\displaystyle \text{STATEMENT BASED QUESTIONS - 1 Mark Each}

\displaystyle \textbf{Question 27: }\text{Let }f(x)\text{ be a function such that }f'(x)=g(x)\text{ and }
\displaystyle f''(x)=-f(x). \ \text{Let }h(x)=\{f(x)\}^2+\{g(x)\}^2.
\displaystyle \text{Then, consider the following statements:}
\displaystyle \text{Statement I: }h'(2024)=0.
\displaystyle \text{Statement II: }h(2)=h\left(\frac12\right).
\displaystyle \text{Which of the statements given above is/are correct?}
\displaystyle \text{(a) I only}\qquad\text{(b) II only}\qquad\text{(c) Both I and II}\qquad\text{(d) Neither I nor II}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f'(x)=g(x)\text{ and }f''(x)=-f(x).
\displaystyle \text{Since }f'(x)=g(x),\text{ we get }g'(x)=f''(x)=-f(x).
\displaystyle h(x)=\{f(x)\}^2+\{g(x)\}^2.
\displaystyle \therefore h'(x)=2f(x)f'(x)+2g(x)g'(x).
\displaystyle =2f(x)g(x)+2g(x)\{-f(x)\}=0.
\displaystyle \therefore h'(x)=0\text{ for all }x.
\displaystyle \text{Hence, }h(x)\text{ is a constant function.}
\displaystyle \text{Therefore, }h'(2024)=0.
\displaystyle \therefore \text{Statement I is true.}
\displaystyle \text{Also, since }h(x)\text{ is constant, }h(2)=h\left(\frac12\right).
\displaystyle \therefore \text{Statement II is true.}
\displaystyle \therefore \text{Both Statement I and Statement II are correct.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Statement I: For any two real numbers }a\text{ and }b,\text{ we define }aRb\text{ if}
\displaystyle \sec^2 a-\tan^2 b=1.\text{ Then, }R\text{ is transitive.}
\displaystyle \text{Statement II: The relation }R\text{ on the set }\{2,3,4\}\text{ defined by }R=\{(2,2)\}\text{ is not symmetric.}
\displaystyle \text{Which of the following options is correct?}
\displaystyle \text{(a) Both the statements are true.}
\displaystyle \text{(b) Both the statements are false.}
\displaystyle \text{(c) Statement I is true, and Statement II is false.}
\displaystyle \text{(d) Statement I is false, and Statement II is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For Statement I, }aRb\Rightarrow \sec^2 a-\tan^2 b=1.
\displaystyle \text{Using }\sec^2 a-1=\tan^2 a,
\displaystyle \tan^2 a=\tan^2 b.
\displaystyle \text{If }aRb\text{ and }bRc,\text{ then }\tan^2 a=\tan^2 b\text{ and }\tan^2 b=\tan^2 c.
\displaystyle \therefore \tan^2 a=\tan^2 c.
\displaystyle \therefore aRc.
\displaystyle \therefore R\text{ is transitive.}
\displaystyle \therefore \text{Statement I is true.}
\displaystyle \text{For Statement II, }R=\{(2,2)\}.
\displaystyle \text{The reverse of }(2,2)\text{ is }(2,2),\text{ which belongs to }R.
\displaystyle \therefore R\text{ is symmetric.}
\displaystyle \therefore \text{Statement II is false.}
\displaystyle \therefore \text{Statement I is true and Statement II is false.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Statement I: }f:R\to R\text{ given by }f(x)=\frac{1}{x-2},\text{ is neither injective}
\displaystyle \text{nor surjective.}
\displaystyle \text{Statement II: }f:Z\to Z\text{ given by }f(x)=\sqrt[3]{x^9},\text{ is neither injective nor surjective.}
\displaystyle \text{(a) Both the statements are true.}
\displaystyle \text{(b) Both the statements are false.}
\displaystyle \text{(c) Statement I is true, and Statement II is false.}
\displaystyle \text{(d) Statement I is false, and Statement II is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For Statement I, }f(x)=\frac{1}{x-2}.
\displaystyle \text{Strictly, }f:R\to R\text{ is not defined at }x=2.
\displaystyle \text{Taking the intended domain as }R-\{2\},\text{ let }f(x_1)=f(x_2).
\displaystyle \frac{1}{x_1-2}=\frac{1}{x_2-2}.
\displaystyle \therefore x_1=x_2.
\displaystyle \therefore f\text{ is injective.}
\displaystyle \text{Also, }f(x)\ne0\text{ for every }x\text{ in its domain.}
\displaystyle \therefore f\text{ is not surjective onto }R.
\displaystyle \therefore \text{Statement I is false.}
\displaystyle \text{For Statement II, }f(x)=\sqrt[3]{x^9}=x^3.
\displaystyle \text{If }f(x_1)=f(x_2),\text{ then }x_1^3=x_2^3.
\displaystyle \therefore x_1=x_2.
\displaystyle \therefore f\text{ is injective.}
\displaystyle \text{However, every integer is not the cube of an integer. For example, }2\text{ has no pre-image in }Z.
\displaystyle \therefore f\text{ is not surjective onto }Z.
\displaystyle \therefore \text{Statement II is false.}
\displaystyle \therefore \text{Both Statement I and Statement II are false.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{In the third-order matrix, }a_{ij}\text{ denotes the element of the }i^{th}\text{ row}
\displaystyle \text{and }j^{th}\text{ column: }a_{ij}=\begin{cases}0,&i\ne j\\1,&i=j\end{cases}
\displaystyle \text{Statement I: }A\text{ is an upper triangular matrix.}
\displaystyle \text{Statement II: The determinant of the matrix is equal to }1.
\displaystyle \text{Which of the above statement/s is/are correct?}
\displaystyle \text{(a) Only I.}\qquad\text{(b) Only II.}\qquad\text{(c) Both I and II.}\qquad\text{(d) Neither I nor II.}
\displaystyle \text{Answer:}
\displaystyle \text{From the given definition,}
\displaystyle A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3.
\displaystyle \text{All the elements below the principal diagonal of }A\text{ are zero.}
\displaystyle \therefore A\text{ is an upper triangular matrix.}
\displaystyle \therefore \text{Statement I is true.}
\displaystyle |A|=\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}=1\times1\times1=1.
\displaystyle \therefore \text{Statement II is true.}
\displaystyle \therefore \text{Both Statement I and Statement II are correct.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Let }y=x^5+x^3,\ x\in R.\text{ Rina, Abha and Saurabh have given their}
\displaystyle \text{opinions about the function in the following statements:}
\displaystyle \text{(i) Rina says that }y'\text{ is an increasing function for all values of }x.
\displaystyle \text{(ii) Abha says that }y'\text{ is an odd function.}
\displaystyle \text{(iii) Saurabh says that }y'\text{ is symmetrical about the origin.}
\displaystyle \text{Related to the above statements, which of the following option is true?}
\displaystyle \text{(a) Rina, Abha and Saurabh are correct.}
\displaystyle \text{(b) Rina and Abha are correct, but Saurabh is wrong.}
\displaystyle \text{(c) Rina and Saurabh are correct, but Abha is wrong.}
\displaystyle \text{(d) Saurabh and Abha are correct, but Rina is wrong.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=x^5+x^3.
\displaystyle \therefore y'=5x^4+3x^2.
\displaystyle \text{Also, }y''=20x^3+6x=2x(10x^2+3).
\displaystyle \text{For }x<0,\ y''<0,\text{ and for }x>0,\ y''>0.
\displaystyle \therefore y'\text{ is not an increasing function for all values of }x.
\displaystyle \therefore \text{Rina's statement is false.}
\displaystyle \text{Now, }y'(-x)=5(-x)^4+3(-x)^2=5x^4+3x^2=y'(x).
\displaystyle \therefore y'\text{ is an even function, not an odd function.}
\displaystyle \therefore \text{Abha's statement is false.}
\displaystyle \text{Since }y'\text{ is even, its graph is symmetrical about the }y\text{ axis.}
\displaystyle \therefore \text{it is not symmetrical about the origin.}
\displaystyle \therefore \text{Saurabh's statement is false.}
\displaystyle \therefore \text{Rina, Abha and Saurabh are all wrong.}
\displaystyle \therefore \text{None of the given options is correct.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Statement I: }f(x)=\begin{cases}x^2\sin\left(\frac1x\right),&x\ne0\\0,&x=0\end{cases}\text{ is continuous}
\displaystyle \text{at }x=0\text{ but }f'(x)\text{ is not continuous at }x=0.
\displaystyle \text{Statement II: The derivative of a continuous function need not be a continuous function.}
\displaystyle \text{(a) Both (I) and (II) are correct and (II) is the correct explanation of (I).}
\displaystyle \text{(b) Both (I) and (II) are correct and (II) is not the correct explanation of (I).}
\displaystyle \text{(c) (I) is correct but (II) is incorrect.}
\displaystyle \text{(d) (II) is correct but (I) is incorrect.}
\displaystyle \text{Answer:}
\displaystyle \text{For Statement I,}
\displaystyle \lim_{x\to0}f(x)=\lim_{x\to0}x^2\sin\left(\frac1x\right).
\displaystyle \text{Since }\left|\sin\left(\frac1x\right)\right|\leq1,
\displaystyle \left|x^2\sin\left(\frac1x\right)\right|\leq x^2.
\displaystyle \therefore \lim_{x\to0}x^2\sin\left(\frac1x\right)=0=f(0).
\displaystyle \therefore f(x)\text{ is continuous at }x=0.
\displaystyle f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}.
\displaystyle =\lim_{h\to0}h\sin\left(\frac1h\right)=0.
\displaystyle \text{For }x\ne0,
\displaystyle f'(x)=2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right).
\displaystyle \text{As }x\to0,\ 2x\sin\left(\frac1x\right)\to0,\text{ but }\cos\left(\frac1x\right)\text{ has no limit.}
\displaystyle \therefore \lim_{x\to0}f'(x)\text{ does not exist.}
\displaystyle \therefore f'(x)\text{ is not continuous at }x=0.
\displaystyle \therefore \text{Statement I is true.}
\displaystyle \text{Statement II is also true, since the derivative of a differentiable function need not be continuous.}
\displaystyle \text{However, Statement II is only a general fact and does not explain the specific discontinuity of }f'(x).
\displaystyle \therefore \text{Statement II is not the correct explanation of Statement I.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Statement I: }\int_{-2}^{2}\frac{x^2}{1+2^x}\,dx=\frac83.
\displaystyle \text{Statement II: }\int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx,\text{ if }f(x)\text{ is an even function.}
\displaystyle \text{(a) Both the statements are true.}
\displaystyle \text{(b) Both the statements are false.}
\displaystyle \text{(c) Statement I is false, and Statement II is true.}
\displaystyle \text{(d) Statement I is true, and Statement II is false.}
\displaystyle \text{Answer:}
\displaystyle \text{For Statement I, let }I=\int_{-2}^{2}\frac{x^2}{1+2^x}\,dx.
\displaystyle \text{Using }\int_{-a}^{a}f(x)\,dx=\int_0^a\{f(x)+f(-x)\}\,dx,
\displaystyle I=\int_0^2\left\{\frac{x^2}{1+2^x}+\frac{x^2}{1+2^{-x}}\right\}dx.
\displaystyle =\int_0^2\left\{\frac{x^2}{1+2^x}+\frac{x^2\,2^x}{1+2^x}\right\}dx.
\displaystyle =\int_0^2x^2\,dx.
\displaystyle =\left[\frac{x^3}{3}\right]_0^2=\frac83.
\displaystyle \therefore \text{Statement I is true.}
\displaystyle \text{For an even function }f(x),
\displaystyle \int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx.
\displaystyle \therefore \text{Statement II is true.}
\displaystyle \therefore \text{Both Statement I and Statement II are true.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{For any given event }A:
\displaystyle \text{Statement I: Event }A\text{ and null event }\phi\text{ are always independent.}
\displaystyle \text{Statement II: Event }A\text{ and sure event }S\text{ are always independent.}
\displaystyle \text{(a) Both the statements are true.}
\displaystyle \text{(b) Both the statements are false.}
\displaystyle \text{(c) Statement I is false, and Statement II is true.}
\displaystyle \text{(d) Statement I is true, and Statement II is false.}
\displaystyle \text{Answer:}
\displaystyle \text{For Statement I, }A\cap\phi=\phi.
\displaystyle \therefore P(A\cap\phi)=P(\phi)=0.
\displaystyle \text{Also, }P(A)P(\phi)=P(A)\times0=0.
\displaystyle \therefore P(A\cap\phi)=P(A)P(\phi).
\displaystyle \therefore A\text{ and }\phi\text{ are independent.}
\displaystyle \therefore \text{Statement I is true.}
\displaystyle \text{For Statement II, }A\cap S=A.
\displaystyle \therefore P(A\cap S)=P(A).
\displaystyle \text{Also, }P(A)P(S)=P(A)\times1=P(A).
\displaystyle \therefore P(A\cap S)=P(A)P(S).
\displaystyle \therefore A\text{ and }S\text{ are independent.}
\displaystyle \therefore \text{Statement II is true.}
\displaystyle \therefore \text{Both Statement I and Statement II are true.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Which of the following statement(s) DOES NOT hold true related to}
\displaystyle \text{regression analysis:}
\displaystyle \text{Statement I: }|r|\text{ is the geometric mean of }b_{yx}\text{ and }b_{xy}.
\displaystyle \text{Statement II: }b_{xy},\ b_{yx}\text{ and }r\text{ all are of the same sign.}
\displaystyle \text{Statement III: The two regression lines do not intersect at }(\bar{x},\bar{y}).
\displaystyle \text{Statement IV: }-1\leq b_{yx}\times b_{xy}\leq1.
\displaystyle \text{(a) III and IV only.}\qquad\text{(b) IV only.}
\displaystyle \text{(c) III only.}\qquad\text{(d) II and III only.}
\displaystyle \text{Answer:}
\displaystyle \text{For Statement I, }b_{yx}b_{xy}=r^2.
\displaystyle \therefore |r|=\sqrt{b_{yx}b_{xy}}.
\displaystyle \therefore |r|\text{ is the geometric mean of }b_{yx}\text{ and }b_{xy}.
\displaystyle \therefore \text{Statement I is true.}
\displaystyle \text{The regression coefficients }b_{yx}\text{ and }b_{xy}\text{ have the same sign as }r.
\displaystyle \therefore \text{Statement II is true.}
\displaystyle \text{The two regression lines always intersect at }(\bar{x},\bar{y}).
\displaystyle \therefore \text{Statement III is false.}
\displaystyle \text{Also, }b_{yx}b_{xy}=r^2\text{ and }0\leq r^2\leq1.
\displaystyle \therefore 0\leq b_{yx}b_{xy}\leq1.
\displaystyle \therefore -1\leq b_{yx}b_{xy}\leq1\text{ is also true.}
\displaystyle \therefore \text{Statement IV is true.}
\displaystyle \therefore \text{only Statement III does not hold true.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

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