IIT JEE Main Maths -Unit 1 -Functions - Exam Style Questions- New Syllabus

Question

Let $ f(x) $ be a real differentiable function such that $ f(0) = 1 $ and $ f(x + y) = f(x)f'(y) + f'(x)f(y) $ for all $ x, y \in \mathbb{R} $. Then $ \sum_{n=1}^{100} \log_e f(n) $ is equal to :
(1) 2384
(2) 2525
(3) 5220
(4) 2406

▶️ Answer/Explanation

Ans. (2)

Sol. $ f(x + y) = f(x)f'(y) + f'(x)f(y) $

Put $ x = y = 0 $

$ f(0) = f(0)f'(0) + f'(0)f(0) $

$ f'(0) = \frac{1}{2} $

Put $ y = 0 $

$ f(x) = f(x)f'(0) + f'(x)f(0) $

$ f(x) = f(x) \times \frac{1}{2} + f'(x) \times 1 $

$ f'(x) = \frac{f(x)}{2} $

$ \frac{dy}{y} = \frac{dx}{2} \Rightarrow \int \frac{dy}{y} = \int \frac{dx}{2} $

$ \Rightarrow \ln y = \frac{x}{2} + c $

$ \because f(0) = 1 \Rightarrow c = 0 $

$ \ln y = \frac{x}{2} \Rightarrow f(x) = e^{x/2} $

$ \ln f(n) = \frac{n}{2} $

$ \sum_{n=1}^{100} \ln f(n) = \sum_{n=1}^{100} \frac{n}{2} = \frac{1}{2} \times \frac{100 \times 101}{2} = \frac{5050}{2} = 2525 $

Question

Let $A = \{1, 2, 3, 4\}$ and $B = \{1, 4, 9, 16\}$. Then the number of many-one functions $f : A \to B$ such that $1 \in f(A)$ is equal to :
(1) 127
(2) 151
(3) 163
(4) 139

▶ Answer/Explanation

Ans. (2)
Sol. Total = $4^4$
One-one = $4!$
Many-one = $256 – 24 = 232$
Many-one which $1 \notin f(A)$
= $3.3.3.3 = 81$
232 – 81 = 151

Question

Let $f(x) = \log_e x$ and $g(x) = \frac{x^4 – 2x^3 + 3x^2 – 2x + 2}{2x^2 – 2x + 1}$. Then the domain of $fog$ is

(1) $\mathbb{R}$
(2) $(0, \infty)$
(3) $[0, \infty)$
(4) $[1, \infty)$

▶️ Answer/Explanation

Ans. (1)

Sol. $f(x) = \ln x$
$g(x) = \frac{x^4 – 2x^3 + 3x^2 – 2x + 2}{2x^2 – 2x + 1}$
$D_g \in \mathbb{R}$
$D_f \in (0, \infty)$
For $D_{fog} \Rightarrow g(x) > 0$
$\Rightarrow x^4 – 2x^3 + 3x^2 – 2x + 2 > 0$
Clearly $x < 0$ satisfies which are included in option (1) only.

IIT JEE Main Maths -Unit 1 -Functions - Exam Style Questions- New Syllabus

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