NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
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 $$g'(63)=$$

Your Input ________

Answer

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).
Your Input ________

Answer

Correct Answer is 2
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 ____________.
Your Input ________

Answer

Correct Answer is 5143

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 ________________.
Your Input ________

Answer

Correct Answer is 256

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.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12