JAMIA 2022 — Mathematics PYQ

Step 1: Analyze the Function $f(x)$

The given function is:

$f(x)=a{x}^7+b{x}^3+cx-5$

Let's define a new function $g(x)$ such that:

$g(x)=a{x}^7+b{x}^3+cx$

Notice that $g(x)$ is an odd function because:

$g(-x)=a(-x{)}^7+b(-x{)}^3+c(-x)=-(a{x}^7+b{x}^3+cx)=-g(x)$

Now, we can write $f(x)$ as:

$f(x)=g(x)-5$

Step 2: Use the Given Information $f(-7)=7$

Substitute $x=-7$ into the equation:

$f(-7)=g(-7)-5=7$

$g(-7)=12$

Since $g(x)$ is an odd function, we know that $g(-7)=-g(7)$. Therefore:

$-g(7)=12⟹g(7)=-12$

Step 3: Find the Value of $f(7)$

Now, calculate $f(7)$ using the value of $g(7)$:

$f(7)=g(7)-5$

$f(7)=-12-5=-17$

Step 4: Determine the Range of $f(7)+17\cos x$

We need the range of:

$y=-17+17\cos x$

We know the range of $\cos x$ is $[-1,1]$:

$-1\le \cos x\le 1$

Multiply the entire inequality by $17$:

$-17\le 17\cos x\le 17$

Add $-17$ to all parts of the inequality:

$-17-17\le -17+17\cos x\le -17+17$

$-34\le y\le 0$

Final Answer:

The range is $[-34,0]$.