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Question

# Let α=∫π40cosxsin3x+cos3xdx; β=limn→∞(∏nr=1(n3+r3)n3n)1n and μ=∫10dx1+x3 Column - I Column - II (P) If μ−kα=0, then k is (1) 0 (Q) If lnβ=lna−3+bμ, then the value of (a+b) is (2) 1 (R) If μ=13(lna+π√b), then the value of 2ba is (3) 2 (S) The value of [ln(4β)] is (where [.] denotes the greatest integer function) (4) 3 (5) 4 (6) 5 Then the correct option is

A
P(4); Q(6); R(2); S(1)
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B
P(2); Q(6); R(4); S(2)
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C
P(2); Q(5); R(4); S(2)
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D
P(2); Q(5); R(2); S(3)
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Solution

## The correct option is B P(2); Q(6); R(4); S(2)(P) α∫π40cosxsin3x+cos3xdx=∫π40sec2x1+tan3xdx let tanx=t, sec2xdx=dt α=∫10dt1+t3=μ⇒μ−α=0 (Q) lnβ=limn→∞1n∑nr=1ln(1+(rn)3) ⇒lnβ=∫10ln(1+x3)dx =xln(1+x3)∣∣10−∫103x21−x3xdx ⇒lnβ=ln2−3∫10x31+x3dx ⇒lnβ=ln2−3[1−∫1011+x3dx] ⇒lnβ=ln2−3+3μ (R) μ=∫10dx(1+x)(x2−x+1) =∫10[13.1x+1−16(2x−1)−3x2−x+1]dx ⇒μ=[13ln(x+1)−16ln(x2−x+1)+1√3tan−1(2x−1√3)]10 ⇒μ=13(ln2+π√3) (S) ∵ lnβ=ln2−3+3μ =ln2−3+(ln2+π√3) ln(4β)=3−π√3 [ln(4β)]=1

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