Relationship Between A.M. and G.M. — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Relationship Between A.M. and G.M. MCQs with step-by-step solutions (20 questions). Part of Sequences and Series. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Relationship Between A.M. and G.M. · easy · theory
For two positive real numbers, the relation between their A.M. and G.M. is:
A. $A.M. < G.M.$
B. $A.M. \le G.M.$
C. $A.M. = G.M.$ always
D. $A.M. \ge G.M.$ ✓ Correct
Solution: For positive reals, $\dfrac{a+b}{2} \ge \sqrt{ab}$, with equality iff $a = b$.
Q2 — Relationship Between A.M. and G.M. · easy · numerical
The A.M. and G.M. of $4$ and $9$ are respectively:
A. $5$ and $6$
B. $6.5$ and $6$ ✓ Correct
C. $6$ and $6.5$
D. $6.5$ and $6.5$
Solution: A.M. $= \tfrac{13}{2} = 6.5$, G.M. $= \sqrt{36} = 6$; indeed A.M. $>$ G.M.
Q3 — Relationship Between A.M. and G.M. · medium · numerical
The A.M. of two positive numbers is $10$ and their G.M. is $8$. The numbers are:
A. $6$ and $14$
B. $2$ and $18$
C. $5$ and $15$
D. $4$ and $16$ ✓ Correct
Solution: Sum $= 20$, product $= 64$; they are roots of $x^2 - 20x + 64 = 0$, i.e. $4$ and $16$.
Q4 — Relationship Between A.M. and G.M. · medium · numerical
The A.M. of two numbers is $25$ and their G.M. is $20$. The numbers are:
A. $10$ and $40$ ✓ Correct
B. $15$ and $35$
C. $5$ and $45$
D. $20$ and $30$
Solution: Sum $= 50$, product $= 400$; roots of $x^2 - 50x + 400 = 0$ are $10$ and $40$.
Q5 — Relationship Between A.M. and G.M. · medium · numerical
For $x > 0$, the minimum value of $x + \dfrac1x$ is:
A. $1$
B. $2$ ✓ Correct
C. $4$
D. $0$
Solution: By A.M.–G.M., $x + \tfrac1x \ge 2\sqrt{x \cdot \tfrac1x} = 2$, attained at $x = 1$.
Q6 — Relationship Between A.M. and G.M. · medium · numerical
For $x > 0$, the minimum value of $x + \dfrac4x$ is:
A. $2$
B. $4$ ✓ Correct
C. $8$
D. $5$
Solution: $x + \tfrac4x \ge 2\sqrt{4} = 4$, attained at $x = 2$.
Q7 — Relationship Between A.M. and G.M. · medium · numerical
If $a + b = 10$ with $a, b > 0$, the maximum value of $ab$ is:
A. $100$
B. $50$
C. $25$ ✓ Correct
D. $20$
Solution: By A.M.–G.M., $ab \le \left(\tfrac{a+b}{2}\right)^2 = 25$, attained at $a = b = 5$.
Q8 — Relationship Between A.M. and G.M. · medium · numerical
For $x > 0$, the minimum value of $9x + \dfrac1x$ is:
A. $9$
B. $10$
C. $6$ ✓ Correct
D. $3$
Solution: $9x + \tfrac1x \ge 2\sqrt{9} = 6$, attained at $x = \tfrac13$.
Q9 — Relationship Between A.M. and G.M. · hard · numerical
For positive $a, b, c$, the minimum value of $(a+b+c)\left(\dfrac1a + \dfrac1b + \dfrac1c\right)$ is:
A. $3$
B. $6$
C. $1$
D. $9$ ✓ Correct
Solution: By A.M.–G.M. on each factor, the product $\ge 9$, with equality when $a = b = c$.
Q10 — Relationship Between A.M. and G.M. · medium · numerical
For $x > 0$, the minimum value of $x^2 + \dfrac{1}{x^2}$ is:
A. $0$
B. $2$ ✓ Correct
C. $1$
D. $4$
Solution: $x^2 + \tfrac{1}{x^2} \ge 2\sqrt{1} = 2$, attained at $x = 1$.
Q11 — Relationship Between A.M. and G.M. · hard · numerical
The minimum value of $\sin^2\theta + \csc^2\theta$ (where $\sin\theta \ne 0$) is:
A. $2$ ✓ Correct
B. $0$
C. $\tfrac12$
D. $1$
Solution: With $t = \sin^2\theta$, $t + \tfrac1t \ge 2$; the minimum $2$ occurs at $\sin^2\theta = 1$.
Q12 — Relationship Between A.M. and G.M. · medium · theory
By applying A.M.–G.M. to $\dfrac{a}{b}$ and $\dfrac{b}{a}$ (positive $a, b$), we get $\dfrac{a}{b} + \dfrac{b}{a} \ge$:
A. $2$ ✓ Correct
B. $4$
C. $0$
D. $1$
Solution: $\tfrac{a}{b} + \tfrac{b}{a} \ge 2\sqrt{\tfrac{a}{b} \cdot \tfrac{b}{a}} = 2$.
Q13 — Relationship Between A.M. and G.M. · hard · numerical
If $x, y, z > 0$ and $x + y + z = 12$, the maximum value of $xyz$ is:
A. $64$ ✓ Correct
B. $81$
C. $27$
D. $48$
Solution: By A.M.–G.M., $xyz \le \left(\tfrac{x+y+z}{3}\right)^3 = 4^3 = 64$, attained at $x = y = z = 4$.
Q14 — Relationship Between A.M. and G.M. · medium · numerical
If two positive numbers have product $36$, the minimum value of their sum is:
A. $9$
B. $12$ ✓ Correct
C. $6$
D. $18$
Solution: By A.M.–G.M., sum $\ge 2\sqrt{36} = 12$, attained when both equal $6$.
Q15 — Relationship Between A.M. and G.M. · medium · numerical
If $a + b + c = 3$ with $a, b, c > 0$, the maximum value of $abc$ is:
A. $1$ ✓ Correct
B. $27$
C. $3$
D. $\tfrac13$
Solution: By A.M.–G.M., $abc \le \left(\tfrac{a+b+c}{3}\right)^3 = 1$, attained at $a = b = c = 1$.
Q16 — Relationship Between A.M. and G.M. · medium · numerical
For $x > 0$, the minimum value of $4x + \dfrac9x$ is:
A. $12$ ✓ Correct
B. $36$
C. $13$
D. $6$
Solution: $4x + \tfrac9x \ge 2\sqrt{36} = 12$, attained at $x = \tfrac32$.
Q17 — Relationship Between A.M. and G.M. · hard · numerical
For $x > 0$, the maximum value of $\dfrac{x}{1 + x^2}$ is:
A. $\tfrac12$ ✓ Correct
B. $\tfrac14$
C. $2$
D. $1$
Solution: Write it as $\dfrac{1}{x + \tfrac1x}$; since $x + \tfrac1x \ge 2$, the maximum is $\tfrac12$ at $x = 1$.
Q18 — Relationship Between A.M. and G.M. · medium · theory
If two positive numbers have A.M. $= A$ and G.M. $= G$, the numbers are:
A. $G \pm \sqrt{A}$
B. $A \pm \sqrt{A^2 - G^2}$ ✓ Correct
C. $\tfrac{A}{G}$ and $AG$
D. $A \pm G$
Solution: The numbers are roots of $x^2 - 2Ax + G^2 = 0$, giving $A \pm \sqrt{A^2 - G^2}$.
Q19 — Relationship Between A.M. and G.M. · medium · numerical
The A.M. of two numbers is $5$ and their G.M. is $4$. The numbers are:
A. $2$ and $8$ ✓ Correct
B. $1$ and $9$
C. $3$ and $7$
D. $4$ and $6$
Solution: Sum $= 10$, product $= 16$; roots of $x^2 - 10x + 16 = 0$ are $2$ and $8$.
Q20 — Relationship Between A.M. and G.M. · hard · numerical
For positive reals $a, b$, if $A.M. - G.M. = \dfrac{(\sqrt{a} - \sqrt{b})^2}{2}$, then $A.M. = G.M.$ exactly when:
A. $a = b$ ✓ Correct
B. $a = 2b$
C. $a + b = 0$
D. $ab = 1$
Solution: The difference $\tfrac{(\sqrt a - \sqrt b)^2}{2}$ is zero iff $\sqrt a = \sqrt b$, i.e. $a = b$.