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Logarithms

Section: Mathematics

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Question : 5 of 11

Marks: +1, -0

The number of integral solutions

x

\log_{\left(x+\frac{7}{2}\right)} \left(\frac{x-7}{2x-3}\right)^2 \ge 0

[11-Apr-2023 shift 1]

5
7
8
6

Solution:

\log_{x+\frac{7}{2}} \left(\frac{x-7}{2x-3}\right)^2 \ge 0

Feasible region:

x+\frac{7}{2} > 0 \Rightarrow x > -\frac{7}{2}

And

x+\frac{7}{2} \neq 1 \Rightarrow x \neq \frac{-5}{2}

Taking intersection:

x \in \left(\frac{-7}{2}, \infty\right) \setminus \left\{-\frac{5}{2}, \frac{3}{2}, 7\right\}

Now

\log_a b \ge 0

if

a > 1

and

b \ge 1

a \in (0,1)

and

b \in (0,1)

C - I ; x+\frac{7}{2} > 1

and

\left(\frac{x-7}{2x-3}\right)^2 \ge 1

x > -\frac{5}{2} ; (2x-3)^2 - (x-7)^2 \le 0

(2x-3+x-7)(2x-3-x+7) \le 0

(3x-10)(x+4) \le 0

x \in \left[-4, \frac{10}{3}\right]

Intersection:

x \in \left(\frac{-5}{2}, \frac{10}{3}\right]

C - \Pi : x+\frac{7}{2} \in (0,1) \text{ and } \left(\frac{x-7}{2x-3}\right)^2 \in (0,1)

0 < x+\frac{7}{2} < 1 ; \left(\frac{x-7}{2x-3}\right)^2 < 1

-\frac{7}{2} < x < \frac{-5}{2} ; (x-7)^2 < (2x-3)^2

x \in (-\infty, -4) \cup \left(\frac{10}{3}, \infty\right)

No common values of

x

.

Hence intersection with feasible region

We get

x \in \left(\frac{-5}{2}, \frac{10}{3}\right] \setminus \left\{\frac{3}{2}\right\}

Integral value of

x

are

\{-2,-1,0,1,2,3\}

No. of integral values

=6

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