    Question

# List IList IIP.The number of polynomials f(x) with non-negative integer coefficients of degree ≤2, satisfying f(0)=0 and 1∫0f(x) dx=1, is 1.8Q.The number of points in the interval [−√13,√13] at which f(x)=sin(x2)+cos(x2) attains its maximum value, is2.2R.2∫−23x21+ex dx equals 3.4S.⎛⎜ ⎜⎝ 1/2∫−1/2cos2xlog(1+x1−x) dx⎞⎟ ⎟⎠⎛⎜⎝ 1/2∫0cos2xlog(1+x1−x) dx⎞⎟⎠ equals 4.0 Which of the following option is correct?

A
(P)(3),(Q)(2)(R)(4),(S)(1)
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B
(P)(2),(Q)(3)(R)(4),(S)(1)
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C
(P)(3),(Q)(2)(R)(1),(S)(4)
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D
(P)(2),(Q)(3)(R)(1),(S)(4)
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Solution

## The correct option is D (P)→(2),(Q)→(3)(R)→(1),(S)→(4)(P) f(0)=0 Let f(x)=ax2+bx 1∫0f(x) dx=a3+b2=1⇒a=0,b=2;a=3,b=0 Two such polynomials is possible. (Q) f(x)=sin(x2)+cos(x2) x2∈[0,13] Maximum value of f(x)=√2 and its attans at sin(x2)=cos(x2) ⇒tan(x2)=1 ⇒x2=π4,9π4<13⇒x=±√π2,±3√π2 Hence, there are four points in x∈[−√13,√13] (R) I=2∫−23x21+ex dx⇒I=2∫−23x21+e−x dx⇒2I=2∫−23x2(1+ex)1+ex dx⇒2I=2∫−23x2 dx⇒I=8 (S) (1/2∫−1/2cos2xlog(1+x1−x) dx)(1/2∫0cos2xlog(1+x1−x) dx) I1=1/2∫−1/2cos2xlog(1+x1−x) dx⇒I1=1/2∫−1/2cos2xlog(1−x1+x) dx⇒2I1=0 Hence, (P)→(2),(Q)→(3)(R)→(1),(S)→(4)  Suggest Corrections  1      Explore more