What is the maximum value of the function a sin-b cos- using Cauchy-Schwarz inequality-

The correct option is A √(a^2+b^2) Cauchy Shwartz inequality is: a_1b_1+a_2b_2+........a_nb_n ≤√(a_1^2+a_2^2+....a_n^2)√(b_1^2+b_2^2+....b_n^2) simila ...

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Standard XII

Mathematics

Local Maxima

What is the m...

Question

What is the maximum value of the function $a sin θ + b cos θ$ using Cauchy-Schwarz inequality.

\sqrt{a^{2} + b^{2}}

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a^{2} + b^{2}

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\sqrt{a + b}

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None of these

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Solution

The correct option is A $\sqrt{a^{2} + b^{2}}$
Cauchy Shwartz inequality is:
$a_{1} b_{1} + a_{2} b_{2} + . . . . . . . . a_{n} b_{n} \leq \sqrt{a_{1}^{2} + a_{2}^{2} + . . . . a_{n}^{2}} \sqrt{b_{1}^{2} + b_{2}^{2} + . . . . b_{n}^{2}}$
similarly , $a s i n θ + b c o s θ \leq \sqrt{a^{2} + b^{2}} \sqrt{s i n^{2} θ + c o s^{2} θ} = \sqrt{a^{2} + b^{2}}$
Therefore Answer is $A$

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Similar questions

If $t a n θ = \frac{a}{b}, t h e n t h e v a l u e o f E = \frac{a s i n θ - b c o s θ}{a s i n θ + b c o s θ} i s -$

If $t a n θ = \frac{a}{b}$ , show that $\frac{a s i n θ - b c o s θ}{a s i n θ + b c o s θ} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

Q. If

t a n θ = \frac{a}{b}

, Prove that

\frac{a sin θ - b cos θ}{a sin θ + b cos θ} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}

Q. If a cos θ − b sin θ = c, then a sin θ + b cos θ =

(a)

\pm \sqrt{a^{2} + b^{2} + c^{2}}

(b)

\pm \sqrt{a^{2} + b^{2} - c^{2}}

(c)

\pm \sqrt{c^{2} - a^{2} - b^{2}}

(d) None of these

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What is the maximum value of the function asinθ+bcosθ using Cauchy-Schwarz inequality.

The correct option is A √a2+b2Cauchy Shwartz inequality is:a1b1+a2b2+........anbn≤√a21+a22+....a2n√b21+b22+....b2nsimilarly , asinθ+bcosθ≤√a2+b2√sin2θ+cos2θ=√a2+b2Therefore Answer is A

What is the maximum value of the function $a sin θ + b cos θ$ using Cauchy-Schwarz inequality.