Question

Consider three functions, $f\left(x\right)=x^3+x^2+x+1,g\left(x\right)=\frac{2x}{x^2+1}$ and $h\left(x\right)={\sin }^{-1}x-{\cos }^{-1}x+{\tan }^{-1}x-{\cot }^{-1}x$ and let $p\left(x\right)$ be a differentiable function on $R$ defined as $p\left(x\right)=\{\begin{array}{cc}a{\int }_0^x\sqrt{p\left(t\right)}dt+b;x>0& \\ & \\ x^2+4x+1;x\le 0& \end{array}$ where, $a,bϵ(0,\infty )$ and tangent drawn to the graph of $p\left(x\right)$ at $x=1$ is $y=mx+c$

$\begin{array}{cccccc}& \text{Column 1}& & \text{Column 2}& & \text{Column 3}\\ \text{(I)}& \text{If range of f(g(x))}\text{is}[l,m],& \left(i\right)& \text{a}& \left(P\right)& 1\\ & \text{then (l+m)=}& & & & \\ \text{(II)}& \text{The number of integers in the}& \left(ii\right)& \text{b}& \left(Q\right)& 3\\ & \text{range of g(f(x)) is equal to}& & & & \\ \text{(III)}& \text{The maximum value of}& \left(iii\right)& |c|& \left(R\right)& 4\\ & \text{g(h(x)) is equal to}& & & & \\ & \text{If the minimum value of}& & & & \\ \text{(IV)}& \text{h(g(f(x))) is}\frac{k\pi }{2},\text{then}|k|\text{is}& \text{(iv)}& \text{(m-7)}& \left(S\right)& 5\\ & \text{equal to}& & & & \end{array}$

Which of the following option is the only correct combination ?

• A. (II), (iii), (Q)
• B. (II), (iii), (P)
• C. (II), (ii), (P)
• D. (II), (ii), (S)

Answer 1

The correct option is A (II), (iii), (Q)

$\frac{x+\frac{1}{x}}{2}\ge 1[\text{Using}AM\ge GM]\Rightarrow \frac{2x}{x^2+1}\le 1\cdots \left(1\right)(x+1{)}^2\ge 0\Rightarrow x^2+1+2x\ge 0\Rightarrow -2x\le x^2+1\Rightarrow \frac{2x}{x^2+1}\ge -1\cdots (2\left)\text{From}\right(1\left)\&\right(2)\text{We get}-1\le g(x)\le 1$

$\text{(I)}f\left(g\right(x\left)\right)=\left[g\right(x){]}^3+[g\left(x\right){]}^2+g\left(x\right)+1\because -1\le g\left(x\right)\le 1f\left(g\right(x\left)\right){|}_{min}=f(-1)=0f\left(g\right(x\left)\right){|}_{max}=f\left(1\right)=4\therefore \text{Range of}f\left(g\right(x\left)\right)=[0,4]\therefore l+m=4$

$\text{(II)}\because -1\le g\left(x\right)\le 1\because -1\le g\left(f\right(x\left)\right)\le 1\therefore \text{Range of}g\left(f\right(x\left)\right)=[-1,1]\therefore \text{Number of integers in the range of}g\left(f\right(x\left)\right)=3$

$\text{(III)}\because -1\le g\left(x\right)\le 1\because -1\le g\left(h\right(x\left)\right)\le 1\therefore g\left(h\right(x\left)\right){|}_{max}=1$

$\text{(IV)}h\left(x\right)={\sin }^{-1}x-{\cos }^{-1}x+{\tan }^{-1}x-{\cot }^{-1}x=\pi -2({\cos }^{-1}x+{\cot }^{-1}x)\therefore \text{Domain of}({\cos }^{-1}x+{\cot }^{-1}x)=[-1,1]\because {\cos }^{-1}x\&{\cot }^{-1}x\text{are decreasing functions in their domain}\therefore ({\cos }^{-1}x+{\cot }^{-1}x){|}_{max}=\left({\cos }^{-1}\right(-1)+{\cot }^{-1}(-1\left)\right)=\pi +\frac{3\pi }{4}=\frac{7\pi }{4}\therefore h\left(x\right){|}_{min}=\pi -\frac{7\pi }{2}=-\frac{5\pi }{2}\therefore h\left(g\right(f\left(x\right)\left)\right){|}_{min}=-\frac{5\pi }{2}\therefore \frac{-5\pi }{2}=\frac{k\pi }{2}\Rightarrow |k|=5$


$p\left(x\right)=\{\begin{array}{cc}a{\int }_0^x\sqrt{p\left(t\right)}dt+b;x>0& \\ & \\ x^2+4x+1;x\le 0& \end{array}\because p\left(x\right)\text{is differentiable at}x=0\therefore LHD=RHD\text{at}x=0\Rightarrow {a\sqrt{p\left(x\right)}|}_{x=0}=2x+4{|}_{x=0}\Rightarrow a\sqrt{p\left(0\right)}=4\Rightarrow a=4[\because p(0)=1]\because p\left(x\right)\text{is differentiable at}x=0\Rightarrow p\left(x\right)\text{is continuous as well at}x=0\therefore LHL=p\left(0\right)\Rightarrow a{\int }_0^0\sqrt{p\left(t\right)}dt+b=1\Rightarrow b=1\therefore \text{For}x>0;p\left(x\right)=4{\int }_0^x\sqrt{p\left(t\right)}dt+1\Rightarrow p^{\prime }\left(x\right)=4\sqrt{p\left(x\right)}\text{Let}p\left(x\right)=y\therefore p^{\prime }\left(x\right)=4\sqrt{p\left(x\right)}\Rightarrow \frac{dy}{dx}=4\sqrt{y}\Rightarrow \frac{dy}{\sqrt{y}}=4x\Rightarrow \sqrt{y}=2x+\lambda \Rightarrow \sqrt{p\left(x\right)}=2x+\lambda \because p\left(0\right)=1\therefore \lambda =1\therefore p\left(x\right)=(2x+1{)}^2;x>0\therefore p(1)=9\&p^{\prime }(1)=12\therefore \text{Tangent}y=mx+c\text{at}x=1\Rightarrow m=12\&c=-3\therefore (m-7)=5\&|c|=3$