Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Inequalities. Click on this to access more such competitive quizzes on Algebra.

Inequalities Quiz 2

mathematics

 The set of values of $x$ which satisfy $4x + 3 < 3x + 8$ and $\dfrac{{x - 2}}{{x + 1}} < 4,$ is

  $4x + 3 < 3x + 8 \\   \Rightarrow 4x - 3x < 8 - 3 \\   \Rightarrow x < 5 \\  \text{And }\dfrac{{x - 2}}{{x + 1}} < 4 \\   \Rightarrow x - 2 < 4x + 4 \\   \Rightarrow  - 6 < 3x \\   \Rightarrow x >  - 2 \\  so,\,x \in ( - 2 - 1) \cup ( - 1,5)  $

 The complete solution of the inequation ${x^2} - 4x < 12\,{\text{ is}}$ 

 &nbsp;${x^2} - 7x <  - 12\, \\   \Rightarrow {x^2} - 7x + 12 < 0 \\   \Rightarrow {x^2} - 3x - 4x + 12 < 0 \\   \Rightarrow x(x - 3) - 4(x - 3) < 0 \\   \Rightarrow (x - 3)(x - 4) < 0 \\  so,\,x \in (3,4) $ 

 Find the value of $x$ for ${\log _4}(x - 1) < {\log _{16}}(x - 1)$ ?

 The inequality is reversed since the base lies between $0\,{\text{and }}1$ .

 First note that $(x - 2)$ must be greater than 0 that $x > 2$ $ \\ $ Now, ${\log _3}(x - 2) < {\log _9}(x - 2)$  $ \\ {\log _{{2^2}}}(x - 1) < {\log _{{2^4}}}(x - 1) \\   \Rightarrow \dfrac{1}{2}{\log _2}(x - 1) < \dfrac{1}{4}{\log _2}(x - 1) \\   \Rightarrow 2{\log _2}(x - 1) < {\log _2}(x - 1) \\   \Rightarrow {(x - 1)^2} > (x - 1) \\  \Rightarrow {(x - 1)^2} - (x - 1) > 0 \\   \Rightarrow (x - 1)(x - 2) > 0 \\ $ Since $x < 1$, the inequality will hold at $x > 2$ that if $x$ lie between $(2,\infty )$ 

 Solve for $x$ when ${\log _{\frac{1}{2}}}(3x) > {\log _{\frac{1}{4}}}(2x + 3)$ ?

 Since the base is $\dfrac{1}{4}$​ , which is less than 1, the given inequality we will reverse its sign when it will take antilog.

&nbsp;${\log _{\frac{1}{2}}}(3x) > {\log _{\frac{1}{4}}}(2x + 3) \\   \Rightarrow {\log _{\frac{1}{2}}}(3x) > {\log _{{{\left( {\dfrac{1}{2}} \right)}^2}}}(2x + 3) \\   \Rightarrow {\log _{\frac{1}{2}}}(3x) > \dfrac{1}{2}{\log _{\frac{1}{2}}}(2x + 3) \\   \Rightarrow 3x < (2x + 3) \\   \Rightarrow 3x - 2x < 3 \\   \Rightarrow x < 3 \\  {\text{and 3}}x > 0 = x > 0 \\  2x + 3 > 0 = x > \dfrac{{ - 3}}{2} \\  So,\,x \in \left( {0, - \dfrac{3}{2}} \right) $ 

 Solve for $x$ in ${3^{{x^2} - 5x}} > 46656$ ?

 Take log on both hand sides. Use a wave-curve method.

&nbsp;${3^{{x^2} - 5x}} > \dfrac{1}{{46656}} \\   \Rightarrow {3^{{x^2} - 5x}} > {3^{ - 6}} \\   \Rightarrow \log {3^{{x^2} - 5x}} > \log {3^{ - 6}} \\   \Rightarrow \left( {{x^2} - 5x} \right)\log 3 >  - 6\log 3 \\  {\text{divide }}\log 3{\text{ from both hands:}} \\   \Rightarrow \left( {{x^2} - 5x} \right)\log 3 >  - 6\log 3 \\   \Rightarrow {x^2} - 5x + 6 > 0 \\   \Rightarrow {x^2} - 2x - 3x + 6 > 0 \\   \Rightarrow x(x - 2) - 3(x - 2) > 0 \\   \Rightarrow (x - 2)(x - 3) > 0 \\  So,\,x \in (2,3) $ 