Of `f(x)=p x^2+q x+r` is a quadratic expression such that `[a^2a1b^2b1c^2c1][f(0)f(1)f(-1)]=[2a+1 2b+1 2c+1]` for three unequal numbers `a ,b ,c` then `p=3/2` (b) `q=-1/2` (c) `r=1` (d) `p+q+r=2`

Of f(x)=p x^2+q x+r is a quadratic expression such that [a^2 a1 b^2 b1 c^2 c1][f(0) f(1) f(-1)]=[2a+1 2b+1 2c+1] for three unequal numbers a ,b ,c then (a) p=3/2 (b) q=-1/2 (c) r=1 (d) p+q+r=2

If f(x) = | x | + [ x ] , then f( -3/2 ) + f( 3/2 ) is equal to : ( a ) 1 ( b ) -1/2 ( c ) 2 ( d ) 1/2

If x^(2)+px+q=0 is the quadratic equation whose roots are a-2 and b-2 where a and b are the roots of x^(2)-3x+1=0, then p-1,q=5 b.p=1,1=-5 c.p=-1,q=1 d.p=1,q=-1

If the roots of the equation 1/(x+p)+1/(x+q)=1/r are equal in magnitude and opposite in sign, then (A) p+q=r (B)p+q=2r (C) product of roots=- 1/2(p^2+q^2) (D) sum of roots=1

The value of p and q(p!=0,q!=0) for which p ,q are the roots of the equation x^2+p x+q=0 are (a) p=1,q=-2 (b) p=-1,q=-2 (c) p=-1,q=2 (d) p=1,q=2

If f(x)=2x^(2)+bx+c and f(0)=3 and f(2)=1 , then f(1) is equal to a)1 b)2 c)0 d) 1/2

(2) If a=x^(q+r)*y^(p) , b=x^(r+p)*y^(q) , c=x^(p+q)y^(r) , show that a^(q-r)b^(r-p)c^(p-q)=1

Let f(x) = (1 - x)2sin2x + x2 for all x ∈ R, and let g(x) = ∫(2(t - 1)/(t + 1) - ln t)f(t)dt for t ∈ [1, x] for all x ∈ (1, ∞). Consider the statements: P: There exists some x ∈ R, such that f(x) + 2x = 2(1 + x2) Q: There exists some x ∈ R, such that 2f(x) + 1 = 2x(1 + x) (A) both P and Q are true (B) P is true and Q is false (C) P is false and Q is true (D) both P and Q are false.

Consider the statements : P : There exists some x IR such that f(x) + 2x = 2(1+x2) Q : There exists some x IR such that 2f(x) +1 = 2x(1+x) f(x)=(1-x)^(2) sin^(2) x+x^(2)" "AA x in R Then (A) both P and Q are true (B) P is true and Q is false (C) P is false and Q is true (D) both P and Q are false.

If the roots of the equation 1/(x+p)+1/(x+q)=1/r (x ne -p, x ne -q, r ne 0) are equal in magnitude but opposite in sign, then p+q is equal to a)r b)2r c) r^2 d) 1/r