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Q39

REVIEWMCQpage 17holdmath reviewREVIEW_MATH_COMPLEXP2UNREVIEWED

type=MCQ · page=17 · marks=2 · content_conf=0.7999999999999999 · math_conf=0.55

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Source / evidence

file: GATE CH 2020 (with Solutions).pdf

pages: 17–17

assets/questions/Q039/residual_577eeb95f302.png

Extracted39 / 67

The maximum value of the function

[VISUAL / REVIEW — NO INVENTED LATEX] Source math clip

, 3.5) is

Options

A.
16
B.
48
C.
0
D.
8

Answer

Solution

.

f'(x) = −5x^2+20x-15

Putting, f'( x )=0 to find critical points.

−5x^2+20x-15 = 0

x^2-4x+3 = 0

x^2-3x-x+3 = 0

(x-3)(x-1) = 0

∴ x =3 and x =1

f"(x) = −10x+20

f"(3) = −10 /text{and} f " (1) = 10

∴ maxima lies at 3 as f"(3)<0 Now checking the value of f ( x) at 3 and boundary value,

f(0.5) = (/frac{−5()}{30.5})^3+10(0.5)^2-15(0.5)+16f(0.5) = 10.792

Similarly, f(3)=16

f(3.5) = 14.542

Hence, the correct option is (A).

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Math review evidence

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Source clip

page=17 bbox=[36,466.94,169.33,480.27]

asset=assets/questions/Q039/residual_577eeb95f302.png

Raw text

f(x)=−5/3 x³ + 10 x² − 15 x+16 in the interval (0.5

Current normalized

f(x) = - 5 / 3 x^3 + 10 x^2 - 15 x + 16 in the interval (0.5

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Reason for review: Complex layout math with partial LaTeX still marked REVIEW_REQUIRED.

Provenance status

Source PDF: GATE CH 2020 (with Solutions).pdf

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