\displaystyle \text{SHORT ANSWER QUESTIONS - 2 Marks Each}

\displaystyle \textbf{Question 71: }\text{The values of two functions }f\text{ and }g\text{ for certain values of } \\ x\text{ are given in the following table:}
\displaystyle \begin{array}{c|ccc}x&-2&0&3\\\hline f(x)&-12&-4&8\\g(x)&0&-12&30\end{array}
\displaystyle \text{(a) Find the value of }f^{-1}(8)?
\displaystyle \text{(b) Given that }f(x)\text{ is a linear function, find }f(x)?
\displaystyle \text{Answer:}
\displaystyle \text{(a) From the table, }f(3)=8.
\displaystyle \therefore f^{-1}(8)=3.
\displaystyle \text{(b) Let }f(x)=mx+c.
\displaystyle \text{Using }(0,-4),\quad c=-4.
\displaystyle \text{Using }(3,8),\quad 8=3m-4.
\displaystyle \therefore m=4.
\displaystyle \therefore f(x)=4x-4.
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{"A function }f\text{ is called a self-inverse function if }f^{-1}(x)=f(x)
\displaystyle \text{ for all values of }x\text{ in the}  \  \text{domain."}
\displaystyle \text{Let }f(x)=\frac{\pi^2}{x},\text{ where }x\ne0,\ x\in R.
\displaystyle \text{Show that }f(x)\text{ is a self-inverse function.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\frac{\pi^2}{x},\quad x\ne0.
\displaystyle f(f(x))=f\left(\frac{\pi^2}{x}\right).
\displaystyle =\frac{\pi^2}{\frac{\pi^2}{x}}=x.
\displaystyle \therefore f(f(x))=x.
\displaystyle \therefore f^{-1}(x)=f(x).
\displaystyle \therefore f(x)\text{ is a self-inverse function.}
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{Rahul sits on a speed boat in an island to do water sports, which is moving}
\displaystyle \text{along a curve } y=\frac{1}{\cos x\sin x}\text{ in water. Find the number of stationary points.}
\displaystyle \text{Answer:}
\displaystyle y=\frac{1}{\cos x\sin x}=\frac{2}{\sin2x}=2\mathrm{cosec}\,2x.
\displaystyle \frac{dy}{dx}=-4\mathrm{cosec}\,2x\cot2x.
\displaystyle \text{For a stationary point, }\frac{dy}{dx}=0.
\displaystyle \therefore \cot2x=0.
\displaystyle 2x=\frac{\pi}{2}+n\pi,\qquad n\in Z.
\displaystyle \therefore x=\frac{\pi}{4}+\frac{n\pi}{2},\qquad n\in Z.
\displaystyle \therefore \text{there are infinitely many stationary points.}
\displaystyle \\

\displaystyle \textbf{Question 74: }\text{The function defined by }f(x)=\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\text{ is differentiable at }
\displaystyle x=0. \ \text{Is this statement true or false? Give reason for your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\tan\theta.
\displaystyle \frac{1-x^2}{1+x^2}=\cos2\theta,\qquad \theta=\tan^{-1}x.
\displaystyle \therefore f(x)=\sin^{-1}(\cos2\theta).
\displaystyle \text{For }x>0,\quad f(x)=\frac{\pi}{2}-2\tan^{-1}x.
\displaystyle \therefore f'_+(0)=\lim_{x\to0^+}\frac{f(x)-f(0)}{x}=-2.
\displaystyle \text{For }x<0,\quad f(x)=\frac{\pi}{2}+2\tan^{-1}x.
\displaystyle \therefore f'_-(0)=\lim_{x\to0^-}\frac{f(x)-f(0)}{x}=2.
\displaystyle \therefore f'_-(0)\ne f'_+(0).
\displaystyle \therefore f(x)\text{ is not differentiable at }x=0.
\displaystyle \therefore \text{the statement is false.}
\displaystyle \\

\displaystyle \textbf{Question 75: }\text{Seema enjoys a roller coaster ride in Ferrari world by first going}
\displaystyle \text{downwards and then upwards to the maximum height. The relation between the distance}
\displaystyle \text{travelled (cm) with respect to the time taken to complete the ride by Seema is given by}
\displaystyle \text{the following equation:}
\displaystyle y=4x-\frac{1}{2}x^2
\displaystyle \text{where }x=\text{time in seconds.}
\displaystyle \text{(a) What is the rate of change of displacement with respect to the time?}
\displaystyle \text{(b) How many seconds it will take her to go to its maximum height?}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }y=4x-\frac{1}{2}x^2.
\displaystyle \therefore \frac{dy}{dx}=4-x.
\displaystyle \therefore \text{the rate of change of displacement is }4-x\text{ cm/s.}
\displaystyle \text{(b) At maximum height, }\frac{dy}{dx}=0.
\displaystyle \therefore 4-x=0.
\displaystyle \therefore x=4\text{ seconds.}
\displaystyle \text{Also, }\frac{d^2y}{dx^2}=-1<0,\text{ confirming a maximum.}
\displaystyle \therefore \text{Seema reaches the maximum height after }4\text{ seconds.}
\displaystyle \\

\displaystyle \textbf{Question 76: }\text{A child is solving an equation }x^2+y^2=1\text{ and finds that this is the}
\displaystyle \text{solution of the following differential equation: }1+yy''+(y')^2=0.\text{ Justify his answer.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x^2+y^2=1.
\displaystyle \text{Differentiating with respect to }x,
\displaystyle 2x+2yy'=0.
\displaystyle \therefore x+yy'=0.
\displaystyle \text{Differentiating again,}
\displaystyle 1+(y')^2+yy''=0.
\displaystyle \therefore 1+yy''+(y')^2=0.
\displaystyle \therefore \text{the child's answer is justified.}
\displaystyle \\

\displaystyle \textbf{Question 77: }\text{Given }2f(x)=\log|x|^a-bx^2+x,\text{ if }f(x)\text{ has extreme values at }
\displaystyle x=-1 \ \text{and }x=2,\text{ find the value of }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle 2f(x)=\log|x|^a-bx^2+x.
\displaystyle \therefore 2f'(x)=\frac{a}{x}-2bx+1.
\displaystyle \text{At an extreme point, }f'(x)=0.
\displaystyle \text{At }x=-1,\quad -a+2b+1=0.
\displaystyle \therefore a=2b+1.\qquad ...(1)
\displaystyle \text{At }x=2,\quad \frac{a}{2}-4b+1=0.
\displaystyle \therefore a=8b-2.\qquad ...(2)
\displaystyle \text{From (1) and (2), }2b+1=8b-2.
\displaystyle \therefore 6b=3\Rightarrow b=\frac12.
\displaystyle \therefore a=2\left(\frac12\right)+1=2.
\displaystyle \therefore a=2,\qquad b=\frac12.
\displaystyle \\

\displaystyle \textbf{Question 78: }\text{The solution of }x\,dy-y\,dx=0\text{ represents a family of straight lines }
\displaystyle \text{passing through the origin. Justify.}
\displaystyle \text{Answer:}
\displaystyle x\,dy-y\,dx=0.
\displaystyle x\frac{dy}{dx}=y.
\displaystyle \therefore \frac{dy}{y}=\frac{dx}{x}.
\displaystyle \text{Integrating both sides,}
\displaystyle \log|y|=\log|x|+\log C.
\displaystyle \therefore y=Cx.
\displaystyle \text{This represents a family of straight lines with slope }C\text{ and }y\text{-intercept }0.
\displaystyle \therefore \text{all the straight lines pass through the origin.}
\displaystyle \\

\displaystyle \textbf{Question 79: }\text{Find the function }f\text{ which satisfies the equation }\frac{df}{dx}=2f, \\ \text{ given that }f(0)=e^3.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{df}{dx}=2f.
\displaystyle \therefore \frac{df}{f}=2\,dx.
\displaystyle \text{Integrating both sides,}
\displaystyle \log|f|=2x+\log C.
\displaystyle \therefore f=Ce^{2x}.
\displaystyle \text{Given, }f(0)=e^3.
\displaystyle \therefore e^3=Ce^0=C.
\displaystyle \therefore f(x)=e^3e^{2x}=e^{2x+3}.
\displaystyle \\

\displaystyle \textbf{Question 80: }\text{Evaluate: }\int 2^{2^{2x}}\cdot2^{2x}\cdot2^x\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=2^x.
\displaystyle \therefore dt=(\log2)2^x\,dx.
\displaystyle \therefore 2^x\,dx=\frac{dt}{\log2}.
\displaystyle \text{Also, }2^{2x}=t^2.
\displaystyle \therefore \int2^{2^{2x}}\cdot2^{2x}\cdot2^x\,dx=\frac{1}{\log2}\int t^2 2^{t^2}\,dt.

\displaystyle \textbf{Question 81: }\text{Evaluate: }\int\{f(ax+b)\}^n\cdot f'(ax+b)\,dx,\ n\ne-1.
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=f(ax+b).
\displaystyle \therefore du=a f'(ax+b)\,dx.
\displaystyle \therefore f'(ax+b)\,dx=\frac{du}{a}.
\displaystyle \therefore \int\{f(ax+b)\}^n f'(ax+b)\,dx=\frac1a\int u^n\,du.
\displaystyle =\frac1a\cdot\frac{u^{n+1}}{n+1}+C,\qquad n\ne-1.
\displaystyle =\frac{\{f(ax+b)\}^{n+1}}{a(n+1)}+C.
\displaystyle \therefore \text{the required integral is }\frac{\{f(ax+b)\}^{n+1}}{a(n+1)}+C.
\displaystyle \\

\displaystyle \textbf{Question 82: }\text{Evaluate: }\int\frac{dx}{x^{1/2}-x^{1/3}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=x^{1/6}.
\displaystyle \therefore x=t^6,\qquad dx=6t^5\,dt.
\displaystyle x^{1/2}-x^{1/3}=t^3-t^2=t^2(t-1).
\displaystyle \therefore \int\frac{dx}{x^{1/2}-x^{1/3}}=6\int\frac{t^3}{t-1}\,dt.
\displaystyle =6\int\left(t^2+t+1+\frac{1}{t-1}\right)dt.
\displaystyle =2t^3+3t^2+6t+6\log|t-1|+C.
\displaystyle =2x^{1/2}+3x^{1/3}+6x^{1/6}+6\log|x^{1/6}-1|+C.
\displaystyle \\

\displaystyle \textbf{Question 83: }\text{Derive a condition for which given, }\overrightarrow{b}=\overrightarrow{c}, \\ \text{ we can say }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{c}\times\overrightarrow{a}?
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\overrightarrow{b}=\overrightarrow{c}.
\displaystyle \therefore \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{a}.
\displaystyle \text{But }\overrightarrow{b}\times\overrightarrow{a}=-(\overrightarrow{a}\times\overrightarrow{b}).
\displaystyle \therefore \overrightarrow{a}\times\overrightarrow{b}=-(\overrightarrow{a}\times\overrightarrow{b}).
\displaystyle \therefore 2(\overrightarrow{a}\times\overrightarrow{b})=\overrightarrow{0}.
\displaystyle \therefore \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{0}.
\displaystyle \therefore \overrightarrow{a}\text{ and }\overrightarrow{b}\text{ must be parallel, or one of them must be the zero vector.}
\displaystyle \\

\displaystyle \textbf{Question 84: }\text{If the angle between }\overrightarrow{a}=-3\widehat{i}+x\widehat{j}+\widehat{k}\text{ and }
\displaystyle \overrightarrow{b}=x\widehat{i}+2x\widehat{j}+\widehat{k}\text{ is acute and the angle} \ \text{between }\overrightarrow{b}\text{ and the }x\text{-axis lies between } \\ \frac{\pi}{2}\text{ and }\pi.\text{ Find the range of values for }x.
\displaystyle \text{Answer:}
\displaystyle \text{Since the angle between }\overrightarrow{b}\text{ and the }x\text{-axis lies between }\frac{\pi}{2}\text{ and }\pi,
\displaystyle \cos\theta=\frac{x}{\sqrt{x^2+4x^2+1}}<0.
\displaystyle \therefore x<0.
\displaystyle \text{Also, the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is acute, so }\overrightarrow{a}\cdot\overrightarrow{b}>0.
\displaystyle (-3)(x)+(x)(2x)+(1)(1)>0.
\displaystyle 2x^2-3x+1>0.
\displaystyle (2x-1)(x-1)>0.
\displaystyle \therefore x<\frac12\text{ or }x>1.
\displaystyle \text{Combining this with }x<0,\text{ we get }x<0.
\displaystyle \therefore \text{the required range is }x<0.
\displaystyle \\

\displaystyle \textbf{Question 85: }\text{To launch a new product in his company, Mr. Rajesh spends Rs. }1\text{ lakh on}
\displaystyle \text{the infrastructure and the variable cost of the product is estimated as Rs. }150\text{ per unit.}
\displaystyle \text{The sale price per unit is fixed as Rs. }200.\text{ Additionally, Mr Rajesh also spends Rs. }0.5
\displaystyle \text{per unit squared for marketing. Find the profit function. Hence, draw an inference regarding}
\displaystyle \text{the breakeven point.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ units be produced and sold.}
\displaystyle \text{Total cost}=100000+150x+\frac12x^2.
\displaystyle \text{Total revenue}=200x.
\displaystyle \therefore P(x)=200x-\left(100000+150x+\frac12x^2\right).
\displaystyle \therefore P(x)=-\frac12x^2+50x-100000.
\displaystyle \text{For the breakeven point, }P(x)=0.
\displaystyle -\frac12x^2+50x-100000=0.
\displaystyle \therefore x^2-100x+200000=0.
\displaystyle \text{Its discriminant}=(-100)^2-4(1)(200000).
\displaystyle =10000-800000=-790000<0.
\displaystyle \therefore \text{there is no real breakeven point.}
\displaystyle \therefore \text{the company incurs a loss for all levels of production.}
\displaystyle \\

\displaystyle \textbf{Question 86: }\text{Anuj and Ashish launched a new fountain pen in their pen-factory which is}
\displaystyle \text{consisting of Rs. }6400\text{ as overheads, Rs. }35\text{ per pen as the cost of material and labour cost,}
\displaystyle \text{Rs. } \frac{x^2}{100}\text{ for }x \ \text{items produced. Find the values of }x\text{ for which average cost is increasing.}
\displaystyle \text{Answer:}
\displaystyle C(x)=6400+35x+\frac{x^2}{100}.
\displaystyle \text{Average cost }=\frac{C(x)}{x}=\frac{6400}{x}+35+\frac{x}{100}.
\displaystyle \frac{d}{dx}(\text{Average cost})=-\frac{6400}{x^2}+\frac1{100}.
\displaystyle \text{For average cost to be increasing,}
\displaystyle -\frac{6400}{x^2}+\frac1{100}>0.
\displaystyle \frac1{100}>\frac{6400}{x^2}.
\displaystyle x^2>640000.
\displaystyle \text{Since }x>0,\quad x>800.
\displaystyle \therefore \text{average cost is increasing for }x>800.
\displaystyle \\

\displaystyle \textbf{Question 87: }\text{If }\tan^{-1}\left(\frac{1}{1+1\cdot2}\right)+\tan^{-1}\left(\frac{1}{7}\right)+\cdots+\tan^{-1}\left(\frac{1}{111}\right)=S,
\displaystyle \text{then find }\tan S.
\displaystyle \text{Answer:}
\displaystyle \tan^{-1}\left(\frac{1}{n}\right)-\tan^{-1}\left(\frac{1}{n+1}\right)
\displaystyle =\tan^{-1}\left(\frac{\frac1n-\frac1{n+1}}{1+\frac1{n(n+1)}}\right).
\displaystyle =\tan^{-1}\left(\frac{1}{n^2+n+1}\right).
\displaystyle \therefore S=\sum_{n=1}^{10}\left\{\tan^{-1}\left(\frac1n\right)-\tan^{-1}\left(\frac1{n+1}\right)\right\}.
\displaystyle =\tan^{-1}(1)-\tan^{-1}\left(\frac1{11}\right).
\displaystyle \therefore \tan S=\frac{1-\frac1{11}}{1+\frac1{11}}.
\displaystyle =\frac{10}{12}=\frac56.
\displaystyle \therefore \tan S=\frac56.
\displaystyle \\

\displaystyle \textbf{Question 88: }\text{The adjacent figure is the graph of function }y=f(x).\text{ Give answers }
\displaystyle \text{to the following questions.}
\displaystyle \text{(a) Name the type of discontinuity of the function.}
\displaystyle \text{(b) Give reason for your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) The function has a jump discontinuity at }x=1.
\displaystyle \text{(b) From the graph, the left-hand and right-hand limits at }x=1\text{ exist but are unequal.}
\displaystyle \therefore \lim_{x\to1^-}f(x)\ne\lim_{x\to1^+}f(x).
\displaystyle \therefore \lim_{x\to1}f(x)\text{ does not exist.}
\displaystyle \therefore \text{the function has a jump discontinuity at }x=1.
\displaystyle \\

\displaystyle \textbf{Question 89: }\text{If }|\overrightarrow{a}-\overrightarrow{b}|=|\overrightarrow{a}+\overrightarrow{b}|\text{ in the given figure, what inference can you draw?} \displaystyle \text{Answer:}
\displaystyle \text{Given, }|\overrightarrow{a}-\overrightarrow{b}|=|\overrightarrow{a}+\overrightarrow{b}|.
\displaystyle \text{Squaring both sides,}
\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}+\overrightarrow{b}|^2.
\displaystyle |\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\overrightarrow{a}\cdot\overrightarrow{b}
\displaystyle =|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2\overrightarrow{a}\cdot\overrightarrow{b}.
\displaystyle \therefore 4\overrightarrow{a}\cdot\overrightarrow{b}=0.
\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}=0.
\displaystyle \therefore \overrightarrow{a}\perp\overrightarrow{b}.
\displaystyle \therefore \text{the parallelogram in the figure is a rectangle.}
\displaystyle \\

\displaystyle \textbf{Question 90: }\text{In Z Square Mall in Kanpur, there is a space to keep }300\text{ cars and the}
\displaystyle \text{entry fees per car is Rs. }20.\text{ It is estimated that if the entry fee is decreased by Rs. }5,
\displaystyle \text{then }50\text{ additional cars can be adjusted in the same parking. Justify that the Marginal}
\displaystyle \text{Revenue (MR) decreases at a higher rate than the Average Revenue (AR).}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the number of cars and }p\text{ be the entry fee per car.}
\displaystyle \frac{\Delta p}{\Delta x}=\frac{-5}{50}=-\frac{1}{10}.
\displaystyle \therefore p=-\frac{x}{10}+c.
\displaystyle \text{When }x=300,\ p=20.
\displaystyle \therefore 20=-30+c\Rightarrow c=50.
\displaystyle \therefore AR=p=50-\frac{x}{10}.
\displaystyle \text{Total Revenue}=x\left(50-\frac{x}{10}\right)=50x-\frac{x^2}{10}.
\displaystyle \therefore MR=\frac{d}{dx}\left(50x-\frac{x^2}{10}\right)=50-\frac{x}{5}.
\displaystyle \frac{d(AR)}{dx}=-\frac{1}{10},\qquad \frac{d(MR)}{dx}=-\frac{1}{5}.
\displaystyle \therefore \text{MR decreases at twice the rate at which AR decreases.}
\displaystyle \\

\displaystyle \textbf{Question 91: }\text{Consider the functions }f(x)=-(x-h)^2+2k,\text{ and }
\displaystyle g(x)=e^{x-2}+k,\text{ where }h,k\in R. \ \text{(a) Find }f'(x).
\displaystyle \text{The graphs of }f\text{ and }g\text{ have a common tangent at }x=3.
\displaystyle \text{(b) Show that: }2h=e+6.
\displaystyle \text{Answer:}
\displaystyle \text{(a) }f(x)=-(x-h)^2+2k.
\displaystyle \therefore f'(x)=-2(x-h)=2(h-x).
\displaystyle \text{(b) }g(x)=e^{x-2}+k.
\displaystyle \therefore g'(x)=e^{x-2}.
\displaystyle \text{Since the graphs have a common tangent at }x=3,\ f'(3)=g'(3).
\displaystyle 2(h-3)=e^{3-2}.
\displaystyle 2h-6=e.
\displaystyle \therefore 2h=e+6.
\displaystyle \\

\displaystyle \textbf{Question 92: }\text{Evaluate: }\int 2^x[f'(x)+f(x)\log2]\,dx.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}[2^xf(x)]=2^xf'(x)+2^xf(x)\log2.
\displaystyle =2^x[f'(x)+f(x)\log2].
\displaystyle \therefore \int 2^x[f'(x)+f(x)\log2]\,dx=2^xf(x)+C.
\displaystyle \\

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