Question

List IList II (A)If limn→∞(n2+1n+1−an)−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,x≤1x2+bx+c,x>1 and f′(x)exists for all x∈R, then the value of c is(R)2(D)If f(x)=x∫0tsin1t 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?(A)→(Q)(B)→(U)(C)→(S)(D)→(T)

Solution

The correct option is B (B)→(U)(A) limn→∞(n2+1n+1−an)−b=0⇒limn→∞((1−a)n2+1−ann+1)−b=0 For the limit to exist, 1−a=0⇒a=1 ⇒limn→∞(1−nn+1)−b=0⇒−1−b=0⇒b=−1 (A)→(S) (B) x2y+y3=2 Differentiating w.r.t. x, 2xy+x2y′+3y2y′=0 If x=1, then y3+y−2=0⇒y=1 Putting x=1,y=1, we get 2+y′+3y′=0⇒y′=−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 2−1−1+y′′+32+3y′′=0⇒y′′=−38∴m=3 (B)→(U) (C) f(x)={x,x≤1x2+bx+c,x>1 f′(x)={1,x≤12x+b,x>1 The function is differentiable, so it is continuous at x=1 1=1+b+c⇒b+c=0     ⋯(2) Also, f′(1+)=f′(1−) ⇒2+b=1⇒b=−1∴c=1     [From (2)] (C)→(Q) (D) f(x)=x∫0tsin1t 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)

Suggest corrections