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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Area Under Curves

NTA Abhyas 2020 · Mathematics · Area Under Curves

The value of the parameter ‘a' such that the area bounded by y = a^...

The value of the parameter ‘a′‘a' such that the area bounded by y=a2x2+ax+1y = a^{2}x^{2} + ax + 1, coordinate axes and the line x=1x = 1 attains its least value, is equal to

Options

  1. A−14- \dfrac{1}{4}
  2. B−12- \dfrac{1}{2}
  3. C−34- \dfrac{3}{4}
  4. D−1- 1

Correct answer

C. −34- \dfrac{3}{4}

Step-by-step solution

a2x2+ax+1isa^{2}x^{2} + ax + 1 is positive for all real values of xx.

∴Area=∫01(a2x2+ax+1)dx∴Area = \mathop{∫}_{0}^{1}\left(a^{2}x^{2} + ax + 1\right)dx

=a23+a2+1= \dfrac{a^{2}}{3} + \dfrac{a}{2} + 1

=16(2a2+3a+6)= \dfrac{1}{6} \left(2a^{2} + 3a + 6\right)

=16(2(a2+32a+916)+6−1816)= \dfrac{1}{6}\left(2 \left(a^{2} + \dfrac{3}{2}a + \dfrac{9}{16}\right) +6 - \dfrac{18}{16}\right)

=16(2(a+34)2+398),= \dfrac{1}{6} \left(2{ \left(a + \dfrac{3}{4}\right)}^{2} + \dfrac{39}{8}\right) ,which is minimum for a=−34a = - \dfrac{3}{4}
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