If loge (x2 – 16) < loge (4x – 11), then(a) – 1 < \(x\) < 5 (b) – InterviewSolution

InterviewSolution

1.

If loge (x2 – 16) < loge (4x – 11), then(a) – 1 < \(x\) < 5 (b) \(x\) < – 1 or \(x\) > 5 (c) 4 < \(x\) < 5 (d) \(x\) < – 4 or \(x\) > 4

Answer»

(a) – 1 < x < 5

log_e (x² – 16) < log_e (4x – 11)

(x² – 16) < 4x – 11

(∵ a > 1, log_a f (x) > log_a(gx) ⇒ f (x) > g(x) > 0)

⇒ x² – 4x – 5 < 0 ⇒ (x – 5) (x + 1) < 0

⇒ – 1 < x < 5

(∵ If a < b, then (x – a) (x – b) < 0 ⇒ a < \(x\) < b)

Discussion

No Comment Found

Related InterviewSolutions

If a, b, c are all positive, a + b + c = 1 and (1 – a) (1 – b) (1 – c) ≥ K(abc), then find the value of K.
The longest side of a triangle is twice the shortest side and the third side is 2cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum length of the shortest side.
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter then the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Prove that for any three positive reals numbers a, b, c, a2 + b2 + c2 ≥ ab + bc + ca.
If a, b, c are the sides of a Δ ABC, then show that \(\frac12<\frac{ab+bc+ca}{a^2+b^2+c^2}\) ≤ 1.
The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum length of the shortest side.
Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is(a) > 2c (b) < √2c (c) > √2c (d) < 2c
If a, b, c be the lengths of the sides of a triangle and (a + b + c)3 ≥ k (a + b – c) (b + c – a) (c + a – b), then k equals(a) 1 (b) 3 (c) 8 (d) 27
Which of the following inequations represents the shaded region ?(a) 2x + y ≤ 4 (b) 2x + y ≥ 4 (c) x + 2y ≤ 4 (d) x + 2y ≥ 4
If x < 5, then A.–x < – 5 B. –x ≤ –5 C. –x > –5 D. –x ≥ –5