IIT JEE 1990 Mathematics Question Paper with Answer and Solution

(D) The logarithmic function $y = \log_a x$ is defined under the following conditions:
$1$. The base $a$ must be a positive real number,i.e.,$a > 0$.
$2$. The base $a$ cannot be equal to $1$,i.e.,$a \neq 1$.
$3$. The argument $x$ must be positive,i.e.,$x > 0$.
Therefore,for the expression to be defined,$a$ must be any positive real number except $1$.

(C) Suppose,if possible,$\log_{2} 7$ is rational,say $p/q$ where $p$ and $q$ are integers,prime to each other.
Then,$\frac{p}{q} = \log_{2} 7 \implies 7 = 2^{p/q} \implies 2^{p} = 7^{q}$.
This is a contradiction because the $L.H.S$ is an even number (power of $2$) and the $R.H.S$ is an odd number (power of $7$).
Since it cannot be expressed as a ratio of two integers,$\log_{2} 7$ is an irrational number.
Obviously,$\log_{2} 7$ is not an integer and hence not a prime number.

(C) Given $z_1 = 10 + 6i$ and $z_2 = 4 + 6i$. Let $z = x + iy$.
The condition $\text{amp}\left( \frac{z - z_1}{z - z_2} \right) = \frac{\pi}{4}$ represents an arc of a circle passing through $z_1$ and $z_2$.
The locus is given by $\frac{(y-6)(x-4) - (y-6)(x-10)}{(x-4)(x-10) + (y-6)^2} = \tan\left(\frac{\pi}{4}\right) = 1$.
Simplifying,we get $(y-6)(x-4 - x + 10) = (x-4)(x-10) + (y-6)^2$.
$6(y-6) = x^2 - 14x + 40 + y^2 - 12y + 36$.
$x^2 - 14x + y^2 - 18y + 76 + 36 = 0 \implies x^2 - 14x + y^2 - 18y + 112 = 0$.
Adding $49 + 81$ to both sides to complete the square:
$(x^2 - 14x + 49) + (y^2 - 18y + 81) = -112 + 49 + 81 = 18$.
$(x - 7)^2 + (y - 9)^2 = 18$.
We need to find $|z - 7 - 9i| = |(x-7) + i(y-9)| = \sqrt{(x-7)^2 + (y-9)^2}$.
Substituting the value from the equation of the circle,we get $\sqrt{18} = 3\sqrt{2}$.

(D) Given that $\log_3 2, \log_3(2^x - 5)$ and $\log_3(2^x - \frac{7}{2})$ are in $A.P.$
By the property of $A.P.$,$2b = a + c$,so:
$2 \log_3(2^x - 5) = \log_3 2 + \log_3(2^x - \frac{7}{2})$
Using the property $\log a + \log b = \log(ab)$:
$\log_3(2^x - 5)^2 = \log_3(2(2^x - \frac{7}{2}))$
$(2^x - 5)^2 = 2^{x+1} - 7$
Let $2^x = y$. Then $(y - 5)^2 = 2y - 7$
$y^2 - 10y + 25 = 2y - 7$
$y^2 - 12y + 32 = 0$
$(y - 8)(y - 4) = 0$
So,$y = 8$ or $y = 4$.
If $2^x = 8$,then $x = 3$.
If $2^x = 4$,then $x = 2$.
Check the domain of the logarithmic terms:
For $x = 2$,$\log_3(2^2 - 5) = \log_3(-1)$,which is undefined.
For $x = 3$,$\log_3(2^3 - 5) = \log_3(3) = 1$ and $\log_3(2^3 - 3.5) = \log_3(4.5)$,which are defined.
Thus,$x = 3$ is the only solution.

(B) The equation of the line $L$ with respect to the original axes is $\frac{x}{a} + \frac{y}{b} = 1$.
When the axes are rotated by an angle $\alpha$,the coordinates $(x, y)$ transform to $(x', y')$ where $x = x'\cos \alpha - y'\sin \alpha$ and $y = x'\sin \alpha + y'\cos \alpha$.
Substituting these into the line equation:
$\frac{x'\cos \alpha - y'\sin \alpha}{a} + \frac{x'\sin \alpha + y'\cos \alpha}{b} = 1$
Rearranging terms:
$x'\left(\frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}\right) + y'\left(\frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}\right) = 1$
Comparing this to the intercept form $\frac{x'}{p} + \frac{y'}{q} = 1$,we get:
$\frac{1}{p} = \frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}$ and $\frac{1}{q} = \frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}$.
Squaring and adding these equations:
$\frac{1}{p^2} + \frac{1}{q^2} = \left(\frac{\cos \alpha}{a} + \frac{\sin \alpha}{b}\right)^2 + \left(\frac{\cos \alpha}{b} - \frac{\sin \alpha}{a}\right)^2$
$= \frac{\cos^2 \alpha}{a^2} + \frac{\sin^2 \alpha}{b^2} + \frac{2\sin \alpha \cos \alpha}{ab} + \frac{\cos^2 \alpha}{b^2} + \frac{\sin^2 \alpha}{a^2} - \frac{2\sin \alpha \cos \alpha}{ab}$
$= \frac{\cos^2 \alpha + \sin^2 \alpha}{a^2} + \frac{\sin^2 \alpha + \cos^2 \alpha}{b^2} = \frac{1}{a^2} + \frac{1}{b^2}$.

(B) Let $L = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}}$.
Since the form is $\frac{0}{0}$,we apply $L$-Hospital's rule:
$L = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\frac{d}{{dx}}(\sqrt 2 \cos x - 1)}}{{\frac{d}{{dx}}(\cot x - 1)}}$
$L = \mathop {\lim }\limits_{x \to \pi /4} \frac{{-\sqrt 2 \sin x}}{{-\csc^2 x}}$
$L = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sqrt 2 \sin x}}{{\csc^2 x}} = \mathop {\lim }\limits_{x \to \pi /4} \sqrt 2 \sin x \sin^2 x = \sqrt 2 \sin^3 x$
Substituting $x = \pi /4$:
$L = \sqrt 2 \left( \frac{1}{{\sqrt 2 }} \right)^3 = \sqrt 2 \cdot \frac{1}{{2\sqrt 2 }} = \frac{1}{2}$.

(C) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}}$,we factor out the highest power of $x$ from each term:
$= \mathop {\lim }\limits_{x \to \infty } \frac{{{{[x(2 + \frac{1}{x})]}^{40}}{{[x(4 - \frac{1}{x})]}^5}}}{{{{[x(2 + \frac{3}{x})]}^{45}}}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{40}}{{(2 + \frac{1}{x})}^{40}} \cdot {x^5}{{(4 - \frac{1}{x})}^5}}}{{{x^{45}}{{(2 + \frac{3}{x})}^{45}}}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{45}}{{(2 + \frac{1}{x})}^{40}}{{(4 - \frac{1}{x})}^5}}}{{{x^{45}}{{(2 + \frac{3}{x})}^{45}}}}$
$= \frac{{{{(2 + 0)}^{40}}{{(4 - 0)}^5}}}{{{{(2 + 0)}^{45}}}}$
$= \frac{{{2^{40}} \cdot {4^5}}}{{{2^{45}}}}$
$= \frac{{{2^{40}} \cdot {{(2^2)}^5}}}{{{2^{45}}}}$
$= \frac{{{2^{40}} \cdot {2^{10}}}}{{{2^{45}}}}$
$= \frac{{{2^{50}}}}{{{2^{45}}}} = {2^5} = 32$

(B) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \times P(B)$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Let $P(B) = x$. Substituting the given values:
$0.8 = 0.3 + x - (0.3 \times x)$
$0.8 - 0.3 = x - 0.3x$
$0.5 = 0.7x$
$x = \frac{0.5}{0.7} = \frac{5}{7}$.
Therefore,$P(B) = \frac{5}{7}$.

(D) By the definition of a logarithm,the base $a$ must be a positive real number such that $a > 0$ and $a \neq 1$.
Therefore,the correct condition for the base $a$ is any positive real number except $1$.

(C) Given equation: $(\cos p-1) x^2+(\cos p) x+\sin p=0$.
Since the roots are real,the discriminant $\Delta \geq 0$.
$\Delta = b^2 - 4ac = (\cos p)^2 - 4(\cos p - 1)(\sin p) \geq 0$.
$\cos^2 p - 4\sin p \cos p + 4\sin p \geq 0$.
For the quadratic equation to exist,the coefficient of $x^2$ must be non-zero: $\cos p - 1 \neq 0 \Rightarrow \cos p \neq 1$.
Since $\cos^2 p \geq 0$ and $(\cos p - 1) < 0$ for all $p \neq 2n\pi$,the condition $\Delta \geq 0$ is satisfied when $\sin p > 0$.
Thus,$p \in (0, \pi)$.
Hence,option $C$ is correct.

(D) Given $d \times b = c \times b$,this implies $(d - c) \times b = 0$.
This means $(d - c)$ is parallel to $b$,so $d - c = \lambda b$ for some scalar $\lambda$.
Thus,$d = c + \lambda b = (4i - 3j + 7k) + \lambda(i + j + k) = (4 + \lambda)i + (-3 + \lambda)j + (7 + \lambda)k$.
We are given $d \cdot a = 0$,where $a = 2i + k$.
Substituting $d$ and $a$: $((4 + \lambda)i + (-3 + \lambda)j + (7 + \lambda)k) \cdot (2i + 0j + k) = 0$.
$2(4 + \lambda) + 0(-3 + \lambda) + 1(7 + \lambda) = 0$.
$8 + 2\lambda + 7 + \lambda = 0$.
$15 + 3\lambda = 0 \implies \lambda = -5$.
Substituting $\lambda = -5$ back into the expression for $d$:
$d = (4 - 5)i + (-3 - 5)j + (7 - 5)k = -i - 8j + 2k$.

(A) For the function $f(x)$ to be continuous at $x = 0$,the left-hand limit,right-hand limit,and the value of the function at $x = 0$ must be equal.
$1$. Left-hand limit $(LHL)$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos 4x}{x^2} = \lim_{x \to 0^-} \frac{2 \sin^2(2x)}{x^2} = \lim_{x \to 0^-} 2 \times \left( \frac{\sin 2x}{2x} \right)^2 \times 4 = 2 \times 1^2 \times 4 = 8$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}$.
Rationalizing the denominator: $\lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}} = \lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4) = \sqrt{16} + 4 = 4 + 4 = 8$.
$3$. Since the function is continuous at $x = 0$,$f(0) = \text{LHL} = \text{RHL}$.
Therefore,$a = 8$.

(B) The domain of $(g \circ f)(x)$ is the set of all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$.
First,find the domain of $g(x) = \sqrt{3 + 4x - 4x^2}$.
For $g(x)$ to be defined,$3 + 4x - 4x^2 \ge 0$,which implies $4x^2 - 4x - 3 \le 0$.
Factoring the quadratic: $(2x - 3)(2x + 1) \le 0$.
This inequality holds for $x \in \left[ -\frac{1}{2}, \frac{3}{2} \right]$.
Now,we require $f(x) \in \left[ -\frac{1}{2}, \frac{3}{2} \right]$,so $-\frac{1}{2} \le \frac{1}{2} - \tan \left( \frac{\pi x}{2} \right) \le \frac{3}{2}$.
Subtracting $\frac{1}{2}$ from all parts: $-1 \le -\tan \left( \frac{\pi x}{2} \right) \le 1$.
Multiplying by $-1$ (reversing the inequality): $-1 \le \tan \left( \frac{\pi x}{2} \right) \le 1$.
Applying the inverse tangent function: $-\frac{\pi}{4} \le \frac{\pi x}{2} \le \frac{\pi}{4}$.
Dividing by $\frac{\pi}{2}$: $-\frac{1}{2} \le x \le \frac{1}{2}$.
Since this interval is within the given domain of $f$ $(-1 < x < 1)$,the domain of $(g \circ f)$ is $\left[ -\frac{1}{2}, \frac{1}{2} \right]$.

(B) Let $I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$.
Substitute $\sqrt{x} = t$.
Then,differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{1}{\sqrt{x}} dx = 2 dt$.
Substituting these into the integral,we get $I = \int \cos(t) \cdot 2 dt = 2 \int \cos(t) dt$.
Integrating $\cos(t)$,we get $I = 2 \sin(t) + c$.
Substituting back $t = \sqrt{x}$,we get $I = 2 \sin \sqrt{x} + c$.

(D) Let $I = \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx$. Multiply numerator and denominator by $e^x$:
$I = \int \frac{4e^{2x} + 6}{9e^{2x} - 4} dx$.
We express the numerator as $4e^{2x} + 6 = m(18e^{2x}) + n(9e^{2x} - 4)$.
Comparing coefficients: $18m + 9n = 4$ and $-4n = 6 \implies n = -\frac{3}{2}$.
$18m + 9(-\frac{3}{2}) = 4 \implies 18m = 4 + \frac{27}{2} = \frac{35}{2} \implies m = \frac{35}{36}$.
Thus,$I = \int \frac{\frac{35}{36}(18e^{2x}) - \frac{3}{2}(9e^{2x} - 4)}{9e^{2x} - 4} dx = \frac{35}{36} \int \frac{18e^{2x}}{9e^{2x} - 4} dx - \frac{3}{2} \int dx$.
$I = \frac{35}{36} \log |9e^{2x} - 4| - \frac{3}{2}x + C$.
Comparing with $Ax + B \log |9e^{2x} - 4| + C$,we get $A = -\frac{3}{2}$ and $B = \frac{35}{36}$.
Since none of the options match these values,the correct answer is $(d)$.

(D) Let $t = \frac{1}{x}$.
Then,$dt = -\frac{1}{x^2} \,dx$,which implies $-\,dt = \frac{1}{x^2} \,dx$.
When $x = \frac{1}{\pi}$,$t = \pi$.
When $x = \frac{2}{\pi}$,$t = \frac{\pi}{2}$.
Substituting these into the integral:
$\int_{1/\pi}^{2/\pi} \frac{\sin(1/x)}{x^2} \,dx = \int_{\pi}^{\pi/2} \sin(t) \cdot (-dt)$
$= \int_{\pi/2}^{\pi} \sin(t) \,dt$
$= [-\cos(t)]_{\pi/2}^{\pi}$
$= -(\cos(\pi) - \cos(\pi/2))$
$= -(-1 - 0) = 1$.

(D) Let $h(x) = [f(x) + f(-x)][g(x) - g(-x)]$.
Now,calculate $h(-x)$:
$h(-x) = [f(-x) + f(-(-x))][g(-x) - g(-(-x))] = [f(-x) + f(x)][g(-x) - g(x)]$.
We can rewrite this as:
$h(-x) = -[f(x) + f(-x)][g(x) - g(-x)] = -h(x)$.
Since $h(-x) = -h(x)$,the function $h(x)$ is an odd function.
For any odd function $h(x)$,the integral over a symmetric interval $[-a, a]$ is zero:
$\int_{-a}^{a} h(x) \, dx = 0$.
Therefore,$\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx = 0$.

(A) We are given the integrals:
$\int_0^1 f(x) \, dx = 1$,
$\int_0^1 x f(x) \, dx = a$,
$\int_0^1 x^2 f(x) \, dx = a^2$.
We need to evaluate the integral $I = \int_0^1 (x - a)^2 f(x) \, dx$.
Expanding the term $(x - a)^2 = x^2 - 2ax + a^2$,we get:
$I = \int_0^1 (x^2 - 2ax + a^2) f(x) \, dx$
$I = \int_0^1 x^2 f(x) \, dx - 2a \int_0^1 x f(x) \, dx + a^2 \int_0^1 f(x) \, dx$
Substituting the given values:
$I = a^2 - 2a(a) + a^2(1)$
$I = a^2 - 2a^2 + a^2$
$I = 0$.