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1

### WB JEE 2010

The value of $${2 \over {3!}} + {4 \over {5!}} + {6 \over {7!}} +$$ ............ is

A
e1/2
B
e$$-$$1
C
e
D
e1/3

## Explanation

$${{3 - 1} \over {3!}} + {{5 - 1} \over {5!}} + {{7 - 1} \over {7!}} + \,\,.......$$

$$= \left( {{1 \over {2!}} - {1 \over {3!}}} \right) + \left( {{1 \over {4!}} - {1 \over {5!}}} \right) + \,\,.......$$

$$= \left( {1 - {1 \over {1!}}} \right) + \left( {{1 \over {2!}} - {1 \over {3!}}} \right) + \left( {{1 \over {4!}} - {1 \over {5!}}} \right) + \,\,.......$$

$$= {e^{ - 1}}$$

2

### WB JEE 2010

If angles A, B and C are in A.P., then $${{a + c} \over b}$$ is equal to

A
$$2\sin \left( {{{A - C} \over 2}} \right)$$
B
$$2\cos \left( {{{A - C} \over 2}} \right)$$
C
$$\cos \left( {{{A - C} \over 2}} \right)$$
D
$$\sin \left( {{{A - C} \over 2}} \right)$$

## Explanation

$$\because$$ A, B, C are in A.P.

$$\therefore$$ $${{A + C} \over 2} = B$$ ...... (i)

Now, $${{a + c} \over b} = {{2R\sin A + 2R\sin C} \over {2R\sin B}}$$

$$= {{2\sin \left( {{{A + C} \over 2}} \right)\cos \left( {{{A - C} \over 2}} \right)} \over {\sin B}} = 2\cos \left( {{{A - C} \over 2}} \right)$$ [Using (i)]

3

### WB JEE 2010

The value of n for which $${{{x^{n + 1}} + {y^{n + 1}}} \over {{x^n} + {y^n}}}$$ is the geometric mean of x and y is

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

## Explanation

$${{{x^{n + 1}} + {y^{n + 1}}} \over {{x^n} + {y^n}}} = \sqrt {x\,.\,y} = {x^{1/2}}\,.\,{y^{1/2}}$$

$$\Rightarrow {x^{n + 1}} + {y^{n + 1}} = {x^{n + {1 \over 2}}}{y^{1/2}} + {x^{1/2}}\,.\,{y^{n + {1 \over 2}}}$$

$$\Rightarrow {x^{n + {1 \over 2}}}({x^{1/2}} - {y^{1/2}}) = {y^{n + {1 \over 2}}}({x^{1/2}} - {y^{1/2}})$$

$$\Rightarrow {x^{n + {1 \over 2}}} = {y^{n + {1 \over 2}}}$$

$$\Rightarrow {\left( {{x \over y}} \right)^{n + {1 \over 2}}} = 1 = {\left( {{x \over y}} \right)^0} \Rightarrow n + {1 \over 2} = 0$$

$$\therefore$$ $$n = - {1 \over 2}$$

4

### WB JEE 2010

G.M. and H.M. of two numbers are 10 and 8 respectively. The numbers are

A
5, 20
B
4, 25
C
2, 50
D
1, 100

## Explanation

Let nos. be a and b

(A = A.M., G = G.M., H = H.M.)

Here, G = 10, H = 8

$$\because$$ G2 = AH $$\Rightarrow$$ 100 = 8A

$$\therefore$$ $$A = {{25} \over 2}$$

$$\therefore$$ ab = 100 & a + b = 25, which are satisfied by option (a) only.

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