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Question

# Match the following items of column I with column II Column Icolumn II(a)If f (x)=∫g(x)0dt√1+t3 where g(x)=(p)2∫cosx0(1+sint2)dt, then the value of f(π2) is(b)If∫ba(2+x−x2)dx is maximum, then (a + b) is(q)−2equal to(c)If limx→0(sin2xx3+a+bx2)=0, then (3a + b)is equal(r)1to(d)Number of points of maxima for f(x)=x1+xtanx in(s)−1xϵ(0,π2)

A
(a) – (r); (b) – (s); (c) – (p); (d) – (q)
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B
(a) – (s); (b) – (r); (c) – (p); (d) – (r)
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C
(a) – (s); (b) – (q); (c) – (p); (d) – (r)
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D
(a) – (r); (b) – (p); (c) – (q); (d) – (s)
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Solution

## The correct option is B (a) – (s); (b) – (r); (c) – (p); (d) – (r)(A) f′(x)=g′(x)√1+(g(x))3 and g′(x)=[1+sin(cos2x)](−sinx) Hence, f′(x)=[1+sin(cos2x)(−sinx)]√1+(g(x))3 f′(π2)=(1+0)(−1)√1+(g(π2))3=−11+0=−1 as g(π2)=0 ∴f′(π2)=−1 (B) f(x)=2+x−x2 Area under the graph is maximum when a=−1 and b=2 ⇒a+b=1 (C) Given, limx→0(sin2xx3+a+bx2)=0 ⇒limx→0sin2x+ax3+bxx3=0 For limit to exist, 2+b=0⇒b=−2 ∴limx→0sin2x+ax3−2xx3=0 Applying L'Hospital Rule, limx→02cos2x+3ax2−23x2=0 ⇒limx→0−4sin2x+6ax6x=0 ⇒limx→0−8cos2x+6a6=0 ⇒−8+6a=0 ⇒a=43 ⇒3a+b=3×43−2=2 (D) f(x)=x1+xtanx f′(x)=1+xtanx−x(tanx+xsec2x)(1+xtanx)2 =1+xtanx−xtanx−x2secx(1+xtanx)2=1−x2sec2x(1+xtanx)2 To find maxima, f′(x)=0 ⇒x2sec2x=1 ⇒x=±cosx f(x) has maxima at x=cosx Hence, number of points of maxima is 1.

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