1.

\(\Leftrightarrow1-2sin^2x+sinx+m=0\)

\(\Leftrightarrow2sin^2x-sinx-1=m\)

Đặt \(sinx=t\Rightarrow t\in\left[-\dfrac{1}{2};\dfrac{\sqrt{2}}{2}\right]\)

Xét hàm \(f\left(t\right)=2t^2-t-1\) trên \(\left[-\dfrac{1}{2};\dfrac{\sqrt{2}}{2}\right]\)

\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-\dfrac{1}{2};\dfrac{\sqrt{2}}{2}\right]\)

\(f\left(-\dfrac{1}{2}\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=-\dfrac{9}{8}\) ; \(f\left(\dfrac{\sqrt{2}}{2}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow-\dfrac{9}{8}\le f\left(t\right)\le0\Rightarrow-\dfrac{9}{8}\le m\le0\)

Có 2 giá trị nguyên của m (nếu đáp án là 3 thì đáp án sai)

2.

ĐKXĐ: \(sin2x\ne1\Rightarrow x\ne\dfrac{\pi}{4}\) (chỉ quan tâm trong khoảng xét)

Pt tương đương:

\(\left(tan^2x+cot^2x+2\right)-\left(tanx+cotx\right)-4=0\)

\(\Leftrightarrow\left(tanx+cotx\right)^2+\left(tanx+cotx\right)-4=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx+cotx=\dfrac{1+\sqrt{17}}{2}\\tanx+cotx=\dfrac{1-\sqrt{17}}{2}\left(loại\right)\end{matrix}\right.\)

Nghiệm xấu quá, kiểm tra lại đề chỗ \(-tanx+...-cotx\) có thể 1 trong 2 cái đằng trước phải là dấu "+"

Biểu diễn được 3 điểm

\(cos^{2015}\left(x-\frac{11\pi}{2}\right)\)

\(=cos^{2015}\left(x+\frac{\pi}{2}-\frac{12\pi}{2}\right)=cos^{2015}\left(x+\frac{\pi}{2}-6\pi\right)\)

\(=cos^{2015}\left(x+\frac{\pi}{2}\right)=-\sin^{2015}x\)

\(cos^{2019}\left(x+\frac{7\pi}{2}\right)\)

\(=cos^{2019}\left(x-\frac{\pi}{2}+4\pi\right)=cos^{2019}\left(x-\frac{\pi}{2}\right)\)

\(=cos^{2019}\left(\frac{\pi}{2}-x\right)=\sin^{2019}x\)

\(\sin^{2019}\left(\frac52\pi-x\right)=\sin^{2019}\left(2\pi+\frac{\pi}{2}-x\right)\)

\(=\sin^{2019}\left(\frac{\pi}{2}-x\right)=cos^{2019}x\)

\(\cot^2\left(x-\frac{\pi}{2}\right)=\frac{1}{\sin^2\left(x-\frac{\pi}{2}\right)}-1=\frac{1}{\sin^2\left(\frac{\pi}{2}-x\right)}-1=\frac{1}{cos^2x}-1\)

\(=\tan^2x\)

TA có: \(B=\frac{\tan\left(\frac{21}{2}\pi-x\right)\cdot cos\left(38\pi-x\right)\cdot\sin\left(x-7\pi\right)}{\sin\left(\frac{13}{2}\pi-x\right)\cdot cos\left(x-2023\pi\right)}\)

\(=\frac{\tan\left(10\pi+\frac{\pi}{2}-x\right)\cdot cos\left(-x\right)\cdot\sin\left(x-\pi\right)}{\sin\left(6\pi+\frac{\pi}{2}-x\right)\cdot cos\left(x-\pi-2022\pi\right)}\)

\(=\frac{\tan\left(\frac{\pi}{2}-x\right)\cdot cosx\cdot\left(-1\right)\cdot\sin\left(\pi-x\right)}{\sin\left(\frac{\pi}{2}-x\right)\cdot cos\left(x-\pi\right)}\)

\(=\frac{\cot x\cdot cosx\cdot\left(-1\right)\cdot\sin x}{cosx\cdot\left(-1\right)\cdot cosx}=\frac{\cot x\cdot\sin x}{cosx}=1\)

\(=\dfrac{tan\left(\dfrac{pi}{2}+x\right)\cdot sin\left(-x\right)\cdot cos\left(x-pi\right)}{cos\left(\dfrac{pi}{2}-x\right)\cdot sin\left(x+pi\right)}\)

\(=\dfrac{-cotx\cdot sin\left(-x\right)\cdot\left(-cosx\right)}{sinx\cdot-sinx}\)

\(=\dfrac{cotx\cdot sinx\left(-1\right)\cdot cosx}{-sinx\cdot sinx}=\dfrac{\dfrac{cosx}{sinx}\cdot cosx}{sinx}=\dfrac{cos^2x}{sin^2x}=cot^2x\)

Cho hàm số y=f(x)y=f(x) có đạo hàm và liên tục trên [0;π2][0;π2]thoả mãn f(x)=f′(x)−2cosxf(x)=f′(x)−2cosx. Biết f(π2)=1f(π2)=1, tính giá trị f(π3)f(π3)

A. √3+1/2         B. √3−1/2          C. 1−√3/2             D. 0

B

Đề.