Download App

STD 11 - Trigonometrical equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - Trigonometrical equations4 Marks
MCQ
The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for
  • A
    $ - \frac{1}{2} \le \alpha \le \frac{1}{2}$
  • B
    $ - 3 \le \alpha \le 1$
  • ✓
    $ - \frac{3}{2} \le \alpha \le \frac{1}{2}$
  • D
    $ - 1 \le \alpha \le 1$
✓

Answer

Correct option: C.
$ - \frac{3}{2} \le \alpha \le \frac{1}{2}$
c
(c) ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$

==> ${({\sin ^2}x + {\cos ^2}x)^2} - 2{\sin ^2}x{\cos ^2}x + \sin 2x + \alpha = 0$

==> ${\sin ^2}2x - 2\sin 2x - 2 - 2\alpha = 0$

Let $sin 2x = y$. Then the given equation becomes

${y^2} - 2y - 2(1 + \alpha ) = 0$,

where $ - 1 \le y \le 1$,  $({\rm{  }} - 1 \le \sin 2x \le 1)$

For real, discriminant

$ \ge 0$$ \Rightarrow $$3 + 2\alpha \ge 0$

$ \Rightarrow $ $\alpha \ge - \frac{3}{2}$

Also $ - 1 \le y \le 1 \Rightarrow - 1 \le 1 - \sqrt {3 + 2\alpha } \,\, \le 1$

$ \Rightarrow $ $3 + 2\alpha \le 4 \Rightarrow \alpha \le \frac{1}{2}$.

Thus $ - \frac{3}{2} \le \alpha \le \frac{1}{2}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a}=\vec{b} \times(\vec{b} \times \vec{c}) .$ If magnitudes of the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are $\sqrt{2}, 1$ and 2 respectively and the angle between $\vec{b}$ and $\vec{c}$ is $\theta\left(0<\theta<\frac{\pi}{2}\right)$, then the value of $1+\tan \theta$ is equal to:
→
The point on the parabola ${y^2} = 18x$, for which the ordinate is three times the abscissa, is
→
Number of values of $x$ satisfying $2sin^22x = 2cos^28x + cos10x$ in $x  \in \left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]$ is-
→
Let $f: R \rightarrow R$ be defined $f(x)=a e^{2 x}+b e^x+c x$. If $f(0)=-1, f^{\prime}\left(\log _e 2\right)=21$ and $\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$, then the value of $|a+b+c|$ equals:
→
A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is received by station $\mathrm{A}$ and then transmitted to station $B$. The probability of each station receiving the signal correctly is $\frac{3}{4}$. If the signal received at station $\mathrm{B}$ is green, then the probability that the original signal was green is
→
For the function $f(x) = x^4 (12 ln x - 7)$
→
Let $y = f (x)$ and $y = g (x)$ be two differentiable function in $[0,2]$ such that  $f(0) = 3,$ $f(2) = 5$ , $g (0) = 1$ and $g(2) = 2$. If there exist atlellst one $c \in \left( {0,2} \right)$ such that $f'(c)=kg'(c)$,then $k$ must be
→
The number of integral values of $\lambda $ for which $x^2 + y^2 + \lambda x + (1 - \lambda )y + 5 = 0$ is the equation of a circle whose radius cannot exceed $5$ , is
→
$\lim _{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to
→
Let for some function $y=f(x), \int_{0}^{x} t f(t) d t=x^{2} f(x)$, $x>0$ and $f(2)=3$. Then $f(6)$ is equal to :
→