If 3(a + 2c) = 4(b + 3d) ≠ 0 then the equation ax^3 + bx^2 + cx + d = 0 will have

Q: If 3(a + 2c) = 4(b + 3d) ≠ 0 then the equation ax3 + bx2 + cx + d = 0 will have

(A) no real solution

(B) at least one real root in (–1, 0)

(C) at least one real root in (0, 1)

(D) none of these

Sol: Let $\large f(x) = \frac{a x^4}{4} + \frac{b x^3}{3}+ \frac{c x^2}{2}+ dx $ ; which is continuous and differentiable.

f(0) = 0 ,

$\large f(-1) = \frac{a}{4} – \frac{b}{3} + \frac{c}{2} – d$

$\large = \frac{1}{4}(a+2c)-\frac{1}{3}(b+3d) = 0 $

so, according to Rolle’s theorem, there exist atleast one root of f'(x)= 0 in (–1, 0).

f : [1 , ∞) → R : f(x) is a monotonic and differentiable function and f(1) = 1, then number of solutions…

Q:  f : [1 , ∞) → R : f(x) is a monotonic and differentiable function and f(1) = 1, then number of solutions of the equation $\large f(f(x)) = \frac{1}{x^2 – 2x + 2} $ is /are

(A) 2

(B) 1

(C) infinite

(D) zero

Ans: (B)

Sol: g (x) = f (f (x))

g'(x) = f'(f (x)) f’ (x) > 0

g (1) = f (f (1)) = f (1) = 1

g (x) ≥ 1

$\large \frac{1}{(x-1)^2 + 1} \le 1 $

Hence (B) is correct

If $ f(x) = \int_{x^2}^{x^3} \frac{dt}{ln t} , x > 0 , \ne 1$ then

Q: If $\large f(x) = \int_{x^2}^{x^3} \frac{dt}{ln t} , x > 0 , \ne 1$ then

(A) f(x) is an increasing function

(B) f(x) has a minima at x = 1

(C) f(x) is a decreasing function

(D) f(x) has a maxima at x = 1

Ans: (A)

Sol: $\large f(x) = \int_{x^2}^{x^3} \frac{dt}{ln t}$

For increasing or decreasing function,

$\large f'(x) = \frac{1}{ln x^3} . 3x^2 – \frac{1}{ln x^2} . 2x $ (using Leibnitz formula)

$\large f'(x) = \frac{1}{3 ln x} . 3x^2 – \frac{1}{2 ln x} . 2x $

$\large = \frac{1}{ln x}(x^2 – x) $

Since f'(x) > 0 for x > 0 , x ≠ 1 hence f(x) is increasing function.

Hence (A) is correct.

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.