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Q.1. The coefficient of x^{99} in the polynomial (x – 1) (x – 2) ....(x – 100) is ................... (1982  2 Marks)
Ans. Sol. Given polynomial :
(x – 1) (x – 2) (x – 3) . . . (x – 100)
Here coeff. of x^{ 99 }= – (1 + 2 + 3 + ... + 100)
= 5050
Q.2. If 2 + i is a root of the equation x^{2} + px +q= 0 , where p and q are real, then (p, q) = ( ................... , ................... ). (1982  2 Marks)
Ans. Sol. As p and q are real;and one root is 2 + i, other should be 2 – i
Then p = – (sum of roots) = – 4,
q = product of roots = 4 + 3 = 7.
Q.3. If the product of the roots of the equation x^{2 }– 3kx + 2 e^{2lnk} – 1 = 0 is 7, then the roots are real fork = ................... (1984  2 Marks)
Ans. Sol. The given equation is x^{2}  3kx + 2e^{2lnk}  1=0
Or x^{ 2 } 3kx + (2k^{2}  1)=0
Here product of roots = 2k^{2} –1
∴ 2k^{ 2}  1 = 7 ⇒ k^{2}=4 ⇒ k= 2, –2
Now for real roots we must have D ≥ 0
⇒ 9k ^{2} – 4(2k^{ 2}  1) ≥ 0 ⇒k^{2} + 4≥0
Which is true for all k. Thus k = 2, – 2
But for k = –2, ln k is not define
∴ Rejecting k = –2, we get k = 2
Q.4. If the quadratic equations x^{2} + ax + b = 0 and x^{2} + bx + a = 0 (a ¹b) have a common root, then the numerical value of a + b is ................... (1986  2 Marks)
Ans. Sol. ∵ x = 1 reduces both the equations to 1 + a + b = 0
∴ 1 is the common root. for a + b = –1
∴ Numerical value of a + b = 1
Q.5. The solution of equation log7 log5 = 0 is ................... (1986  2 Marks)
Ans. Sol.
NOTE THIS STEP
⇒ ⇒ x+5 = 25+ x
⇒ 2 = ⇒ x=4 which satisfies the given equation.
Q.6. If x < 0, y < 0, x + y + and (x + y) , then x = ........ and y = ...... (1990  2 Marks)
Ans. Sol. Given x < 0, y < 0
and
Let x + y = a and .... (1)
∴ We get a + b = and ab = –
Solving these two, we get
⇒ 2a^{2} a  1=0 ⇒ a = 1, – 1/2 ⇒ b = – 1/2, 1
∴ (1) ⇒ x + y = 1 and
or x + y = = 1 But x, y < 0
∴ x + y < 0 ⇒ x + y = = 1
On solving, we get x = – 1/4 and y = –1/4.
Q.7. Let n and k be positive such that . The number of solutions
all integers, satisfying x_{1} + x_{2 }+ ....+ x_{k }= n, is ................... (1996  2 Marks)
Ans. Sol. We have x_{1} + x_{2} + ...............+ x_{k} = n .... (1)
where all integers
Let y_{1} = x_{1}  1, y_{2}= x_{2}  2.................. y_{k }= x_{k } k
so that y_{1}, y_{2 },.........,y_{k} ≥0
Substituting the values of x_{1}, x_{2} , ............,x_{k} in equation .. (1)
We get y_{1} + y_{2} +........ y_{k} = n  (1 + 2 + 3K+k)
.... (2)
Now keeping in mind that number of solutions of the equation
for α, β, γ , .........θ∈ I and each is ≥ 0, is given by coeff of x^{n} in
We find that no. of solutions of equation (2)
Q. 8. The sum of all the real roots of the equation  x – 2 ^{2} +  x – 2  – 2 = 0 is ................... (1997  2 Marks)
Ans. Sol. x– 2^{2}+ x22=0
Case 1. x ≥ 2
⇒ (x  2)^{2} + (x  2)  2= 0
⇒ x^{2}  3x = 0 ⇒ x(x  3)=0
⇒ x = 0, 3 (0 is rejected as x ≥ 2)
⇒ x = 3 ....(1)
Case 2. x < 2
{( x  2)}  (x 2) 2 = 0
⇒ x^{2} + 4  4x x=0 ⇒ (x 1)(x  4)= 0
⇒ x = 1, 4 (4 is rejected as x < 2)
⇒ x = 1 ....(2)
Therefore, the sum of the roots is 3 + 1= 4.
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