Question

$\begin{array}{cccc}\left(a\right)& \text{If [.] denotes the greatest integer function and}& \left(p\right)& 1\\ & f\left(x\right)=\{\begin{array}{cc}3\left[x\right]-\frac{5|x|}{x};x\ne 0& \\ 2;x=0& \end{array}\text{then}& & \\ & {\int }_{\frac{-3}{2}}^{2}f\left(x\right)dx\text{is equal to}& & \\ \left(b\right)& \text{The value of}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{cosx}{1+e^x}dx\text{is}& \left(q\right)& -\frac{11}{2}\\ \left(c\right)& \text{If}{I}_1={\int }_1^{\sin \theta }\frac{x}{1+x^2}dx\text{and}{I}_2={\int }_1^{\csc \theta }\frac{1}{x(x^2+1)}dx\text{then}& \left(r\right)& 9\\ & \text{the value of}\left|\begin{array}{ccc}{I}_1& I_1^2& I_2\\ e^{I_1+I_2}& I_1^2& -1\\ 1& I_1^2+I_2^2& -1\end{array}\right|& & \\ \left(d\right)& \text{Let}f\left(x\right)\text{be a polynomial of degree 2 satisfying}& \left(s\right)& 0\\ & f\left(0\right)=1,f^{\prime }\left(0\right)=-2\text{and}{f}^{\prime \prime }\left(0\right)=6,\text{then}{\int }_{-1}^{2}f\left(x\right)dx& & \\ & \text{is equal to}& & \end{array}$

• A. $\left(a\right)\to \left(p\right);\left(b\right)\to \left(q\right);\left(c\right)\to \left(r\right);\left(d\right)\to \left(s\right)$
• B. $\left(a\right)\to \left(q\right);\left(b\right)\to \left(s\right);\left(c\right)\to \left(p\right);\left(d\right)\to \left(r\right)$
• C. $\left(a\right)\to \left(q\right);\left(b\right)\to \left(p\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(r\right)$
• D. $\left(a\right)\to \left(p\right);\left(b\right)\to \left(q\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(r\right)$

Answer 1

The correct option is C $\left(a\right)\to \left(q\right);\left(b\right)\to \left(p\right);\left(c\right)\to \left(s\right);\left(d\right)\to \left(r\right)$

(a) We have, $f\left(x\right)=\{\begin{array}{c}3\left[x\right]-5,x>0\\ 2,z=0\\ 3\left[x\right]+5,x<0\end{array}\therefore {\int }_{-\frac{3}{2}}^{2}f\left(x\right)dx={\int }_{-\frac{3}{2}}^{-1}\left(3\right[x]+5)dx+{\int }_{-1}^0\left(3\right[x]+5)dx+{\int }_0^1\left(3\right[x]-5)dx+{\int }_1^2\left(3\right[x]-5)dx={\int }_{-\frac{3}{2}}^{-1}(-6+5)dx+{\int }_{-1}^0(-3+5)dx+{\int }_0^1(-5)dx+{\int }_1^2(3-5)dx=-(-1+\frac{3}{2})+2(0+1)-5(1-0)+(-2)(2-1)=-\frac{1}{2}+2-5-2=-\frac{11}{2}$

(b)
$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\cos x}{1+e^x}dx\Rightarrow I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\cos x}{1+e^{-x}}dx\Rightarrow 2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos xdx\Rightarrow 2I=1-(-1)\Rightarrow I=1$

(c) We have, $I_2={\int }_1^{\csc \theta }\frac{x}{x^2(x^2+1)}dx=-{\int }_1^{\sin \theta }\frac{t}{1+t^2}dt\text{where}t=\frac{1}{x}\Rightarrow I_2=-I_1\therefore \left|\begin{array}{ccc}{I}_1& I_1^2& I_2\\ e^{I_1+I_2}& I_1^2& -1\\ 1& I_1^2+I_2^2& -1\end{array}\right|=\left|\begin{array}{ccc}{I}_1& I_1^2& -I_1\\ e^0& I_1^2& -1\\ 1& 2I_1^2& -1\end{array}\right|=\left|\begin{array}{ccc}{I}_1& I_1^2& -I_1\\ 1& I_1^2& -1\\ 0& I_1^2& 0\end{array}\right|[\text{Applying}{R}_3\to R_3-R_2]=-I_1^2(-I_1+I_1)\left[\text{Expanding along}{R}_3\right]=0$

(d) Let $f\left(x\right)=a{x}^2+bx+c$
Then $f\left(0\right)=1\Rightarrow c=1$
Now, $f^{{}^{\prime }}\left(x\right)=2ax+b$ and $f^{\prime \prime }\left(x\right)=2a$
$\because f^{{}^{\prime }}\left(0\right)=-2$ and $f^{\prime \prime }\left(0\right)=6$
$\Rightarrow b=-2$ and $a=3$
$\therefore f\left(x\right)=3x^2-2x+1\Rightarrow {\int }_{-1}^{2}f\left(x\right)dx={\int }_{-1}^2(3x^2-2x+1)dx=[x^3-x^2+x{]}_{-1}^2=(8-4+2)-(-1-1-1)=9$