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1

JEE Advanced 2016 Paper 2 Offline

MCQ (More than One Correct Answer)
Let
$$f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {\left( {{{{n^n}\left( {x + n} \right)\left( {x + {n \over 2}} \right)...\left( {x + {n \over n}} \right)} \over {n!\left( {{x^2} + {n^2}} \right)\left( {{x^2} + {{{n^2}} \over 4}} \right)....\left( {{x^2} + {{{n^2}} \over {{n^2}}}} \right)}}} \right)^{{x \over n}}},$$ for

all $$x>0.$$ Then
A
$$f\left( {{1 \over 2}} \right) \ge f\left( 1 \right)$$
B
$$f\left( {{1 \over 3}} \right) \le f\left( {{2 \over 3}} \right)$$
C
$$\,f'\left( 2 \right) \le 0$$
D
$$\,{{f'\left( 3 \right)} \over {f\left( 3 \right)}} \ge {{f'\left( 2 \right)} \over {f\left( 2 \right)}}$$

Explanation

$$f(x) = \mathop {\lim }\limits_{n \to \infty } {\left\{ {{{{n^{2n}}\left( {{x \over n} + 1} \right)\left( {{x \over n} + {1 \over 2}} \right)...\left( {{x \over n} + {1 \over n}} \right)} \over {n!{n^{2n}}\left( {{{{x^2}} \over {{n^2}}} + 1} \right)\left( {{{{x^2}} \over {{n^2}}} + {1 \over {{2^2}}}} \right)....\left( {{{{x^2}} \over {{n^2}}} + {1 \over {{n^2}}}} \right)}}} \right\}^{{x \over n}}}$$

By taking logarithm we have

$$\ln f(x) = \mathop {\lim }\limits_{n \to \infty } {x \over n}\left\{ {\sum\limits_{r = 1}^n {\ln \left( {1 + {{rx} \over n}} \right) - \sum\limits_{r = 1}^n {\ln \left( {1 + {{{r^2}{x^2}} \over {{n^2}}}} \right)} } } \right\}$$

By definite integration, we can write it as

$$\ln f(x) = x\int\limits_0^1 {\ln (1 + xy)dy - x\int\limits_0^1 {\ln (1 + {x^2}{y^2})dy} } $$

Let xy = u

Then, $$\ln f(x) = \int\limits_0^x {\ln (1 + u)du - \int\limits_0^x {\ln (1 + {u^2})du} } $$

Using Newton-Leibnitz formula, we get

$${1 \over {f(x)}}\,.\,f'(x) = log\left( {{{1 + x} \over {1 + {x^2}}}} \right)$$ ...... (i)

Here, at x = 1,

$${{f'(1)} \over {f(1)}} = \log (1) = 0$$

$$\therefore$$ $$f'(1) = 0$$

Now, sign scheme of f'(x) is shown below

$$\therefore$$ At x = 1, function attains maximum. Since, f(x) increases on (0, 1).

$$\therefore$$ f(1) > f(1/2)

$$\therefore$$ Option (a) is incorrect.

f(1/3) < f(2/3)

$$\therefore$$ Option (b) is correct.

Also, f'(x) < 0, when x > 1

$$\Rightarrow$$ f'(2) < 0

$$\therefore$$ Option (c) is correct.

Also, $${{f'(x)} \over {f(x)}} = \log \left( {{{1 + x} \over {1 + {x^2}}}} \right)$$

$$\therefore$$ $${{f'(3)} \over {f(3)}} - {{f'(2)} \over {f(2)}} = \log \left( {{4 \over {10}}} \right) - \log \left( {{3 \over 5}} \right)$$

$$ = \log (2/3) < 0$$

$$ \Rightarrow {{f'(3)} \over {f(3)}} < {{f'(2)} \over {f(2)}}$$

$$\therefore$$ Option (d) is incorrect.

2

JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $$F:R \to R$$ be a thrice differentiable function. Suppose that
$$F\left( 1 \right) = 0,F\left( 3 \right) = - 4$$ and $$F\left( x \right) < 0$$ for all $$x \in \left( {{1 \over 2},3} \right).$$ Let $$f\left( x \right) = xF\left( x \right)$$ for all $$x \in R.$$

The correct statement(s) is (are)

A
$$f'\left( 1 \right) < 0$$
B
$$f\left( 2 \right) < 0$$
C
$$f'\left( x \right) \ne 0$$ for any $$x \in \left( {1,3} \right)$$
D
$$f'\left( x \right) = 0$$ for some $$x \in \left( {1,3} \right)$$

Explanation

Given,

F(1) = 0, F(3) = 4

F'(x) < 0 for all x $$\in$$(1, 3)

and f(x) = xF(x)

Now, f'(x) = F(x) + xF'(x)

$$\Rightarrow$$ f'(1) = F(1) + 1 . F'(1)

$$\Rightarrow$$ f'(1) = 0 + F'(1) ....... (1)

As F'(x) < 0 for all x $$\in$$ (1, 3)

$$\therefore$$ F'(1) < 0

From (1), we get

f'(1) = F'(1) < 0

From Lagrange theorem

$$F'(2) = {{F(3) - F(1)} \over {3 - 1}}$$

$$ = {{ - 4 - 0} \over 2}$$

= $$-$$2

As f(x) = xF(x)

$$\therefore$$ f(3) = 3 . F(3) = 3 . ($$-$$4) = $$-$$12

and f(1) = 1 . F(1) = 0

$$\therefore$$ $$f'(2) = {{f(3) - f(1)} \over {3 - 1}}$$

$$ = {{ - 12} \over 2} = - 6$$

As, f'(x) = F(x) + x . F'(x)

$$\therefore$$ f'(2) = F(2) + 2 . F'(2)

$$ \Rightarrow - 6 = {{f(2)} \over 2} + 2.\,( - 2)$$

$$ \Rightarrow {{f(2)} \over 2} = - 2$$

$$\Rightarrow$$ f(2) = $$-$$4

$$\therefore$$ f(2) < 0

$$f'(x) = {{f(3) - f(1)} \over {3 - 1}}$$ when x $$\in$$ (1, 3)

f(3) = $$-$$12

and f(1) = 0

$$\therefore$$ $$f'(x) = {{ - 12} \over 2} = - 6$$

$$\therefore$$ f'(x) $$\ne$$ 0 for any x $$\in$$(1, 3)

3

JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $$F:R \to R$$ be a thrice differentiable function. Suppose that
$$F\left( 1 \right) = 0,F\left( 3 \right) = - 4$$ and $$F\left( x \right) < 0$$ for all $$x \in \left( {{1 \over 2},3} \right).$$ Let $$f\left( x \right) = xF\left( x \right)$$ for all $$x \in R.$$

If $$\int_1^3 {{x^2}F'\left( x \right)dx = - 12} $$ and $$\int_1^3 {{x^3}F''\left( x \right)dx = 40,} $$ then the correct expression(s) is (are)

A
$$9f'\left( 3 \right) + f'\left( 1 \right) - 32 = 0$$
B
$$\int_1^3 {f\left( x \right)dx = 12} $$
C
$$9f'\left( 3 \right) - f'\left( 1 \right) + 32 = 0$$
D
$$\int_1^3 {f\left( x \right)dx = -12} $$

Explanation

Given, $$\int_1^3 {{x^2}F'(x)dx = - 12} $$

$$ \Rightarrow [{x^2}F(x)]_1^3 - \int_1^3 {2x\,.\,F(x)dx = - 12} $$

$$ \Rightarrow 9F(3) - F(1) - 2\int_1^3 {f(x)dx = - 12} $$

[$$\because$$ $$xF(x) = f(x)$$ given]

$$ \Rightarrow - 36 - 0 - 2\int_1^3 {f(x)dx = - 12} $$

$$\therefore$$ $$\int_1^3 {f(x)dx = - 12} $$

and $$\int_1^3 {{x^3}F''(x)dx = 40} $$

$$ \Rightarrow [{x^3}F'(x)]_1^3 - \int_1^3 {3{x^2}F'(x)dx = 40} $$

$$ \Rightarrow [{x^2}(xF'(x)]_1^3 - 3 \times ( - 12) = 40$$

$$ \Rightarrow \{ {x^2}\,.\,[f'(x) - F(x)]\} _1^3 = 4$$

$$ \Rightarrow 9[f'(3) - F(3)] - [f'(1) - F(1)] = 4$$

$$ \Rightarrow 9[f'(3) + 4] - [f'(1) - 0] = 4$$

$$ \Rightarrow 9f'(3) - f'(1) = - 32$$

4

JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
The option(s) with the values of a and $$L$$ that satisfy the following equation is (are) $$${{\int\limits_0^{4\pi } {{e^t}\left( {{{\sin }^6}at + {{\cos }^4}at} \right)dt} } \over {\int\limits_0^\pi {{e^t}\left( {{{\sin }^6}at + {{\cos }^4}at} \right)dt} }} = L?$$$
A
$$a = 2,L = {{{e^{4\pi }} - 1} \over {{e^\pi } - 1}}$$
B
$$a = 2,L = {{{e^{4\pi }} + 1} \over {{e^\pi } + 1}}$$
C
$$a = 4,L = {{{e^{4\pi }} - 1} \over {{e^\pi } - 1}}$$
D
$$a = 4,L = {{{e^{4\pi }} + 1} \over {{e^\pi } + 1}}$$

Explanation

Let $${I_1} = \int_0^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $$

$$ = \int_0^\pi {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_\pi ^{2\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_{2\pi }^{3\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_{3\pi }^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} } } } $$

$$\therefore$$ $${I_1} = {I_2} + {I_3} + {I_4} + {I_5}$$ ...... (i)

Now, $${I_3} = \int_\pi ^{2\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $$

Put $$t = \pi + t \Rightarrow dt = dt$$

$$\therefore$$ $${I_3} = \int_0^\pi {{e^{\pi + t}}.\,({{\sin }^6}at + {{\cos }^6}at)dt} $$

$$ = {e^t}\,.\,{I_2}$$ ..... (ii)

Now, $${I_4} = \int_{2\pi }^{3\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $$

Put $$t = 2\pi + t \Rightarrow dt = dt$$

$$\therefore$$ $${I_4} = \int_0^\pi {{e^{t + 2\pi }}({{\sin }^6}at + {{\cos }^6}at)dt} $$

$$ = {e^{2\pi }}.{I_2}$$ ...... (iii)

and $${I_5} = \int_{3\pi }^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} $$

Put $$t = 3\pi + t$$

$$\therefore$$ $${I_5} = \int_0^\pi {{e^{3\pi + t}}({{\sin }^6}at + {{\cos }^6}at)dt} $$

$$ = {e^{3\pi }}.\,{I_2}$$ ...... (iv)

From Eqs. (i), (ii), (iii) and (iv), we get

$${I_1} = {I_2} + {e^\pi }.\,{I_2} + {e^{2\pi }}.\,{I_2} + {e^{3\pi }}.\,{I_2}$$

$$ = (1 + {e^\pi } + {e^{2\pi }} + {e^{3\pi }}){I_2}$$

$$\therefore$$ $$L = {{\int_0^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} } \over {\int_0^\pi {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} }}$$

$$ = (1 + {e^\pi } + {e^{2\pi }} + {e^{3\pi }})$$

$$ = {{1.\,({e^{4\pi }} - 1)} \over {{e^\pi } - 1}}$$ for $$a \in R$$

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