The numerator of a fraction is 4 less than the denominator.
Question: The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction. Solution: Let the numerator and denominator of the fraction be $x$ and $y$ respectively. Then the fraction is $\frac{x}{y}$ The numerator of the fraction is 4 less the denominator. Thus, we have $x=y-4$ $\Rightarrow x-y=-4$ If the numerator is decreased by 2 and denominator is increased by...
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Question: $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$ Solution: $I=\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$ ...(1) $\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \cos ^{2}\left(\frac{\pi}{2}-x\right) d x$ $\left(\int_{0}^{0} f(x) d x=\int_{0}^{0} f(a-x) d x\right)$ $\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x$ ...(2) Adding (1) and (2), we obtain $2 I=\int_{2}^{\frac{\pi}{2}}\left(\sin ^{2} x+\cos ^{2} x\right) d x$ $\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} 1 d x$ $\Rightarrow 2 I=[x]_{0}^{\...
Read More →The domain and range of the real function of defined by f(x)=
Question: The domain and range of the real function of defined by $f(x)=\frac{4-x}{x-4}$ is given by (a) Domain =R, Range ={1, 1} (b) Domain =R{1}, Range =R (c) Domain =R{4}, Range = {1} (d) Domain =R{4}, Range = {1, 1} Solution: $f(x)=\frac{4-x}{x-4}$ Domain :- for $f(x)=\frac{4-x}{x-4}$ $x-4 \neq 0$ i.e $x \neq 4$ Domain isR {4} Range:- Since $f(x)=\frac{-(x-4)}{x-4}=-1$ Range is {1} Hence, the correct answer is option C....
Read More →Draw the graphs of the following linear equations on the same graph paper:
Question: Draw the graphs of the following linear equations on the same graph paper: 2x + 3y = 12, x - y = 1 Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis. Also, find the area of the triangle. Solution: We are given, 2x +3y =12 We get, $y=\frac{12-2 x}{3}$ Now, substituting x = 0 in $y=\frac{12-2 x}{3}$ we get y = 4 Substituting x = 6 in $y=\frac{12-2 x}{3}$ we get y = 0 Thus, we have the following table exhibiting the abscissa and ordinates...
Read More →Seven times a two-digit number is equal to four times the number obtained by reversing the digits.
Question: Seven times a two-digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number. Solution: Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$. The difference between the two digits of the number is 3 . Thus, we have $x-y=\pm 3$ After interchanging the digits, the number becomes $10 x+y$. Seven times the number is equal to four times the number ...
Read More →If f(x) = ax + b, where, a and b are integers,
Question: Iff(x) =ax+b, where,aandbare integers,f(1) = 5 andf(x) = 3, thenaandbare equal (a)a= 3,b= 1 (b)a= 2,b= 3 (c)a= 0,b= 2 (d)a= 2,b= 3 Solution: f(x) =ax+b Givenf(1) = 5 andf(x) = 3...
Read More →Domain of f(x)=
Question: Domain of $f(x)=\sqrt{a^{2}-x^{2}}, a0$ is (a) (a,a) (b) [a,a] (c) [0,a] (d) (a, 0] Solution: $f(x)=\sqrt{a^{2}-x^{2}} \cdot a0$ Sincea2x2 0 i.e x2 a2 i.ex2a2 i.e |x|a i.e axa i.ex [a,a] $\therefore$ Domain of $f(x)$ is $[-a, a]$ Hence, the correct answer is option B....
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Question: If $f(x)=\int_{0}^{x} t \sin t d t$, then $f^{\prime}(x)$ is A. $\cos x+x \sin x$ B. $x \sin x$ C. $x \cos x$ D. $\sin x+x \cos x$ Solution: $f(x)=\int_{0}^{x} t \sin t d t$ Integrating by parts, we obtain $f(x)=t \int_{0}^{x} \sin t d t-\int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \int \sin t d t\right\} d t$ $=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cos t) d t$ $=[-t \cos t+\sin t]_{0}^{x}$ $=-x \cos x+\sin x$ $\Rightarrow f^{\prime}(x)=-[\{x(-\sin x)\}+\cos x]+\cos x$ $=x \sin x-\cos...
Read More →Draw the graph of y = |x| + 2.
Question: Draw the graph of y = |x| + 2. Solution: We are given, y = |x| + 2 Substituting x = 0 we get y = 2 Substituting x = 1, we get y = 3 Substituting x = - 1, we get y = 3 Substituting x = 2, we get y = 4 Substituting x = - 2, we get y = 4 For every value of x, whether positive or negative, we get y as a positive number and the minimum value of y is equal to 2 units....
Read More →The domain of the function
Question: The domain of the function $f(x) \sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$ is equal to (a) (, 1) (1, 4) (b) (, 1] (1, 4] (c) (, 1) [1, 4] (d) (, 1) [1, 4) Solution: $f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$ for $\sqrt{4-x}$ 4x 0 i.e 4x i.ex 4 and $\frac{1}{\sqrt{x^{2}-1}}$ Since $x^{2}-10 \quad\left(\because x^{2}-1=0\right.$ is not defined for $\left.\frac{1}{\sqrt{x^{2}-1}}\right)$ i.e x2 1 i.e x 1 orx 1 for $\sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$, Domain is $(-\infty,-1) \cup(1,4]$...
Read More →The domain of the function
Question: The domain of the function $f(x) \sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$ is equal to (a) (, 1) (1, 4) (b) (, 1] (1, 4] (c) (, 1) [1, 4] (d) (, 1) [1, 4) Solution: $f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$ for $\sqrt{4-x}$ 4x 0 i.e 4x i.ex 4 and $\frac{1}{\sqrt{x^{2}-1}}$ Since $x^{2}-10 \quad\left(\because x^{2}-1=0\right.$ is not defined for $\left.\frac{1}{\sqrt{x^{2}-1}}\right)$ i.e x2 1 i.e x 1 orx 1 for $\sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$, Domain is $(-\infty,-1) \cup(1,4]$...
Read More →Draw the graph of y = |×|.
Question: Draw the graph of y = ||. Solution: We are given, y = |x| Substituting x = 1, we get y = 1 Substituting x = - 1, we get y = 1 Substituting x = 2, we get y = 2 Substituting x = - 2, we get y = 2 For every value of x, whether positive or negative, we get y as a positive number....
Read More →Draw the graph of the equation x/3 + y/4 = 1.
Question: Draw the graph of the equation x/3 + y/4 = 1. Also, find the area of the triangle formed by 3/4 the line and the coordinates axes. Solution: We are given.x/3 + y/4 = 1 4x + 3y =12 $y=\frac{12-4 x}{3}$ Now, substituting x = 0 in $\mathrm{y}=\frac{12-4 \mathrm{x}}{3}$ we get y = 4 Substituting x = 3 in $y=\frac{12-4 x}{3}$ we get y = 0 Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation The region bounded by the ...
Read More →The sum of the digits of a two-digit number is 9.
Question: The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. Solution: Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$. The sum of the two digits of the number is 9 . Thus, we have $x+y=9$ After interchanging the digits, the number becomes $10 x+y$. Also, 9 times the number is equal to twice the number obtained by rev...
Read More →The range of f(x)=
Question: The range of $f(x)=\frac{1}{1-2 \cos x}$ is (a) [1/3, 1] (b) [1, 1/3] (c) (, 1) [1/3, ) (d) [1/3, 1] Solution: We know that 1 cosx 1 for allxR. Now, $-1 \leq \cos x \leq 1$ $\Rightarrow-1 \leq-\cos x \leq 1$ $\Rightarrow-2 \leq-2 \cos x \leq 2$ $\Rightarrow-1 \leq 1-2 \cos x \leq 3$ (Adding 1 to each term) But, $\cos x \neq \frac{1}{2}$ $\Rightarrow 1-2 \cos x \in[-1,3]-\{0\}$ $\Rightarrow \frac{1}{1-2 \cos x} \in(-\infty,-1] \cup\left[\frac{1}{3}, \infty\right)$ $\therefore$ Range of ...
Read More →The range of f(x)=
Question: The range of $f(x)=\frac{1}{1-2 \cos x}$ is (a) [1/3, 1] (b) [1, 1/3] (c) (, 1) [1/3, ) (d) [1/3, 1] Solution: We know that 1 cosx 1 for allxR. Now, $-1 \leq \cos x \leq 1$ $\Rightarrow-1 \leq-\cos x \leq 1$ $\Rightarrow-2 \leq-2 \cos x \leq 2$ $\Rightarrow-1 \leq 1-2 \cos x \leq 3$ (Adding 1 to each term) But, $\cos x \neq \frac{1}{2}$ $\Rightarrow 1-2 \cos x \in[-1,3]-\{0\}$ $\Rightarrow \frac{1}{1-2 \cos x} \in(-\infty,-1] \cup\left[\frac{1}{3}, \infty\right)$ $\therefore$ Range of ...
Read More →The difference between two numbers is 26 and one number is three times the other.
Question: The difference between two numbers is 26 and one number is three times the other. Find them. Solution: Let the numbers are $x$ and $y$. One of them must be greater than or equal to the other. Let us assume that $x$ is greater than or equal to $y$. The difference between the two numbers is 26 . Thus, we have $x-y=26$ One of the two numbers is three times the other number. Here, we are assuming that $x$ is greater than or equal to $y$. Thus, we have $x=3 y$ So, we have two equations $x-y...
Read More →Draw the graph of the equation 2x + y = 6. Shade the region bounded
Question: Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region. Solution: We are given, 2x + y = 6 We get, y = 6 - 2x Now, substituting x = 0 in y = 6 - 2x, we get y = 6 Substituting x = 3 in y = 6 - 2x, we get y = 0 Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation The region bounded by the graph is ABC which forms a triangle...
Read More →A two-digit number is such that the product of its digits is 20. If 9 is added to the number,
Question: A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number. Solution: Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$. The product of the two digits of the number is 20 . Thus, we have $x y=20$ After interchanging the digits, the number becomes $10 x+y$. If 9 is added to the number, the digits interchange their places. Thus, we hav...
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Question: The value of the integral $\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ is A. 6 B. 0 C. 3 D. 4 Solution: Let $I=\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ Also, let $x=\sin \theta \Rightarrow d x=\cos \theta d \theta$ When $x=\frac{1}{3}, \theta=\sin ^{-1}\left(\frac{1}{3}\right)$ and when $x=1, \theta=\frac{\pi}{2}$ $\Rightarrow I=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \frac{\left(\sin \theta-\sin ^{3...
Read More →Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:
Question: Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes: (i) 6x 3y = 12 (ii) - x + 4y = 8 (iii) 2x + y = 6 (iv) 3x + 2y + 6 = 0 Solution: (i) We are given, 6x - 3y = 12 We get, y = (6x -12)/3 Now, substituting x = 0 in y = - (6x 12)/3 we get y = - 4 Substituting x = 2 in y = (- 6x -12)/3, we get y = 0 Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by...
Read More →A two-digit number is 4 times the sum of its digits and twice the product of the digits.
Question: A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number. Solution: Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$. The number is 4 times the sum of the two digits. Thus, we have $10 y+x=4(x+y)$ $\Rightarrow 10 y+x=4 x+4 y$ $\Rightarrow 4 x+4 y-10 y-x=0$ $\Rightarrow 3 x-6 y=0$ $\Rightarrow 3(x-2 y)=0$ $\Rightarrow x-2 y=0$ $\Rightarrow x=2 y$ After interchanging the di...
Read More →If
Question: If $[x]^{2}-5[x]+6=0$, where [.] denotes the greatest integer function, then (a)x [3, 4] (b)x (2, 3] (c)x [2, 3] (d)x [2, 4) Solution: The given equation is $[x]^{2}-5[x]+6=0$. $[x]^{2}-5[x]+6=0$ $\Rightarrow[x]^{2}-3[x]-2[x]+6=0$ $\Rightarrow[x]([x]-3)-2([x]-3)=0$ $\Rightarrow([x]-2)([x]-3)=0$ $\Rightarrow[x]-2=0$ or $[x]-3=0$ $\Rightarrow[x]=2$ or $[x]=3$ $\Rightarrow x \in[2,3)$ or $x \in[3,4)$ $\Rightarrow x \in[2,4)$ Hence, the correct answer is option (d)....
Read More →Let f(x)=
Question: Let $f(x)=\sqrt{x^{2}+1}$. Then, which of the following is correct? (a) $f(x y)=f(x) f(y)$ (b) $f(x y) \geq f(x) f(y)$ (c) $f(x y) \leq f(x) f(y)$ (d) none of these Solution: Given: $f(x)=\sqrt{x^{2}+1}$ ....(1) Replacingxbyyin (1), we get $f(y)=\sqrt{y^{2}+1}$ $\therefore f(x) f(y)=\sqrt{x^{2}+1} \sqrt{y^{2}+1}$ $=\sqrt{\left(x^{2}+1\right)\left(y^{2}+1\right)}$ $=\sqrt{x^{2} y^{2}+x^{2}+y^{2}+1}$ Also, replacingxbyxyin (1), we get $f(x y)=\sqrt{x^{2} y^{2}+1}$ Now, $x^{2} y^{2}+1 \le...
Read More →Let f(x)=
Question: Let $f(x)=\sqrt{x^{2}+1}$. Then, which of the following is correct? (a) $f(x y)=f(x) f(y)$ (b) $f(x y) \geq f(x) f(y)$ (c) $f(x y) \leq f(x) f(y)$ (d) none of these Solution: Given: $f(x)=\sqrt{x^{2}+1}$ ....(1) Replacingxbyyin (1), we get $f(y)=\sqrt{y^{2}+1}$ $\therefore f(x) f(y)=\sqrt{x^{2}+1} \sqrt{y^{2}+1}$ $=\sqrt{\left(x^{2}+1\right)\left(y^{2}+1\right)}$ $=\sqrt{x^{2} y^{2}+x^{2}+y^{2}+1}$ Also, replacingxbyxyin (1), we get $f(x y)=\sqrt{x^{2} y^{2}+1}$ Now, $x^{2} y^{2}+1 \le...
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