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1

### JEE Main 2022 (Online) 25th June Morning Shift [ Memory Based ]

Numerical

$$f(x) = {x^3} + x - 5$$ and $$f\left( {g(x)} \right) = x$$ and $$g'(63)=$$

Correct Answer is $${1 \over {49}}$$

## Explanation

Given, $$f\left( {g(x)} \right) = x$$

$$g(x) = {f^{ - 1}}(x)$$

$$\Rightarrow {d \over {dx}}g(x) = {d \over {dx}}{f^{ - 1}}(x)$$

$$\Rightarrow {d \over {dx}}{f^{ - 1}}{(x)_{at\,x = 63}} = {1 \over {{d \over {dx}}f{{(x)}_{at\,x = 4}}}}$$

$$= {1 \over {{{(3{x^2} + 1)}_{at\,x = 4}}}}$$

$$= {1 \over {49}}$$

2

### JEE Main 2022 (Online) 25th June Morning Shift [ Memory Based ]

Numerical
If f(x) = $${\left\{ {{{\left( {2\left( {1 - {x \over 2}} \right)} \right)}^{25}}{{\left( {2 + x} \right)}^{25}}} \right\}^{{1 \over {50}}}}$$ and

g(x) = f(f(f((x))) + f(f(x)) then [g(1)] is ________ (where [x] is greatest integer less than or equal to x).

3

### JEE Main 2021 (Online) 31st August Evening Shift

Numerical
The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________.

## Explanation

A = 4-digit numbers divisible by 3

A = 1002, 1005, ....., 9999.

9999 = 1002 + (n $$-$$ 1)3

$$\Rightarrow$$ (n $$-$$ 1)3 = 8997 $$\Rightarrow$$ n = 3000

B = 4-digit numbers divisible by 7

B = 1001, 1008, ......., 9996

$$\Rightarrow$$ 9996 = 1001 + (n $$-$$ 1)7

$$\Rightarrow$$ n = 1286

A $$\cap$$ B = 1008, 1029, ....., 9996

9996 = 1008 + (n $$-$$ 1)21

$$\Rightarrow$$ n = 429

So, no divisible by either 3 or 7

= 3000 + 1286 $$-$$ 429 = 3857

total 4-digits numbers = 9000

required numbers = 9000 $$-$$ 3857 = 5143
4

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
If A = {x $$\in$$ R : |x $$-$$ 2| > 1},
B = {x $$\in$$ R : $$\sqrt {{x^2} - 3}$$ > 1},
C = {x $$\in$$ R : |x $$-$$ 4| $$\ge$$ 2} and Z is the set of all integers, then the number of subsets of the
set (A $$\cap$$ B $$\cap$$ C)c $$\cap$$ Z is ________________.

## Explanation

A = ($$-$$$$\infty$$, 1) $$\cup$$ (3, $$\infty$$)

B = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ (2, $$\infty$$)

C = ($$-$$$$\infty$$, 2] $$\cup$$ [6, $$\infty$$)

So, A $$\cap$$ B $$\cap$$ C = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ [6, $$\infty$$)

z $$\cap$$ (A $$\cap$$ B $$\cap$$ C)' = {$$-$$2, $$-$$1, 0, $$-$$1, 2, 3, 4, 5}

Hence, no. of its subsets = 28 = 256.

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