Phrases Function Previous Year Question Paper and Solution with PDF

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Previous Year Questions
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Phrases Previous Year Questions (PYQs)

Phrases Function PYQ

Qus : 1

The graph of function

$f(x)=\log _e({x}^3+\sqrt[]{{x}^6+1})$ is symmetric about:

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Solution

Qus : 2

If f(x) is a polynomial of degree 4, f(n) = n + 1 & f(0) = 25, then find f(5) = ?

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Solution

Correct Shortcut Method — Find $f(5)$

Step 1: Define a helper polynomial:

$g(x) = f(x) - (x + 1)$

Given: $f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5 \Rightarrow g(1) = g(2) = g(3) = g(4) = 0$

So,

$g(x) = A(x - 1)(x - 2)(x - 3)(x - 4) \quad \Rightarrow \quad f(x) = A(x - 1)(x - 2)(x - 3)(x - 4) + (x + 1)$

Step 2: Use $f(0) = 25$ to find A:

$f(0) = A(-1)(-2)(-3)(-4) + (0 + 1) = 24A + 1 = 25 \Rightarrow A = 1$

Step 3: Compute $f(5)$ :

$f(5) = (5 - 1)(5 - 2)(5 - 3)(5 - 4) + (5 + 1) = 4 \cdot 3 \cdot 2 \cdot 1 + 6 = 24 + 6 = \boxed{30}$

✅ Final Answer: $\boxed{f(5) = 30}$

Qus : 3

The maximum value of

$f(x) = (x – 1)^2 (x + 1)^3$ is equal to

$\frac{2^p3^q}{3125}$ then the ordered pair of (p, q) will be

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Solution

Maximum Value of $f(x) = (x - 1)^2(x + 1)^3$

Step 1: Let’s define the function:

$f(x) = (x - 1)^2 (x + 1)^3$

Step 2: Take derivative to find critical points

Use product rule:
Let $u = (x - 1)^2$ , $v = (x + 1)^3$

$f'(x) = u'v + uv' = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2$

$f'(x) = (x - 1)(x + 1)^2 [2(x + 1) + 3(x - 1)]$

$f'(x) = (x - 1)(x + 1)^2 (5x - 1)$

Step 3: Find critical points

Set $f'(x) = 0$ :

$(x - 1)(x + 1)^2 (5x - 1) = 0 \Rightarrow x = 1,\ -1,\ \frac{1}{5}$

Step 4: Evaluate $f(x)$ at these points

$f(1) = 0$
$f(-1) = 0$
$f\left(\frac{1}{5}\right) = \left(\frac{1}{5} - 1\right)^2 \left(\frac{1}{5} + 1\right)^3 = \left(-\frac{4}{5}\right)^2 \left(\frac{6}{5}\right)^3$

$f\left(\frac{1}{5}\right) = \frac{16}{25} \cdot \frac{216}{125} = \frac{3456}{3125}$

Step 5: Compare with given form:

It is given that maximum value is $\frac{3456}{3125} = 2^p \cdot 3^q / 3125$

Factor 3456:

$3456 = 2^7 \cdot 3^3 \Rightarrow \text{So } p = 7, \quad q = 3$

✅ Final Answer: $\boxed{(p, q) = (7,\ 3)}$

Qus : 4

$| x - 6|= | x - 4x | -| x^2- 5x +6 |$ , where x is a real variable

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Qus : 5

A real valued function f is defined as

$f(x)=\begin{cases}{-1} & {-2\leq x\leq0} \\ {x-1} & {0\leq x\leq2}\end{cases}$ . Which of the following statement is FALSE?

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Qus : 6

Number of onto (surjective) functions from A to B if n(A)=6 and n(B)=3, is

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Qus : 7

Let

$X_i, i = 1,2,.. , n$ be n observations and

$w_i = px_i +k, i = 1,2, ,n$ where p and k are constants. If the mean of

$x_i 's$ is 48 and the standard deviation is 12, whereas the mean of

$w_i 's$ is 55 and the standard deviation is 15, then the value of p and k should be

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Qus : 8

Let S be the set

$\{a\in Z^+:a\leq100\}$ .If the equation

$[tan^2 x]-tan x - a = 0$ has real roots (where [ . ] is the greatest integer function), then the number of elements is S is

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Solution

Qus : 9

The number of one - one functions f: {1,2,3} → {a,b,c,d,e} is

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Solution

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Step 1: One-one (injective) function means no two elements map to the same output.

We choose 3 different elements from 5 and assign them to 3 inputs in order.

So, total one-one functions = $P(5,3) = 5 \times 4 \times 3 = 60$

✅ Final Answer: $\boxed{60}$

Qus : 10

The domain of the function

$f(x)=\frac{{\cos }^{-1}x}{[x]}$ is

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Solution

Qus : 11

The function

$f(x)=\log (x+\sqrt[]{{x}^2+1})$ is

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Qus : 12

The value of

$f(1)$ for

$f\Bigg{(}\frac{1-x}{1+x}\Bigg{)}=x+2$ is

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Solution

Given:

$f\left(\frac{1 - x}{1 + x}\right) = x + 2$

To Find: $f(1)$

Let $\frac{1 - x}{1 + x} = 1 \Rightarrow x = 0$

Then, $f(1) = f\left(\frac{1 - 0}{1 + 0}\right) = 0 + 2 = 2$

Answer:

$\boxed{2}$

Qus : 13

If f(x)=cos[

$\pi$ ^2]x+cos[-

$\pi$ ^2]x, where [.] stands for greatest integer function, then

$f(\pi/2)$ =

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Solution

? Function with Greatest Integer and Cosine

Given:

$f(x) = \cos\left([\pi^2]x\right) + \cos\left([-\pi^2]x\right)$

Find:

$f\left(\frac{\pi}{2}\right)$

Step 1: Estimate Floor Values

$\pi^2 \approx 9.8696 \Rightarrow [\pi^2] = 9,\quad [-\pi^2] = -10$

Step 2: Plug into the Function

$f\left(\frac{\pi}{2}\right) = \cos\left(9 \cdot \frac{\pi}{2}\right) + \cos\left(-10 \cdot \frac{\pi}{2}\right) = \cos\left(\frac{9\pi}{2}\right) + \cos(-5\pi)$

Step 3: Simplify

$\cos\left(\frac{9\pi}{2}\right) = 0,\quad \cos(-5\pi) = -1$

✅ Final Answer:

$\boxed{-1}$

Qus : 14

The function

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Solution

Qus : 15

Which of the following function is the inverse of itself?

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Solution

Qus : 16

If the graph of y = (x – 2)² – 3 is shifted by 5 units up along y-axis and 2 units to the right along the x-axis, then the equation of the resultant graph is

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Solution

When y= f (x) is shifted by k units to the right along x

– axis, it become y= f (x - k )

Hence, new equation of

graph is y = (x - 4)² + 2

Qus : 17

A function

$f : (0,\pi) \to R$ defined by

$f(x) = 2 sin x + cos 2x$ has

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Qus : 18

The number of one-to-one functions from {1, 2, 3} to {1, 2, 3, 4, 5} is

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Solution

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Phrases Previous Year Questions (PYQs)

Phrases Function PYQ

Solution

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Correct Shortcut Method — Find f(5) f(5)

Solution

Maximum Value of f(x)=(x−1)2(x+1)3 f(x) = (x - 1)^2(x + 1)^3

Solution

Solution

Solution

Solution

Solution

Solution

Given: A one-one function from set {1,2,3}\{1,2,3\} to set {a,b,c,d,e}\{a,b,c,d,e\}

Solution

Solution

Solution

Solution

? Function with Greatest Integer and Cosine

Step 1: Estimate Floor Values

Step 2: Plug into the Function

Step 3: Simplify

✅ Final Answer:

Solution

Solution

Solution

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Correct Shortcut Method — Find $f(5)$

Maximum Value of $f(x) = (x - 1)^2(x + 1)^3$

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$