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Standard XI

Mathematics

Equations with Solutions at Boundary Values

A function y ...

Question

A function y = f(x) satisfies the condition $f (x) s i n x + f (x) c o s x = 1$ , $f (x)$ being bounded when $x \to 0$ . If $| = \frac{π}{2} \int 0 f (x) d x$ then

\frac{π}{2} < | < \frac{π^{2}}{4}

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\frac{π}{4} < | < \frac{π^{2}}{2}

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1 < | < \frac{π}{2}

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0 < | < 1

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Solution

The correct option is B $\frac{π}{4} < | < \frac{π^{2}}{2}$

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Similar questions

A function $y = f (x)$ satisfies the condition $f^{'} (x) s i n x + f (x) c o s x = 1, f (x)$ being bounded when $x \to 0$ . if $I = \int_{0}^{\frac{π}{2}} f (x) d x,$ then

Q. Let

f

be a function defined as

f (x) = \frac{1}{1 + (tan x)^{m}},

where

m

is a constant. Then

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, & - π \leq x \leq \frac{- π}{2} a cos (x) + b, & \frac{- π}{2} < x < 0 - 2 + cos x, & 0 \leq x \leq \frac{π}{2} c sin x, & \frac{π}{2} < x \leq π \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

is continuous in

[- π, π],

then number of points at which

f (x)

is not differentiable is/are

Q. Let

f (x) = \frac{1 - \tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{2}] .

If f(x) is continuous in

[0, \frac{π}{2}]

, then

f (\frac{π}{4}) =

_________.

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A function y = f(x) satisfies the condition f(x)sinx+f(x)cosx=1, f(x) being bounded when x→0. If |=π2∫0f(x)dx then

The correct option is B π4<|<π22

A function y = f(x) satisfies the condition $f (x) s i n x + f (x) c o s x = 1$ , $f (x)$ being bounded when $x \to 0$ . If $| = \frac{π}{2} \int 0 f (x) d x$ then

The correct option is B $\frac{π}{4} < | < \frac{π^{2}}{2}$