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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

MCQ (Single Correct Answer)
The set of points where $$f\left( x \right) = {x \over {1 + \left| x \right|}}$$ is differentiable is
A
$$\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)$$
B
$$\left( { - \infty ,1} \right) \cup \left( { - 1,\infty } \right)$$
C
$$\left( { - \infty ,\infty } \right)$$
D
$$\left( {0,\infty } \right)$$

Explanation

$$f\left( x \right) = \left\{ {\matrix{ {{x \over {1 - x}},} & {x < 0} \cr {{x \over {1 + x}},} & {x \ge 0} \cr } } \right.$$

$$ \Rightarrow f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right.$$

$$\therefore$$ $$f'\left( x \right)$$ exist at everywhere.
2

AIEEE 2005

MCQ (Single Correct Answer)
If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals
A
$$-2$$
B
$$3$$
C
$$2$$
D
$$1$$

Explanation

Let $$\alpha ,\,\,\alpha + 1\,\,$$ be roots

Then $$\alpha + \alpha + 1 = b = $$ sum of -

roots $$\alpha \left( {\alpha + 1} \right) = c$$

$$=$$ product of roots

$$\therefore$$ $${b^2} - 4c $$

$$ = {\left( {2\alpha + 1} \right)^2} - 4\alpha \left( {\alpha + 1} \right)$$

$$ = 1.$$
3

AIEEE 2005

MCQ (Single Correct Answer)
The value of $$a$$ for which the sum of the squares of the roots of the equation $${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$ assume the least value is
A
$$1$$
B
$$0$$
C
$$3$$
D
$$2$$

Explanation

$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$

$$ \Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)$$

$${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$

$$ = {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5$$

For min. value of $${\alpha ^2} + {\beta ^2}$$ where $$\alpha $$ is an integer

$$ \Rightarrow \,\,a = 1.$$
4

AIEEE 2005

MCQ (Single Correct Answer)
Let $$f:R \to R$$ be a differentiable function having $$f\left( 2 \right) = 6$$,
$$f'\left( 2 \right) = \left( {{1 \over {48}}} \right)$$. Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} $$ equals
A
$$24$$
B
$$36$$
C
$$12$$
D
$$18$$

Explanation

$$\mathop {\lim }\limits_{x \to 2} \int\limits_0^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt$$

$$ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$$

Applying $$L'$$ Hospital rule

$$\mathop {\lim }\limits_{x \to 2} {{\left[ {4f{{\left( x \right)}^3}f'\left( x \right)} \right]} \over 1}$$

$$ = 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)$$

$$ = 4 \times {6^3} \times {1 \over {48}} = 18$$

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