−2 is already ruled out (as the function is undefined atx=−2), therefore,x= 7 is the solution to the given equation. 6