For all integers a and b- a-b2-a2-b2 and a-b2-a2-b2-

Proving this by taking examplesLet us assume (a+b)^2≥a^2+b^2, for all integers a and bNow if a= 4 and b=6[( 4)+6]^2≥( 4)^2+(6)^2[2]^2≥ 16+364≥ 52Which is not tr ...

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Byju's Answer

Standard XIII

Mathematics

nCr Definitions and Properties

For all integ...

Question

For all integers $a$ and $b$ , ${(a + b)}^{2} \geq a^{2} + b^{2}$ and ${(a - b)}^{2} \leq a^{2} + b^{2}$ .

True

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False

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Solution

The correct option is B False
Proving this by taking examples
Let us assume ${(a + b)}^{2} \geq a^{2} + b^{2}$ , for all integers $a$ and $b$
Now if $a = - 4 a n d b = 6$
${[(- 4) + 6]}^{2} \geq {(- 4)}^{2} + {(6)}^{2}$
${[2]}^{2} \geq 16 + 36$
$4 \geq 52$
Which is not true
Hence, the above statement is false.

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Similar questions

Q. If AB = A and BA = B, where A and B are square matrices, then
(a) B² = B and A² = A
(b) B² ≠ B and A² = A
(c) A² ≠ A, B² = B
(d) A² ≠ A, B² ≠ B

Q. If θ =

\frac{a}{b}

, then

\frac{(a sinθ - b cosθ)}{(a \sin θ + b cosθ)}

= ?
(a)

\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}

(b)

\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}

(c)

\frac{a^{2}}{(a^{2} + b^{2})}

(d)

\frac{b^{2}}{(a^{2} + b^{2})}

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

Q. Integers

a, b, c

satisfy

a + b - c = 1

and

a^{2} + b^{2} - c^{2} = - 1

. What is the sum of all possible values of

a^{2} + b^{2} + c^{2}

\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}

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For all integers a and b, (a+b)2≥a2+b2 and (a-b)2≤a2+b2.

The correct option is B FalseProving this by taking examplesLet us assume (a+b)2≥a2+b2, for all integers a and bNow if a=-4andb=6-4+62≥-42+6222≥16+364≥52Which is not trueHence, the above statement is false.

For all integers $a$ and $b$ , ${(a + b)}^{2} \geq a^{2} + b^{2}$ and ${(a - b)}^{2} \leq a^{2} + b^{2}$ .