Lakshya Education MCQs

Question: The value of limxx+cos xx+sin xis
Options:
A.-1
B.0
C.1
D.2
Answer: Option C
: C

limxx+cosxx+sinx[Puttingx=1h;asx,h0]=limh01h+cos(1h)1h+sin(1h)=limh01+hcos1h1+hcos1h=1+01+0⎢ ⎢ ⎢1sin1h1and1cos1h1,whereh0,hcos1h0andhsin1h0⎥ ⎥ ⎥=1

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More Questions on This Topic :

Question 1. The value oflimx01cos3xxsin xcosx
  1.    2/5
  2.    3/5
  3.    3/2
  4.    3/4
Answer: Option C
: C

limx01cos3xxsinxcosx=limx0(1cosx)(1+cosx+cos2x)xsinxcosx=limx02sin2(x2)2sin(x2)cos(x2).x×(1+cosx+cos2x)cosx=limx0sin(x2)2(x2)×1+cosx+cos2xcos(x2)cosx=12×3=32
Question 2. If α is a repeated root of ax2+bx+c=0 then limxαsin(ax2+bx+c)(xα)2 is 
Answer: Option C
: B

limxαsin(ax2+bx+c)(xα)2=limxαsina(xα)(xα)(xα)2
=limxαsina(xα)2a(xα)2×a=a
Question 3. If f(x)={x23,2<x<32x+5,3<x<4, the equation whose roots are limx3f(x) and limx3+f(x) is
  1.    x2−12x+36=0
  2.    x2−26x+66=0
  3.    x2−17x+66=0
  4.    x2−22x+121=0
Answer: Option C
: C

f(x)={x23,2<x<32x+5,3<x<4
limx3f(x)=limx3(x23)=6
and limx3+f(x)=limx3+(2x+5)=11
Hence, the required equation will be
x2 - (sum of roots)x + (Products of roots) = 0
Question 4. If the derivative of the function f(x)=bx2+ax+4;x1ax2+b;x<1, is everywhere continuous, then
  1.    a = 2, b = 3
  2.    a = 3, b = 2
  3.    a = - 2, b = - 3
  4.    a = - 3, b = - 2
Answer: Option A
: A

Wehave,f(x)={ax2+b,x<1bx2+ax+4,x1f(x)={2ax,<12bx+a,x1Since,f(x)isdifferentiableatx=1,thereforeitiscontinuousatx=1andhence,limx1f(x)=limx1+f(x)a+b=ba+4a=2andalso,limx1f(x)=limx1+f(x)2a=2b+a3a=2bb=3(a=2)Hence,a=2,b=3
Question 5. The value of limxπ2[sin1sinx],[x] is the greatest integer function of x, is
  1.    1
  2.    π2
  3.    0
  4.    12
Answer: Option A
: A

limxπ2[sin1sinx]
= limxπ2[x]
= 1

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