Mathematics Test - 35

Here, \(f(x)=x^{2} \ln x\)
$$
\begin{array}{l}
\Rightarrow f^{\prime}(x)=2 x \ln x+x^{2} \times \frac{1}{x} \\
\Rightarrow f^{\prime}(x)=2 x \ln x+x
\end{array}
$$
Now, \(f^{\prime}(x)=0\)
$$
\begin{array}{l}
\Rightarrow \quad x(2 \ln x+1)=0 \\
\Rightarrow \quad x=0 \quad \text { or } \quad x=e^{-1 / 2} \\
\Rightarrow \quad x=0 \quad \text { or } \quad x=\frac{1}{\sqrt{e}}
\end{array}
$$
both \(0+\frac{1}{\sqrt{e}}\) is less than 1
But given \(x \in[1, e]\) in this ranges \(f^{\prime}(x)>0\)
Hence in
i. \(e, \quad f(x)\) is increasing
so maximum value at \(x=e\)
$$
=f(e)=e^{2} \ln e=e^{2}
$$
minimum value at \(x=1\) and
$$
=f(1)=1^{2} \ln 1=0
$$
Hence, Option ( C) is the correct answer.

Given equation is \(\left(1-a^{2}\right) x^{2}+2 a x-1=\)
Its discriminant \(D=4\) and roots are \(\frac{1}{a-1}, \frac{1}{a+1}\)
Given, \(0<\frac{1}{a-1}<1,0<\frac{1}{a+1}<1\)
Now, \(\frac{1}{a-1}>0 \Rightarrow a>1\)
\(\frac{1}{a-1}<1 \Rightarrow \frac{1}{a-1}-1<0\)
\(\Rightarrow \quad \frac{2-a}{a-1}<0 \Rightarrow a<1\) or \(a>2\)
\(\therefore a>2 \ldots\)
\(\frac{1}{a+1}>0 \Rightarrow a>-1\)
and \(\frac{1}{a+1}<1 \Rightarrow-\frac{a}{a+1}<0\)
\(\Rightarrow a<-1\) or \(a>0\)
\(\therefore a>0\)
From (1) and \((2), \quad a>2\)
Hence, the correct option is (C)

Given equations, \((2 a+b) x+(a+3 b) y+(b-3 a)=0\) and \(m x+2 y+6=0\) are concurrent for all real values of a and \(\mathrm{b},\) so they must represent the same line for same values of a and \(\mathrm{b}\).
Then, \(\frac{2 a+b}{m}=\frac{(a+3 b)}{2}=\frac{b-3 a}{6}\)
Taking last two ratios, \(\frac{a+3 b}{2}=\frac{-3 a+b}{6} \Rightarrow b=-\frac{3}{4} a\)
Taking first two ratios, \(\mathrm{m}=\frac{2(2 \mathrm{a}+\mathrm{b})}{\mathrm{a}+3 \mathrm{b}}=\frac{2(2 \mathrm{a}-(3 / 4) \mathrm{a})}{\mathrm{a}+3(-3 / 4) \mathrm{a}}=\frac{10}{-5}=-2\)
KEY CONCEPTS
Family of straight lines The general equation of family of straight line will be written in one parameter. The equation of straight line which passes through point of intersection of two given lines \(L_{1}\) and \(L_{2}\) can be taken as \(\mathrm{L}_{1}+\lambda \mathrm{L}_{2}=0\)
Hence, the correct option is (A)

\((1+x)^{21}+(1+x)^{22}+\ldots .+(1+x)^{30}\)
Coefficient of \(x^{5}=21 C_{5}+22 C_{5}+\ldots . .+{ }^{30} C_{5}\)
\(={ }^{21} C_{6}+{ }^{21} C_{5}+{ }^{22} C_{5}+\ldots . .+{ }^{30} C_{5}-{ }^{21} C_{6}\)
\(={ }^{22} C_{6}+{ }^{22} C_{5}+\ldots . .+{ }^{30} C_{5}-{ }^{21} C_{6}\)
\(={ }^{31} C_{6}-{ }^{21} C_{6}\)
Hence, the correct option is (D)

Orthogonality Condition of two circles If two circles \(\mathrm{S} \equiv \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) and
\(\mathrm{S}^{\prime} \equiv \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{g}^{\prime} \mathrm{x}+2 \mathrm{f}^{\prime} \mathrm{y}+\mathrm{c}^{\prime}=0\) intersect each other orthogonally, then
\(2 \mathrm{gg}^{\prime}+2 \mathrm{ff}^{\prime}=\mathrm{c}+\mathrm{c}^{\prime}\)
Hence, the correct option is (C)

\(\begin{array}{l}\overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{AC}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+(\mathrm{y}-1) \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ 0 & 0 & \mathrm{y}-1\end{array}\right| \\ \Rightarrow \quad(\mathrm{y}-1)(\mathrm{i}-\mathrm{j}) \\ |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{2(\mathrm{y}-1)^{2}} \\ \Rightarrow \quad & \text { Given } \frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\operatorname{cosec} \frac{\pi}{4} \\ \Rightarrow \quad \frac{1}{2} \sqrt{2(\mathrm{y}-1)^{2}}=\pm \sqrt{2} \\ \Rightarrow \quad \mathrm{y}-1=\pm 2 \Rightarrow \mathrm{y}=3,-1\end{array}\)
Hence, the correct option is (A)

Required area \(=4 \times 2=8\) sq. unit
\(\because \quad x>0\)
\(\therefore \quad \sin x\(\Rightarrow \quad \frac{\sin x}{x}<1\)
or \(\quad \int_{0}^{\pi / 4} \frac{\sin x}{x} d x<\int_{0}^{\pi / 4} 1 . d x\)
\(=\frac{\pi}{4}\)
or \(\int_{0}^{\pi / 4} \frac{\sin x}{x} d x<\frac{\pi}{4}\)
Hence, the correct option is (B)

Line is parallel to the normal of the plane

x-2 y-3 z=7

\(\therefore\) Equation of line through (1,1,-1)

\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z+1}{-3}

KEY CONCEPTS
Line and Plane Equation of Plane Through a Given Line
(i) If equation of the line is given in symmetrical form as \(\frac{x-x_{1}}{1}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n},\) then equation of plane is \(\left(a_{1} x+b_{1} y+c_{1} x+d_{1}\right)+\lambda\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0\)
(ii) If equation of line is given in general form as \(a_{1} x+b_{1} y+c_{1} z+d_{1}=0=a_{2} x+b_{2} y+c_{2} z+\) \(\mathrm{d}_{2}\), then the equation of plane passing through this line is \(\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+\left(a_{2} x+b_{2} y+\right.\) \(\left.c_{2} z+d_{2}\right)=0\)
(iii) If the plane pass through parallel lines \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{c}}+\mu \overrightarrow{\mathrm{b}}\) then
equation of the required plane is \(\mid \overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}} \mathrm{b}]=0\)
Hence, the correct option is (C)