Question:

What is the least value of k such that the function x$^2$ + kx + 1 is strictly increasing on (1,2)

Updated On: Apr 19, 2024
  • 1
  • -1
  • 2
  • -2
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The Correct Option is D

Solution and Explanation

Let f(x) = x$^2$ + kx + 1 f '(x) = 2x + k f(x) is strictly increasing on (1, 2) if f '(x) > 0 for x $\in $(1, 2) $\Rightarrow $ 2x + k > 0 for x $\in $(1, 2) $\Rightarrow$ k > -2x for x $\in$??(1, 2) Now, 1 < x < 2 $\Rightarrow$ 2 < 2x < 4 $\Rightarrow$ -2 > -2x > -4 $\Rightarrow$ - 4 < -2x < -2 $\Rightarrow \, \, k \ge \, \, -2$ Hence least value of k = -2.
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Concepts Used:

Cartesian Products of Sets

Cartesian products of sets here are explained with the help of an example. Consider A and B to be the 2 sets such that A is a set of 3 colors of tables and B is a set of 3 colors of chairs objects, i.e.,

A = {red, blue, purple}

B = {brown, green, yellow},

Now let us find the number of pairs of colored objects that we can make from a set of tables and chairs in various combinations. They can be grouped as given below:

(red, brown), (red, green), (red, yellow), (blue, brown), (blue, green), (blue, yellow), (purple, brown), (purple, green), (purple, yellow)

There are 9 such pairs in the Cartesian product since 3 elements are there in each of the defined sets A and B. The above-ordered pairs shows the definition for the Cartesian product of sets given. This product is resembled by “A × B”.