Question 1

Let _ x 2 ( (x-2))(rx^2 + (p-2)x - 2p) (x-2)^2 = 5 for some r, p R . If the set of all possible values of q, such that the roots of the equation rx^2 - px + q = 0 lie in (0, 2), be the interval ( , ],

Accepted Answer

Given limit is _ x 2 (x-2) x-2 rx^2 + (p-2)x - 2p x-2 = 5 Since _ x 2 (x-2) x-2 = 1, we have: _ x 2 rx^2 + (p-2)x - 2p x-2 = 5 For the limit to exist, the numerator must be zero at x = 2: r(2)^2 + (p-2)(2) - 2p = 0 4r + 2p - 4 - 2p = 0 4r = 4 r = 1 Substituting r = 1 into the limit expression: _ x 2 x^2 + (p-2)x - 2p x-2 = 5 _ x 2 (x-2)(x+p) x-2 = 5 _ x 2 (x+p) = 5 2 + p = 5 p = 3 The quadratic equation is rx^2 - px + q = 0, which becomes x^2 - 3x + q = 0. For both roots of f(x) = x^2 - 3x + q =

Question 2

The value of _ x 0 ( x^2 ^2 x x^2 - ^2 x ) is:

Accepted Answer

Let L = _ x 0 x^2 ^2 x x^2 - ^2 x Dividing the numerator and the denominator by x^4, we get: L = _ x 0 ( x x )^2 x^2 - ^2 x x^4 Factorizing the denominator: L = _ x 0 ( x x )^2 ( x - x x^3 ) ( x + x x ) Using the standard limits: _ x 0 x x = 1 _ x 0 x - x x^3 = 1 6 (using Taylor series expansion x = x - x^3 3! + ) _ x 0 x + x x = _ x 0 (1 + x x ) = 1 + 1 = 2 Substituting these limits into the expression: L = 1^2 ( 1 6 ) 2 = 1 1 3 = 3 Answer: 3

Question 3

Let f(x) = _ y 0 (1 - (xy)) (xy) y^3 . Then the number of solutions of the equation f(x) = x, x R is :

Accepted Answer

The given function is f(x) = _ y 0 (1 - (xy)) (xy) y^3 . Multiplying and dividing by x^3, we get: f(x) = _ y 0 1 - (xy) (xy)^2 (xy) xy x^3 Using the standard limits _ t 0 1 - t t^2 = 1 2 and _ t 0 t t = 1, we obtain: f(x) = 1 2 1 x^3 = x^3 2 We need to find the number of solutions for the equation f(x) = x, which is x^3 2 = x. Let g(x) = x^3 2 - x. Differentiating with respect to x: g'(x) = 3x^2 2 - x g''(x) = 3x + x For x > 0, g''(x) > 0, which implies that g'(x) is strictly increasing on (0, )

Question 4

The product of all possible values of , for which _ x 0 ( 1 - ( x) (( +1)x) (( +2)x) ^2(( +1)x) ) = 2, is:

Accepted Answer

Using the standard expansions for small x, 1 - ^2 2 and . The given limit can be written as: _ x 0 1 - (1 - ^2 x^2 2 ) (1 - ( +1)^2 x^2 2 ) (1 - ( +2)^2 x^2 2 ) (( +1)x)^2 Neglecting higher powers of x, the numerator simplifies to: 1 - (1 - x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) ) = x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) Substituting this back into the limit: _ x 0 x^2 2 ( ^2 + ( +1)^2 + ( +2)^2 ) ( +1)^2 x^2 = ^2 + ( +1)^2 + ( +2)^2 2( +1)^2 We are given that this limit is equal to 2. Therefore: ^2 + ( +1)

Question 5

If _ x 2 (x^3 - 5x^2 + ax + b) ( x-1 - 1) _e(x-1) = m, then a + b + m is equal to :

Accepted Answer

Let x - 2 = t. As x 2, t 0. The given limit can be written as: _ t 0 ((t+2)^3 - 5(t+2)^2 + a(t+2) + b) ( t+1 - 1) _e(t+1) = m The denominator can be approximated for small t as: _ t 0 t+1 - 1 t _e(t+1) t t^2 = 1 2 1 t^2 = t^2 2 For the limit to be finite, the argument of the sine function in the numerator must have t^2 as its lowest degree term in its expansion. Let P(t) = (t+2)^3 - 5(t+2)^2 + a(t+2) + b P(t) = t^3 + 6t^2 + 12t + 8 - 5(t^2 + 4t + 4) + at + 2a + b P(t) = t^3 + t^2 + (a - 8)t + (2

Question 6

The value of _ x 0 _ e ( (e x) (e^ 2 x ) (e^ 10 x ) ) e^ 2 -e^ 2 x is equal to

Accepted Answer

Numerator: _e( (ex) (e^2x) (e^ 10 x)) = _ k=1 ^ 10 _e( (e^kx)). Using (u) 1 + u^2 2 for small u, we get _e( (e^kx)) e^ 2k x^2 2 . Thus the numerator is x^2 2 _ k=1 ^ 10 e^ 2k = x^2 e^2(e^ 20 -1) 2(e^2-1) . Denominator: Using x 1 - x^2 2 , we have e^ 2 x e^2(1-x^2), so e^2 - e^ 2 x e^2 x^2. Therefore: _ x 0 = x^2 e^2(e^ 20 -1) 2(e^2-1) e^2 x^2 = e^ 20 -1 2(e^2-1) .

Question 7

If _ x 0 e ^ ( a -1) x +2 ~b x+( c -2) e ^ -x x x- _ e (1+x) =2, then a ^ 2 + b ^ 2 + c ^ 2 is equal to :

Accepted Answer

_ x 0 (1+(a-1) x+ (a-1)^2 x^2 2! )+2 (1- b^2 x^2 2! )+(c-2) (1-x+ x^2 2! ) x (1- x^2 2! )- (x- x^2 2 ) =2 _ x 0 (1+2+c-2)+x(a-1-c+2)+x^2 ( (a-1)^2 2 -b^2+ ( c-2 2 ) ) x^2 2 - x^3 2! + =2 For which c+1=0 c=-1 a-c=-1 a=-2 (a-1)^2 2 -b^2+ ( c-2 2 )=1 9 2 -b^2- 3 2 =1 b^2=2 a^2+b^2+c^2=4+2+1=7

Question 8

Let [ ] denote the greatest integer function and f(x)= _ n 1 n ^ 3 _ k =1 ^ n [ k ^ 2 3^ x ]. Then 12 _ j =1 ^ f( j ) is equal to \_\_\_\_.

Accepted Answer

Let [·] denote the greatest integer function. We have f(x) = _ n 1 n^3 _ k=1 ^n [ k^2 3^x ]. For large n: _ k=1 ^n [ k^2 3^x ] _ k=1 ^n k^2 3^x = 1 3^x n(n+1)(2n+1) 6 n^3 3 3^x Therefore: f(x) = _ n 1 n^3 n^3 3 3^x = 1 3^ x+1 Computing the infinite series: _ j=1 ^ f(j) = _ j=1 ^ 1 3^ j+1 = 1 3^2 + 1 3^3 + 1 3^4 + = 1 9 (1 + 1 3 + 1 9 + ) = 1 9 1 1 - 1/3 = 1 9 3 2 = 1 6 Therefore: 12 _ j=1 ^ f(j) = 12 1 6 = 2

Question 9

Let f: R (0, ) be a twice differentiable function such that f(3)=18, f^ (3)=0 and f^ (3)=4. Then _ x 1 ( _ e ( f(2+x) f(3) )^ 18 (x-1)^ 2 ) is equal to :

Accepted Answer

Let T = _ x 1 ( f(x+2) f(3) )^ 18 (x-1)^2 ; 1^ form T = e^ _ x 1 18 (x-1)^2 ( f(x+2) - f(3) f(3) ) T = e^ _ x 1 18 (x-1)^2 ( f(x+2) - f(3) 18 ) T = e^ _ x 1 ( f(x+2) - f(3) (x-1)^2 ) 0 0 form, apply L'pital T = e^ _ x 1 f'(x+2) 2(x-1) ; 0 0 form, apply L'pital T = e^ _ x 1 f''(x+2) 2 = e^ 4 2 = e^2 _e(T) = 2

Limits — Previous Year Questions & Solutions