The least value of the expression x^2 + 4y^2 + 3z^2 – 2x – 12y – 6z + 14 is

Q : The least value of the expression x2 + 4y2 + 3z2 – 2x – 12y – 6z + 14 is

(A) 0

(B) 1

(C) no least value

(D) none of these

Ans: (B)

Sol. Let f(x, y, z) = x2 + 4y2 + 3z2 – 2x –12y – 6z + 14

= (x – 1)2 + (2y – 3)2 + 3(z – 1)2 + 1

For least value of f(x, y, z)

x – 1 = 0; 2y – 3 = 0 and z – 1 = 0

x = 1; y = 3/2; z = 1

Hence least value of f(x, y, z) is f( 1, 3/2, 1) = 1

If tanA, tanB, tanC are the solutions of the equation x^3 – k^2x^2 – px + 2k + 1 = 0

Q : If tanA, tanB, tanC are the solutions of the equation x3 – k2 x2 – px + 2k + 1 = 0, then  ΔABC is

(A) an isosceles triangle

(B) an equilateral triangle

(C) a right angled triangle

(D) none of these

Ans: (D)

Sol. tanA + tanB + tanC = k2 and tanA tanB tanC = -2k – 1

In a  ΔABC

tanA + tanB + tanC = tanA tanB tanC

⇒ k2 = -2k-1 ⇒ k = -1

⇒ tanA + tanB + tanC = tanA tanB tanC = 1

which is not possible.

The values of α and β such that equation x^2 + 2x + 2 + e^α – sinβ = 0 have a real solution is

Q : The values of α and β such that equation x2 + 2x + 2 + eα – sinβ = 0 have a real solution is

(A) $\large \alpha , \beta \in R $

(B) $\large \alpha \in (0 , 1) \; , \beta \in (\pi/2 , 2 \pi) $

(C) $\large \alpha \in (0 , \infty) \; and \beta \in (\pi/2 , \pi) $

(D) None of these

Ans: (D)

Solution: x2 + 2x + 2 + eα – sinβ = 0 has real roots if D ≥ 0

⇒  1 – 2 – eα + sinβ  ≥ 0

⇒ sinβ  ≥  1 + eα

Hence no real values of α and β are possible.

The quadratic equation whose roots are A.M. and H.M. between the roots of the equation…

Q : The quadratic equation whose roots are A.M. and H.M. between the roots of the equation ax2 + bx + c = 0 is

(A) abx2 + (b2 + ac)x + bc = 0

(B) 2abx2 + (b2 + 4ac)x + 2bc = 0

(C) 2abx2 + (b2 + ac)x + bc = 0

(D) none of these

Ans: (B)

Solution: Let (α , β) be the roots of the given equation, then

$\large \alpha + \beta = -\frac{b}{a} \; , \alpha . \beta = \frac{c}{a} $

Required equation is

$\large x^2 – (\frac{\alpha + \beta}{2} + \frac{2 \alpha . \beta}{\alpha + \beta})x + (\frac{\alpha + \beta}{2}) (\frac{2 \alpha . \beta}{\alpha + \beta})) = 0 $

2abx2 + (b2 + 4ac)x + 2bc = 0