Match List I with the List II and select the correct answer using the code given below the lists : List IList II (A)The least positive integral value of x satisfying the inequality(P)1tan−1(x3+5x−2)>tan−1(4−6x+6x2), is(B)If f(x)=acosx−cosbxx2,x≠0 and f(0)=4 is continuous at x=0, then(Q)2|a+b| can be(C)Let f(x)=limn→∞xn(a+sin(xn))+(b−sin(xn))(1+xn)sec(tan−1(xn+x−n)) be continuous at x=1. (R)3Then (a+b+1) is(D)Let f(x)=limt→0sin−1(ext−1t). Then limt→06(f(x)−xx3) is(S)4(T)6Which of the following is the only CORRECT combination?

Match List I with the List II and select the correct answer using the

1.	Match List I with the List II and select the correct answer using the code given below the lists : List IList II (A)The least positive integral value of x satisfying the inequality(P)1tan−1(x3+5x−2)>tan−1(4−6x+6x2), is(B)If f(x)=acosx−cosbxx2,x≠0 and f(0)=4 is continuous at x=0, then(Q)2\|a+b\| can be(C)Let f(x)=limn→∞xn(a+sin(xn))+(b−sin(xn))(1+xn)sec(tan−1(xn+x−n)) be continuous at x=1. (R)3Then (a+b+1) is(D)Let f(x)=limt→0sin−1(ext−1t). Then limt→06(f(x)−xx3) is(S)4(T)6Which of the following is the only CORRECT combination?
Answer» Match $List I$ with the $List II$ and select the correct answer using the code given below the lists : $\begin{matrix} List I & List II (A) & The least positive integral value of x satisfying the inequality & (P) & 1 {tan}^{- 1} (x^{3} + 5 x - 2) > {tan}^{- 1} (4 - 6 x + 6 x^{2}), is (B) & If f (x) = \frac{a cos x - cos b x}{x^{2}}, x \neq 0 and f (0) = 4 is continuous at x = 0, then & (Q) & 2 \| a + b \| can be (C) & Let f (x) = lim n \to \infty \frac{x^{n} (a + sin (x^{n})) + (b - sin (x^{n}))}{(1 + x^{n}) sec ({tan}^{- 1} (x^{n} + x^{- n}))} be continuous at x = 1. & (R) & 3 Then (a + b + 1) is (D) & Let f (x) = lim t \to 0 {sin}^{- 1} (\frac{e^{x t} - 1}{t}) . Then lim t \to 0 6 (\frac{f (x) - x}{x^{3}}) is & (S) & 4 (T) & 6 \end{matrix}$ Which of the following is the only CORRECT combination?