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JEE Advanced · Mathematics · 27. Definite Integration

Let fx= 192x32+sin4πx for all x R with f12=0.  If m  121fxdx M, then the possible values of m and M are

  1. A m=13, M=24
  2. B m=14, M=12
  3. C m= -11, M=0
  4. D m=1, M=12
Verified Solution

Answer & Solution

Correct Answer

(D) m=1, M=12

Step-by-step Solution

Detailed explanation

Assuming f(x) to more increasing or less increasing
192x33<fx<192x32
64x3<fx<96x3
64x3 dx< fxdx<96x3dx


16x4-1<fx<24x4-32
12116x4-1dx<121fxdx< 12124x4-32dx
2610<121fxdx<7820
Hence option m=1, M= 12 is correct.
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