WB JEE 2009 | Trigonometric Functions & Equations Question 8 | Mathematics

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WB JEE 2009

MCQ (Single Correct Answer)

-0.25

General solution of $$\sin x + \cos x = \mathop {\min }\limits_{a \in IR} \{ 1,{a^2} - 4a + 6\} $$ is

$${{n\pi } \over 2} + {( - 1)^n}{\pi \over 4}$$

$$2n\pi + {( - 1)^n}{\pi \over 4}$$

$$n\pi + {( - 1)^{n + 1}}{\pi \over 4}$$

$$n\pi + {( - 1)^n}{\pi \over 4} - {\pi \over 4}$$

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

The value of $$\left( {1 + \cos {\pi \over 6}} \right)\left( {1 + \cos {\pi \over 3}} \right)\left( {1 + \cos {{2\pi } \over 3}} \right)\left( {1 + \cos {{7\pi } \over 6}} \right)$$ is

$${3 \over {16}}$$

$${3 \over {8}}$$

$${3 \over {4}}$$

$${1 \over {2}}$$

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

$$P = {1 \over 2}{\sin ^2}\theta + {1 \over 3}{\cos ^2}\theta $$, then

$${1 \over 3} \le P \le {1 \over 2}$$

$$P \ge {1 \over 2}$$

$$2 \le P \le 3$$

$$ - {{\sqrt {13} } \over 6} \le P \le {{\sqrt {13} } \over 6}$$

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

A positive acute angle is divided into two parts whose tangents are 1/2 and 1/3. Then the angle is

$$\pi$$/4

$$\pi$$/5

$$\pi$$/3

$$\pi$$/6

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