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1

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

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

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

2

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 $$\tan {P \over 2}.\,\tan {Q \over 2} = {c \over a}$$]

3

WB JEE 2009

MCQ (Single Correct Answer)

In triangle ABC, if $$\sin A\sin B = {{ab} \over {{c^2}}}$$, then the triangle is

A
equilateral
B
isosceles
C
right angled
D
obtuse angled

Explanation

$$\sin A\sin B = {{ab} \over {{c^2}}}$$

$$ \Rightarrow \sin A\sin B = {{(k\sin A)(k\sin B)} \over {{k^2}{{\sin }^2}C}}$$

$$ \Rightarrow {\sin ^2}C = 1 \Rightarrow \sin C = 1$$

$$ \Rightarrow 1 = \sin {\pi \over 2} \Rightarrow C = {\pi \over 2}$$ ($$\because$$ $$\sin C \ne -1$$)

$$ \Rightarrow \angle C = 90^\circ $$, $$\Delta$$ is right angled.

4

WB JEE 2009

MCQ (Single Correct Answer)

If a = 2$$\sqrt2$$, b = 6, A = 45$$^\circ$$, then

A
no triangle is possible
B
one triangle is possible
C
two triangles are possible
D
either no triangle or two triangles are possible

Explanation

Given, a = 2$$\sqrt2$$, b = 6, A = 45$$^\circ$$

We know, $${a \over {\sin A}} = {b \over {\sin B}}$$

$${{2\sqrt 2 } \over {\sin 45^\circ }} = {6 \over {\sin B}}$$ $$\therefore$$ $$\sin B = {{6 \times {1 \over {\sqrt 2 }}} \over {2\sqrt 2 }} = {6 \over 4} = {3 \over 2}$$

Which is not possible. $$\therefore$$ No triangle is possible.

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