Application of DerivativesmediumFree

Range of an Integral Function on [3,6] and the Value of 3a + b/9 | JEE

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Question
Let f(x)f(x) be a function defined by
f(x)=∫3xt(t2−7t+12)dt,3≤x≤6.f(x) = \int_{3}^{x} t\left(t^{2}-7t+12\right)dt , \qquad 3 \le x \le 6 .
 If the range of f(x)f(x) is [a, b][a,\ b], then the value of \left(3a + \dfrac {9}\right) is
Solution
Answer: 1
Step 1: Differentiate using the fundamental theorem of calculus - the integrand evaluated at the upper limit.
f′(x)=x(x2−7x+12)=x(x−3)(x−4).f'(x) = x\left(x^{2}-7x+12\right) = x(x-3)(x-4) .
 Step 2: Find the sign of f′f' on [3,6][3,6]. Here x>0x > 0 and x−3≥0x-3 \ge 0, so the sign is that of x−4x-4:
f′(x)<0 on (3,4),f′(x)>0 on (4,6).f'(x) < 0 \ \text{on } (3,4), \qquad f'(x) > 0 \ \text{on } (4,6) .
 Step 3: So ff falls then rises: the minimum is at x=4x = 4, and the maximum is at one of the endpoints x=3x = 3 or x=6x = 6. Step 4: Write down an antiderivative. Expanding the integrand as t3−7t2+12tt^{3}-7t^{2}+12t,
F(t)=t44−7t33+6t2.F(t) = \frac{t^{4}}{4} - \frac{7t^{3}}{3} + 6t^{2} .
 Step 5: Evaluate FF at the three points needed.
F(3)=814−63+54=814−9=454,F(3) = \frac{81}{4} - 63 + 54 = \frac{81}{4} - 9 = \frac{45}{4} ,
F(4)=64−4483+96=160−4483=480−4483=323,F(4) = 64 - \frac{448}{3} + 96 = 160 - \frac{448}{3} = \frac{480-448}{3} = \frac{32}{3} ,
F(6)=324−504+216=36.F(6) = 324 - 504 + 216 = 36 .
 Step 6: Convert to values of ff, remembering f(x)=F(x)−F(3)f(x) = F(x) - F(3).
f(3)=0,f(3) = 0 ,
f(4)=323−454=128−13512=−712,f(4) = \frac{32}{3} - \frac{45}{4} = \frac{128-135}{12} = -\frac{7}{12} ,
f(6)=36−454=144−454=994.f(6) = 36 - \frac{45}{4} = \frac{144-45}{4} = \frac{99}{4} .
 Step 7: Identify the range. The least value is f(4)f(4) and the greatest is f(6)f(6):
a=−712,b=994.a = -\frac{7}{12}, \qquad b = \frac{99}{4} .
 Step 8: Compute the required combination.
3a+b9=3(−712)+994×9=−74+114=44=1.3a + \frac{b}{9} = 3\left(-\frac{7}{12}\right) + \frac{99}{4 \times 9} = -\frac74 + \frac{11}{4} = \frac{4}{4} = 1 .
 Answer: 11.
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