As x becomes \(\leq 1\) and tends to zero then it crosses \(\pi, 2\pi, 3\pi, ….\).can you draw the graph?

Can you now finish the problem ……….

If we draw the graph then we can see that the function \(f(x) = sin(1/x^3)\) crosses many times. Therefore number of sign changes is infinite.

Therefore option \((d)\) is correct…..

Given that \(P_n = (1 +1/a_1)(1 + 1/a_2)….(1 + 1/a_n)\)

Therefore \(P_n=\frac{a_1 +1}{a_1}.\frac{a_2 +1}{a_2}.\frac{a_3 +1}{a_3}…..\frac{a_n +1}{a_n}\)

Now \(a_n = n(a_{n-1} + 1)\)

Put \(n=2\), we will get \(a_1+1=\frac{a_2}{2}\)

\(a_2+1=\frac{a_3}{3}\)……………….

………………………..

…………………………

\(a_n+1=\frac{a_n}{n}\)

Therefore \(P_n=\frac{a_1 +1}{a_1}.\frac{a_2 +1}{a_2}.\frac{a_3 +1}{a_3}…..\frac{a_n +1}{a_n}\)

\(\Rightarrow {P_n}= \frac{a_2}{2a_1}.\frac{a_3}{3a_2}.\frac{a_4}{4a_3}……..\frac{a_{n+1}}{(n+1).{a_n}}\)

\(\Rightarrow {P_n}=\frac{a_{n+1}}{{a_1}\{2.3.4………..(n+1)\}}\)

\(\Rightarrow {P_n}=\frac{a_{n+1}}{\{1.2.3.4………..(n+1)\}}\) (as \(a_1=1\))

\(\Rightarrow {P_n}=\frac{a_{n+1}}{(n+1)!}\)

\(\Rightarrow {P_n}=\frac{(n+1)(a_n +1)}{(n+1)!}\)

\(\Rightarrow {P_n}=\frac{(a_n +1)}{n!}\)

\(\Rightarrow {P_n}=\frac{a_n}{n!} +\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{n(a_{n-1}+1)}{n!}+\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{a_{n-1}+1}{(n-1)!}+\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{a_2}{2!}+\frac{1}{2!}+\frac{1}{3!}+…….+\frac{1}{n!}\)

\(\Rightarrow {P_n}=\frac{2(1+1)}{2!}+\frac{1}{2!}+….+\frac{1}{n!}\)

\(\Rightarrow {P_n}=1+\frac{1}{1!}+…..+\frac{1}{n!}\)

Can you now finish the problem ……….

Now we have to find out \(\lim\limits_{x \to \infty} {P_n}\)

we know that \(e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+……….+\infty\)

So,\(e^1=1+\frac{1}{1!}+\frac{1^2}{2!}+……….+\infty\)

\(\lim\limits_{x \to \infty} {P_n}\)=\(1+\frac{1}{1!}+\frac{1^2}{2!}+……….+\infty\)

\(\lim\limits_{x \to \infty} {P_n}\)=\(e\)

Therefore option (b) is correct…..

At first we have to check all the options…..

1. For \(y = 2e^x\), this is not possibe because \(y = 2e^x\) expression represent only positive side’s of x axis
2. For \(y = 2e^-x\),This expression also represents one side of x-axis
3. For \(y = e^x + e^-x\).This expression represents both sides of the x-axise.as we put \(x=0\) then \(y=2\) i.e the curve not starts from the origin
4. \(y = e^x – e^-x + 2\), this expression also represents both sides of the x-axis

Can you now finish the problem ……….

Therefore ,the correct ans is (c)

Draw the figure of given equation \(e^x = x\) and check all the points that are in the options…..

Can you now finish the problem ……….

The given equation is \(e^x = x\)

as \(x \)increases exponential function increases more rapidly than any polynomial function say \(x\) . Hence from graph we can say they never intersect . So, there is no solution for the equation . Therefore \(n=0\) and \(p=0.\)

Therefore option \((d)\) is correct…..

We take the each functions and express it in intercept form.we expand the mod i,e take the value once positive and once negetive .so we will get two equations and solve them,we will get the intersecting point also and draw the graph……..

Can you now finish the problem ……….

\(f_1(x) = |x – 1| – 1\)

\((x-1)-1=y\)

\(\Rightarrow x-y=2\)……………..(1)

\(\Rightarrow \frac{x}{2} +\frac{y}{-2}=0\)\(\Rightarrow (2,0) ,(0,-2)\)

And

\(-(x-1)-1=y\)

\(\Rightarrow x+y=0\)……..(2)

\(\frac{x}{1}+\frac{y}{1}=0\)\(\Rightarrow (1,0),(0,1)\)

Now if we draw the graph of (1) & (2) we will get the figure \(G_2\) and the intersecting point is \((1,-1)\)

Similarly we can draw the graphs for other functions………….

The second function is \(f_2(x) = ||x –1| – 1|\) i.e \(x-y=2\),\(x=y\),\(x+y=1\)…which represents the two figure as given in \(G_3\).

The third function \(f_3(x) = |x| – 1\) which gives \(x-y=1\) & \(x+y=-1\)..if we solve this two equations as first function then we will get \(G_1\)

The third function will gives the \(G_4\) graph

Similarly we will draw the graph for all given functions….

Therefore ,the correct ans is (c)

There are four options ,at first we have to check each options…..

If k is divisible by 24 then cos(kπ/4) = cos(kπ/6) = 1
\(\Rightarrow\) The limit exists and equal to RHS i.e. 0
If k is not divisible by 4 or 6 then cos(kπ/4), cos(kπ/6) both <1

Can you now finish the problem ……….

Therefore ,

lim cosn(kπ/4), cosn(kπ/6) = 0. so we may say that
\(\Rightarrow \)The equation holds.