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List IList II (A)If limn(n2+1n+1an)b=0,thenthe value of b is(P)0(B)If x2y+y3=2 and the value of d2ydx2 at x=1 is m8, then the value of m is(Q)1(C)If f(x)={x,x1x2+bx+c,x>1 and f(x)exists for all xR, then the value of c is(R)2(D)If f(x)=x0tsin1t dt, then the number ofpoint(s) of discontinuity of f(x) in (0,π) is(S)1(T)4(U)3

Which of the following is the only CORRECT combination?
  1. (A)(Q)
  2. (B)(U)
  3. (C)(S)
  4. (D)(T)


Solution

The correct option is B (B)(U)
(A)
limn(n2+1n+1an)b=0limn((1a)n2+1ann+1)b=0
For the limit to exist,
1a=0a=1
limn(1nn+1)b=01b=0b=1
(A)(S)

(B)
x2y+y3=2
Differentiating w.r.t. x,
2xy+x2y+3y2y=0
If x=1, then y3+y2=0y=1
Putting x=1,y=1, we get
2+y+3y=0y=12     (1)
Again differentiating w.r.t. x,
2y+2xy+2xy+x2y′′+6y(y)2+3y2y′′=0
Putting x=1,y=1 and using equation (1), we get
211+y′′+32+3y′′=0y′′=38m=3
(B)(U)

(C)
f(x)={x,x1x2+bx+c,x>1

f(x)={1,x12x+b,x>1
The function is differentiable, so it is continuous at x=1
1=1+b+cb+c=0     (2)
Also, f(1+)=f(1)
2+b=1b=1c=1     [From (2)]
(C)(Q)

(D)
f(x)=x0tsin1t dt
In the interval (0,π),xsin1x is continuous.
f(x)=xsin1x
So, f(x) is finite and exists for all x(0,π), therefore f(x) is continuous at all points.
So, the number of points of discontinuity of f(x) in (0,π) is 0.
(D)(P)

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