The value of $k$ for which $(sin x + cos x)^{2} + k sin x . cos x - 1 = 0$ for $x \in R$ , is

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $- 2$
Now,
$(sin x + cos x)^{2} + k sin x . cos x - 1 = 0$
or, ${sin}^{2} + {cos}^{2} x + 2 sin x cos x + k sin x . cos x - 1 = 0$
or, $1 + (2 - k) sin x . cos x - 1 = 0$ [ Since ${sin}^{2} x + {cos}^{2} x = 1$ ]
or, $(2 + k) sin x . cos x = 0$
or, $(2 + k) = 0$ [ As the above equation is true for all $x \in R]$
or, $k = - 2$ .

Suggest Corrections

Similar questions

l i m x \to 0 \frac{32}{x^{8}} (1 - cos \frac{x^{2}}{2} - cos \frac{x^{2}}{4} + c o s \frac{x^{2}}{2} cos \frac{x^{2}}{4})

k

(k - 2) x^{2} + 8 x + k + 4 > 0

x ϵ R

lim x \to 0 {\frac{1}{x^{8}} (1 - cos \frac{x^{2}}{2} - cos \frac{x^{2}}{4} + cos \frac{x^{2}}{2} cos \frac{x^{2}}{4})} = 2^{- k}

k

k

(k - 2) x^{2} + 8 x + k + 4 > 0

x \in R,

k

(k - 2) x^{2} + 8 x + k + 4 > 0

\forall x \in R

Join BYJU'S Learning Program