The number of integral values of k for which | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation is of the form $a \cos x + b \sin x = c$,which has a solution if and only if $-\sqrt{a^2 + b^2} \leq c \leq \sqrt{a^2 + b^2}$.Here,$a

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The equation is of the form $a \cos x + b \sin x = c$,which has a solution if and only if $-\sqrt{a^2 + b^2} \leq c \leq \sqrt{a^2 + b^2}$.Here,$a = 7$,$b = 5$,and $c = 2k + 1$.Thus,$-\sqrt{7^2 + 5^2} \leq 2k + 1 \leq \sqrt{7^2 + 5^2}$.$-\sqrt{49 + 25} \leq 2k + 1 \leq \sqrt{49 + 25}$.$-\sqrt{74} \leq 2k + 1 \leq \sqrt{74}$.Since $\sqrt{74} \approx 8.602$,we have $-8.602 \leq 2k + 1 \leq 8.602$.Subtracting $1$ from all sides: $-9.602 \leq 2k \leq 7.602$.Dividing by $2$: $-4.801 \leq k \leq 3.801$.The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.The total number of such integral values is $8$.

The number of integral values of $k$ for which the equation $7 \cos x + 5 \sin x = 2k + 1$ has a solution is:

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin^2 x}}{{\sqrt{2} - \sqrt{1 + \cos x}}}$ equals

If $\alpha+\beta=-2$ and $\alpha^3+\beta^3=-56$,then the quadratic equation whose roots are $\alpha$ and $\beta$ is

$A$ conducting rod of length $2l$ is rotating with constant angular speed $\omega$ about its perpendicular bisector. $A$ uniform magnetic field $\overrightarrow{B}$ exists parallel to the axis of rotation. The e.m.f. induced between the two ends of the rod is

Lactose is a disaccharide of

Vedclass Test Series

Exam Paper Generator

Online Exam Module