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1

WB JEE 2010

MCQ (Single Correct Answer)

In a $$\Delta$$ABC, $$2ac\sin \left( {{{A - B + C} \over 2}} \right)$$ is equal to

A
a2 + b2 $$-$$ c2
B
c2 + a2 $$-$$ b2
C
b2 $$-$$ a2 $$-$$ c2
D
c2 $$-$$ a2 $$-$$ b2

Explanation

$$2ac\sin \left( {{{A - B + C} \over 2}} \right) = 2ac\,.\,\sin \left( {{{\pi - B - B} \over 2}} \right)$$

$$ = 2ac\cos B = 2ac\,.\,{{{c^2} + {a^2} - {b^2}} \over {2ca}} = {c^2} + {a^2} - {b^2}$$

2

WB JEE 2010

MCQ (Single Correct Answer)

If $${{\cos A} \over 3} = {{\cos B} \over 4} = {1 \over 5}, - {\pi \over 2} < A < 0, - {\pi \over 2} < B < 0$$, then value of $$2\sin A + 4\sin B$$ is

A
4
B
$$-$$ 2
C
$$-$$ 4
D
0

Explanation

$$\because$$ $$ - {\pi \over 2} < A,B < 0$$

$$\therefore$$ $$\sin A < 0,\sin B < 0$$

$$\because$$ $${{\cos A} \over 3} = {{\cos B} \over 4} = {1 \over 5}$$

$$\therefore$$ $$\cos A = {3 \over 5},\cos B = {4 \over 5}$$

$$ \Rightarrow \sin A = - \sqrt {1 - {{\cos }^2}A} = - \sqrt {1 - {9 \over {25}}} = - {4 \over 5}$$

$$ \Rightarrow \sin B = - \sqrt {1 - {{\cos }^2}B} = - \sqrt {1 - {{16} \over {25}}} = - {3 \over 5}$$

$$\therefore$$ $$2\sin A + 4\sin B = - {8 \over 5} - {{12} \over 5} = - 4$$

3

WB JEE 2010

MCQ (Single Correct Answer)

In a right-angled triangle, the sides are a, b and c, with c as hypotenuse, and c $$-$$ b $$\ne$$ 1, c + b $$\ne$$ 1. Then the value of $$({\log _{c + b}}a + {\log _{c - b}}a)/(2{\log _{c + b}}a \times {\log _{c - b}}a)$$ will be

A
2
B
$$-$$ 1
C
$${1 \over 2}$$
D
1

Explanation

Given expression $$ = {1 \over 2}\left[ {{1 \over {{{\log }_{c - b}}a}} + {1 \over {{{\log }_{c + b}}a}}} \right]$$

$$ = {1 \over 2}[{\log _a}(c - b) + {\log _a}(c + b)]$$

$$ = {1 \over 2}{\log _a}({c^2} - {b^2}) = {1 \over 2}{\log _a}{a^2} = 1$$ [$$\because$$ $${c^2} = {a^2} + {b^2}$$]

4

WB JEE 2010

MCQ (Single Correct Answer)

In a triangle PQR, $$\angle$$R = $$\pi$$/2. If $$\tan \left( {{P \over 2}} \right)$$ and $$\tan \left( {{Q \over 2}} \right)$$ are roots of ax2 + bx + c = 0, where a $$\ne$$ 0, then which one is true?

A
c = a + b
B
a = b + c
C
b = a + c
D
b = c

Explanation

$$\angle R = {\pi \over 2} \Rightarrow P + Q = {\pi \over 2}$$

$$ \Rightarrow \tan \left( {{P \over 2} + {Q \over 2}} \right) = \tan {\pi \over 4} \Rightarrow {{\tan {P \over 2} + \tan {Q \over 2}} \over {1 - \tan {P \over 2} + \tan {Q \over 2}}} = 1$$

$$ \Rightarrow - {b \over a} = 1 - {c \over a}$$ $$\therefore$$ $$c = a + b$$

[$$\because$$ $$\tan {P \over 2},\,\tan {Q \over 2}$$ are roots of $$a{x^2} + bx + c = 0$$

$$\therefore$$ $$\tan {P \over 2} + \tan {Q \over 2} = - {b \over a}$$ and $$\tan {P \over 2}.\,\tan {Q \over 2} = {c \over a}$$]

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