For real a- b and x-a2-b2- a sin x-b cos x -a2-b2

The correct option is A TruePut a=r cos α, b=r sin α where r=√((a^2 + b^2)) ∴ a sin x + b cos x =r sin (x+α) since 1 ≤ sin θ≤ 1 therefore r ≤ sin(x + α) ≤ r e ...

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Standard XI

Mathematics

Inequality

For real a, b...

Question

For real a, b and x $- \sqrt{(a^{2} + b^{2})} \leq a s i n x + b c o s x \leq \sqrt{(a^{2} + b^{2})}$

True

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False

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Solution

The correct option is A True
Put $a = r c o s α, b = r s i n α$ where $r = \sqrt{(a^{2} + b^{2})}$
$∴ a s i n x + b c o s x = r s i n (x + α)$
since $- 1 \leq s i n θ \leq 1$ therefore $- r \leq s i n (x + α) \leq r$ etc

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Similar questions

Q. For real a, b and x

- \sqrt{(a^{2} + b^{2})} \leq a s i n x + b c o s x \leq \sqrt{(a^{2} + b^{2})}

a sin x = b cos x = \frac{2 c tan x}{1 - {tan}^{2} x}

and

(a^{2} - b^{2}) = k c^{2} (a^{2} + b^{2})

then

k =

Q. The value of

\int \frac{d x}{a sin x + b cos x} = A [ln (tan \frac{x}{2} - \frac{a - \sqrt{a^{2} + b^{2}}}{b}) - ln (tan \frac{x}{2} - \frac{a + \sqrt{a^{2} + b^{2}}}{b})]

where

A

Q. The equation

a sin x + b cos x = c

, where

| c | > \sqrt{a^{2} + b^{2}}

has

Q. Assertion :The largest and smallest value of

f (θ)

differs by

\frac{1}{\sqrt{2}}

Reason:

a sin x + b cos x + c \in [c - \sqrt{a^{2} + b^{2}}, c + \sqrt{a^{2} + b^{2}}] \forall x \in R

where

a, b, c \in R

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For real a, b and x −√(a2+b2)≤a sin x+b cos x≤√(a2+b2)

The correct option is A TruePut a=r cos α, b=r sin α where r=√(a2+b2) ∴ a sin x+b cos x=r sin (x+α) since −1≤sin θ≤1 therefore −r≤sin(x+α)≤r etc

For real a, b and x $- \sqrt{(a^{2} + b^{2})} \leq a s i n x + b c o s x \leq \sqrt{(a^{2} + b^{2})}$

The correct option is A True
Put $a = r c o s α, b = r s i n α$ where $r = \sqrt{(a^{2} + b^{2})}$
$∴ a s i n x + b c o s x = r s i n (x + α)$
since $- 1 \leq s i n θ \leq 1$ therefore $- r \leq s i n (x + α) \leq r$ etc